# HG changeset patch # User Robert McIntyre # Date 1319785597 25200 # Node ID 44d3dc936f6a5e5c4f7b5932820603ebea9e2d4b # Parent b4de894a1e2e6d6b64abf5f2f1b05a877c46fe97 moved backup files diff -r b4de894a1e2e -r 44d3dc936f6a bk/bk.org --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/bk/bk.org Fri Oct 28 00:06:37 2011 -0700 @@ -0,0 +1,88 @@ +#+TITLE: Bugs in Quantum Mechanics +#+AUTHOR: Dylan Holmes +#+SETUPFILE: ../../aurellem/org/setup.org +#+INCLUDE: ../../aurellem/org/level-0.org + +#Bugs in the Quantum-Mechanical Momentum Operator + + +I studied quantum mechanics the same way I study most subjects\mdash{} +by collecting (and squashing) bugs in my understanding. One of these +bugs persisted throughout two semesters of +quantum mechanics coursework until I finally found +the paper +[[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum +mechanics/]], which helped me stamp out the bug entirely. I decided to +write an article about the problem and its solution for a number of reasons: + +- Although the paper was not unreasonably dense, it was written for + teachers. I wanted to write an article for students. +- I wanted to popularize the problem and its solution because other + explanations are currently too hard to find. (Even Shankar's + excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it.) +- I wanted to check that the bug was indeed entirely + eradicated. Attempting an explanation is my way of making + sure. + +* COMMENT + I recommend the +paper not only for students who are learning +quantum mechanics, but especially for teachers interested in debugging +them. + +* COMMENT +On my first exam in quantum mechanics, my professor asked us to +describe how certain measurements would affect a particle in a +box. Many of these measurement questions required routine application +of skills we had recently learned\mdash{}first, you recall (or +calculate) the eigenstates of the quantity +to be measured; second, you write the given state as a linear +sum of these eigenstates\mdash{} the coefficients on each term give +the probability amplitude. + +* The infinite square well potential +There is a particle in a one-dimensional potential well that has +infinitely high walls and finite width \(a\). This means that the +particle exists in a potential[fn:coords][fn:infinity] + + +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for +}\;x<0\text{ or }x>a.\end{cases}\) + +The Schr\ouml{}dinger equation describes how the particle's state +\(|\psi\rangle\) will change over time in this system. + +\(\begin{eqnarray} +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) + +This is a differential equation whose solutions are the physically +allowed states for the particle in this system. Like any differential +equation, + + +Like any differential equation, the Schr\ouml{}dinger equation +#; physically allowed states are those that change in physically +#allowed ways. + + +** Boundary conditions +Because the potential is infinite everywhere except within the well, +a realistic particle must be confined to exist only within the +well\mdash{}its wavefunction must be zero everywhere beyond the walls +of the well. + + +[fn:coords] I chose my coordinate system so that the well extends from +\(0a.\end{cases}\) + +The Schr\ouml{}dinger equation describes how the particle's state +\(|\psi\rangle\) will change over time in this system. + +\(\begin{eqnarray} +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) + +This is a differential equation whose solutions are the physically +allowed states for the particle in this system. Physically allowed +states are those that change in physically allowed ways. Like any +differential equation, the Schr\ouml{}dinger equation can be +accompanied by /boundary conditions/\mdash{}conditions that +further restrict which states qualify as physically allowed. + +Whenever possible, physicists impose these boundary conditions: +- The state should be a /continuous function of/ \(x\). This means + that if a particle is very likely to be /at/ a particular location, + it is also very likely to be /near/ that location. +- + +#; physically allowed states are those that change in physically +#allowed ways. + + +** Boundary conditions +Because the potential is infinite everywhere except within the well, +a realistic particle must be confined to exist only within the +well\mdash{}its wavefunction must be zero everywhere beyond the walls +of the well. + + +[fn:coords] I chose my coordinate system so that the well extends from +\(0a \\ \end{cases}\) +#0,&\text{for }x\lt{}0\text{ or }x\gt{}a\end{cases}\) + - no longer has an eigenstate for each value + of $p$. Instead, only values of $p$ that are integer multiples of + $\pi a/\hbar$ are physically realistic. + + + +* COMMENT: + +** Eigenstates with different eigenvalues are orthogonal + +#+begin_quote +*Theorem:* Eigenstates with different eigenvalues are orthogonal. +#+end_quote + +** COMMENT : +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$ +and $|b\rangle$ are eigenstates of $\Lambda$. This means that + + +\( +\begin{eqnarray} +\Lambda |a\rangle&=& a|a\rangle,\\ +\Lambda|b\rangle&=& b|b\rangle.\\ +\end{eqnarray} +\) + +If we take the difference of these eigenstates, we find that + +\( +\begin{eqnarray} +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle +\qquad \text{(because $\Lambda$ is linear.)}\\ +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and +$|b\rangle$ are eigenstates of $\Lambda$)} +\end{eqnarray}\) + + +which means that $a\neq b$. + +** Eigenvectors of hermitian operators span the space of solutions + +#+begin_quote +*Theorem:* If $\Omega$ is a hermitian operator, then every physically + allowed state can be written as a linear sum of eigenstates of + $\Omega$. +#+end_quote + + + +** Momentum and energy are hermitian operators +This ought to be true because hermitian operators correspond to +observable quantities. Since we expect momentum and energy to be +measureable quantities, we expect that there are hermitian operators +to represent them. + + +** Momentum and energy eigenstates in vacuum +An eigenstate of the momentum operator $P$ would be a state +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\). + +** Momentum and energy eigenstates in the infinitely deep well + + + +* Can you measure momentum in the infinite square well? + + + +** COMMENT Momentum eigenstates + +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\). + +In the infinitely deep potential well, the Hamiltonian is the same but +there is a new condition in order for states to qualify as physically +allowed: the states must not exist anywhere outside of well, as it +takes an infinite amount of energy to do so. + +Notice that the momentum eigenstates defined above do /not/ satisfy +this condition. + + + +* COMMENT +For each physical system, there is a Schr\ouml{}dinger equation that +describes how a particle's state $|\psi\rangle$ will change over +time. + +\(\begin{eqnarray} +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) + +This is a differential equation; each solution to the +Schr\ouml{}dinger equation is a state that is physically allowed for +our particle. Here, physically allowed states are +those that change in physically allowed ways. However, like any differential +equation, the Schr\ouml{}dinger equation can be accompanied by +/boundary conditions/\mdash{}conditions that further restrict which +states qualify as physically allowed. + + + + +** Eigenstates of momentum + + + + +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger + +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\) + + + + + + + +* COMMENT + +#* The infinite square well potential + +A particle exists in a potential that is +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the +particle exists in a potential[fn:coords][fn:infinity] + + +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for +}\;x<0\text{ or }x>a.\end{cases}\) + +The Schr\ouml{}dinger equation describes how the particle's state +\(|\psi\rangle\) will change over time in this system. + +\(\begin{eqnarray} +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) + +This is a differential equation; each solution to the +Schr\ouml{}dinger equation is a state that is physically allowed for +our particle. Here, physically allowed states are +those that change in physically allowed ways. However, like any differential +equation, the Schr\ouml{}dinger equation can be accompanied by +/boundary conditions/\mdash{}conditions that further restrict which +states qualify as physically allowed. + + +Whenever possible, physicists impose these boundary conditions: +- A physically allowed state ought to be a /smoothly-varying function of position./ This means + that if a particle in the state is likely to be /at/ a particular location, + it is also likely to be /near/ that location. + +These boundary conditions imply that for the square well potential in +this problem, + +- Physically allowed states must be totally confined to the well, + because it takes an infinite amount of energy to exist anywhere + outside of the well (and physically allowed states ought to have + only finite energy). +- Physically allowed states must be increasingly unlikely to find very + close to the walls of the well. This is because of two conditions: the above + condition says that the particle is /impossible/ to find + outside of the well, and the smoothly-varying condition says + that if a particle is impossible to find at a particular location, + it must be unlikely to be found nearby that location. + +#; physically allowed states are those that change in physically +#allowed ways. + + +#** Boundary conditions +Because the potential is infinite everywhere except within the well, +a realistic particle must be confined to exist only within the +well\mdash{}its wavefunction must be zero everywhere beyond the walls +of the well. + + +[fn:coords] I chose my coordinate system so that the well extends from +\(0a. \\ \end{cases}\) + + Physically realistic states must vary smoothly throughout + space. This means that if a particle in some state is very unlikely to be + /at/ a particular location, it is also very unlikely be /near/ + that location. Combining this requirement with the above + requirement, we find that the momentum operator no longer has + an eigenstate for each value of $p$; instead, only values of + $p$ that are integer multiples of $\pi a/\hbar$ are physically + realistic. Similarly, the energy operator no longer has an + eigenstate for each value of $E$; instead, the only energy + eigenstates in the infinitely deep well + are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$. + +* COMMENT: + +** Eigenstates with different eigenvalues are orthogonal + +#+begin_quote +*Theorem:* Eigenstates with different eigenvalues are orthogonal. +#+end_quote + +** COMMENT : +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$ +and $|b\rangle$ are eigenstates of $\Lambda$. This means that + + +\( +\begin{eqnarray} +\Lambda |a\rangle&=& a|a\rangle,\\ +\Lambda|b\rangle&=& b|b\rangle.\\ +\end{eqnarray} +\) + +If we take the difference of these eigenstates, we find that + +\( +\begin{eqnarray} +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle +\qquad \text{(because $\Lambda$ is linear.)}\\ +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and +$|b\rangle$ are eigenstates of $\Lambda$)} +\end{eqnarray}\) + + +which means that $a\neq b$. + +** Eigenvectors of hermitian operators span the space of solutions + +#+begin_quote +*Theorem:* If $\Omega$ is a hermitian operator, then every physically + allowed state can be written as a linear sum of eigenstates of + $\Omega$. +#+end_quote + + + +** Momentum and energy are hermitian operators +This ought to be true because hermitian operators correspond to +observable quantities. Since we expect momentum and energy to be +measureable quantities, we expect that there are hermitian operators +to represent them. + + +** Momentum and energy eigenstates in vacuum +An eigenstate of the momentum operator $P$ would be a state +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\). + +** Momentum and energy eigenstates in the infinitely deep well + + + +* Can you measure momentum in the infinitely deep well? +In summary, I thought I knew: +1. For any hermitian operator: eigenstates with different eigenvalues + are orthogonal. +2. For any hermitian operator: any physically realistic state can be + written as a linear sum of eigenstates of the operator. +3. The momentum operator and energy operator are hermitian, because + momentum and energy are observable quantities. +4. (The form of the momentum and energy eigenstates in the vacuum potential) +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) + +Additionally, I understood that because the infinitely deep potential +well is not realistic, states of such a system are not necessarily +physically realistic. Instead, I understood +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically +unrealistic Schr\ouml{}dinger equation and its boundary conditions. + +With that final caveat, here is the problem: + +According to (5), the momentum eigenstates in the well are + +\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\) + +However, /these/ states are not orthogonal, which contradicts the +assumption that (3) the momentum operator is hermitian and (2) +eigenstates of a hermitian are orthogonal if they have different eigenvalues. + +#+begin_quote +*Problem 1. The momentum eigenstates of the well are not orthogonal* + +/Proof./ If $p_1\neq p_2$, then + +\(\begin{eqnarray} +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)dx\\ +&=& \int_0^a p_1^*(x)p_2(x)dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ +outside the well.}\\ +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)dx}} +\end{eqnarray}\) +$\square$ + +#+end_quote + + + +** COMMENT Momentum eigenstates + +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\). + +In the infinitely deep potential well, the Hamiltonian is the same but +there is a new condition in order for states to qualify as physically +allowed: the states must not exist anywhere outside of well, as it +takes an infinite amount of energy to do so. + +Notice that the momentum eigenstates defined above do /not/ satisfy +this condition. + + + +* COMMENT +For each physical system, there is a Schr\ouml{}dinger equation that +describes how a particle's state $|\psi\rangle$ will change over +time. + +\(\begin{eqnarray} +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) + +This is a differential equation; each solution to the +Schr\ouml{}dinger equation is a state that is physically allowed for +our particle. Here, physically allowed states are +those that change in physically allowed ways. However, like any differential +equation, the Schr\ouml{}dinger equation can be accompanied by +/boundary conditions/\mdash{}conditions that further restrict which +states qualify as physically allowed. + + + + +** Eigenstates of momentum + + + + +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger + +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\) + + + + + + + +* COMMENT + +#* The infinite square well potential + +A particle exists in a potential that is +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the +particle exists in a potential[fn:coords][fn:infinity] + + +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for +}\;x<0\text{ or }x>a.\end{cases}\) + +The Schr\ouml{}dinger equation describes how the particle's state +\(|\psi\rangle\) will change over time in this system. + +\(\begin{eqnarray} +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) + +This is a differential equation; each solution to the +Schr\ouml{}dinger equation is a state that is physically allowed for +our particle. Here, physically allowed states are +those that change in physically allowed ways. However, like any differential +equation, the Schr\ouml{}dinger equation can be accompanied by +/boundary conditions/\mdash{}conditions that further restrict which +states qualify as physically allowed. + + +Whenever possible, physicists impose these boundary conditions: +- A physically allowed state ought to be a /smoothly-varying function of position./ This means + that if a particle in the state is likely to be /at/ a particular location, + it is also likely to be /near/ that location. + +These boundary conditions imply that for the square well potential in +this problem, + +- Physically allowed states must be totally confined to the well, + because it takes an infinite amount of energy to exist anywhere + outside of the well (and physically allowed states ought to have + only finite energy). +- Physically allowed states must be increasingly unlikely to find very + close to the walls of the well. This is because of two conditions: the above + condition says that the particle is /impossible/ to find + outside of the well, and the smoothly-varying condition says + that if a particle is impossible to find at a particular location, + it must be unlikely to be found nearby that location. + +#; physically allowed states are those that change in physically +#allowed ways. + + +#** Boundary conditions +Because the potential is infinite everywhere except within the well, +a realistic particle must be confined to exist only within the +well\mdash{}its wavefunction must be zero everywhere beyond the walls +of the well. + + +[fn:coords] I chose my coordinate system so that the well extends from +\(0a.\end{cases}\) + +This is apparently a perfectly respectable state: it is normalized ($A$ is a +normalization constant), it is zero +everywhere outside of the well, and it is moreover continuous. + +Even so, we will find a problem if we attempt to calculate the average +energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)). + +** First method + +For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv +H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a +function of $x$ because we know how to express $H$ and $\psi$ in terms +of $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and +$\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$ +is constant. + +Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the +following way. + +\(\begin{eqnarray} +\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H +\psi\rangle\\ +&=& \langle \psi H | H\psi \rangle\\ +&=& \langle \bar\psi | \bar\psi \rangle\\ +&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\ +&=& \frac{A^2\hbar^4 a}{m^2}\\ +\end{eqnarray}\) + +For future reference, observe that this value is nonzero +(which makes sense). + +** Second method +We can also calculate the average energy-squared of $|\psi\rangle$ in the +following way. + +\begin{eqnarray} +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ +&=& \langle \psi |H \bar\psi \rangle\\ +&=&\int_0^a Ax(x-a) +\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\ +&=& 0\quad (!)\\ +\end{eqnarray} + +The second-to-last term must be zero because the second derivative +of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero. + +* What is the problem? + +To recap: We used two different methods to calculate the average +energy-squared of a state $|\psi\rangle$. For the first method, we +used the fact that $H$ is a hermitian operator, replacing \(\langle +\psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H +\psi\rangle\). Using this substitution rule, we calculated the answer. + +For the second method, we didn't use the fact that $H$ was hermitian; +instead, we used the fact that we know how to represent $H$ and $\psi$ +as functions of $x$: $H$ is a differential operator +\(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic +function of $x$. By applying $H$ to $\psi$, we took several +derivatives and arrived at our answer. + +These two methods gave different results. In the following sections, +I'll describe and analyze the source of this difference. + +** Physical operators only act on physical wavefunctions + :PROPERTIES: + :ORDERED: t + :END: +#In quantum mechanics, an operator is a function that takes in a +#physical state and produces another physical state as ouput. Some +#operators correspond to physical quantities such as energy, +#momentum, or position; the mathematical properties of these operators correspond to +#physical properties of the system. + +#Eigenstates are an example of this correspondence: an + +Physical states are represented as wavefunctions in quantum +mechanics. Just as we disallow certain physically nonsensical states +in classical mechanics (for example, we consider it to be nonphysical +for an object to spontaneously disappear from one place and reappear +in another), we also disallow certain wavefunctions in quantum +mechanics. + +For example, since wavefunctions are supposed to correspond to +probability amplitudes, we require wavefunctions to be normalized +\((\langle \psi |\psi \rangle = 1)\). Generally, we disallow +wavefunctions that do not satisfy this property (although there are +some exceptions[fn:2]). + +As another example, we generally expect probability to vary smoothly\mdash{}if +a particle is very likely or very unlikely to be found at a particular +location, it should also be somewhat likely or somewhat unlikely to be +found /near/ that location. In more precise terms, we expect that for +physically meaningful wavefunctions, the probability +\(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of +$x$ and, again, we disallow wavefunctions that do not satisfy this +property because we consider them to be physically nonsensical. + +So, physical wavefunctions must satisfy certain properties +like the two just described. Wavefunctions that do not satisfy these properties are +rejected for being physically nonsensical: even though we can perform +calculations with them, the mathematical results we obtain do not mean +anything physically. + +Now, in quantum mechanics, an *operator* is a function that converts +states into other states. Some operators correspond to +physical quantities such as energy, momentum, or position, and as a +result, the mathematical properties of these operators correspond to +physical properties of the system. Physical operators are furthermore +subject to the following rule: they are only allowed to operate on +#physical wavefunctions, and they are only allowed to produce +#physical wavefunctions[fn:why]. +the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn:: + + If you require a hermitian operator to have physical + eigenstates, you get a very strong result: you guarantee that the + operator will convert /every/ physical wavefunction into another + physical wavefunction: + + For any linear operator $\Omega$, the eigenvalue equation is +\(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an +eigenstate $|\omega\rangle$ is a physical wavefunction, the +eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a +physical wavefunction as well. To elaborate, if the eigenstates of +$\Omega$ are physical functions, then $\Omega$ is guaranteed to +convert them into other physical functions. Even more is true if the +operator $\Omega$ is also hermitian: there is a theorem which states +that \ldquo{}If \Omega is hermitian, then every physical wavefunction +can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This +theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates +of \Omega are physically allowed/, then \Omega is guaranteed to +convert every physically allowed wavefunction into another physically +allowed wavefunction.]. + +In fact, this rule for physical operators is the source of our +problem, as we unknowingly violated it when applying our second +method! + +** The violation + +I'll start explaining this violation by being more specific about the +infinitely deep well potential. We have said already that physicists +require wavefunctions to satisfy certain properties in order to be +deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the +infinitely deep well +- Must be *normalizable*, because they correspond to + probability amplitudes. +- Must have *smoothly-varying probability*, because if a particle is very + likely to be at a location, it ought to be likely to be /near/ + it as well. +- Must *not exist outside the well*, because it + would take an infinite amount of energy to do so. + +Additionally, by combining the second and third conditions, some +physicists reason that wavefunctions in the infinitely deep well + +- Must *become zero* towards the edges of the well. + + + + +You'll remember we had + +\( +\begin{eqnarray} +\psi(x) &=& A\;x(x-a)\\ +\bar\psi(x)&=& \frac{-A\hbar^2}{m}\\ +&&\text{for }0\lt{}x\lt{}a\\ +\end{eqnarray} +\) + +In our second method, we wrote + + +\(\begin{eqnarray} +\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ +&=& \langle \psi |H \bar\psi \rangle\\ +& \vdots&\\ +&=& 0\\ +\end{eqnarray}\) + +However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction +$|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar +\psi\rangle$ is a nonphysical state: in the infinite square well, +physical wavefunctions must approach zero at the edges of the well, +which the constant function $|\bar\psi\rangle$ does not do. By +feeding $H$ a nonphysical wavefunction, we obtained nonsensical +results. + +Second, we claimed that $H$ was a physical operator\mdash{}that $H$ +was hermitian. According to the rule, this means $H$ must convert physical states into other +physical states. But $H$ converts the physical state $|\psi\rangle$ +into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some +physical states into nonphysical states, it cannot be a hermitian operator. + +** Boundary conditions affect hermiticity +We have now discovered a flaw: when applied to the state +$|\psi\rangle$, the second method violates the rule that physical +operators must only take in physical states and must only produce +physical states. This suggests that the problem was with the state +$|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem +is more serious still: the state $|\psi\rangle + +** COMMENT Re-examining physical constraints + +We have now discovered a flaw: when applied to the state +$|\psi\rangle$, the second method violates the rule that physical +operators must only take in physical states and must only produce +physical states. Let's examine the problem more closely. + +We have said already that physicists require wavefunctions to satisfy +certain properties in order to be deemed \ldquo{}physical\rdquo{}. To +be specific, wavefunctions in the infinitely deep well +- Must be *normalizable*, because they correspond to + probability amplitudes. +- Must have *smoothly-varying probability*, because if a particle is very + likely to be at a location, it ought to be likely to be /near/ + it as well. +- Must *not exist outside the well*, because it + would take an infinite amount of energy to do so. + +We now have discovered an important flaw in the second method: when +applied to the state $|\bar\psi\rangle$, the second method violates +the rule that physical operators must only take in +physical states and must only produce physical states. The problem is +even more serious, however + + + +[fn:1] I'm defining a new variable just to make certain expressions + look shorter; this cannot affect the content of the answer we'll + get. + +[fn:2] For example, in vaccuum (i.e., when the potential of the + physical system is $V(x)=0$ throughout all space), the momentum + eigenstates are not normalizable\mdash{}the relevant integral blows + up to infinity instead of converging to a number. Physicists modify + the definition of normalization slightly so that + \ldquo{}delta-normalizable \rdquo{} functions like these are included + among the physical wavefunctions. + + + +* COMMENT: What I thought I knew + +The following is a list of things I thought were true of quantum +mechanics; the catch is that the list contradicts itself. + +1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. +2. For any hermitian operator: Any physically allowed state can be + written as a linear sum of eigenstates of the operator. +3. The momentum operator and energy operator are hermitian, because + momentum and energy are measureable quantities. +4. In the vacuum potential, the momentum and energy operators have these eigenstates: + - the momentum operator has an eigenstate + \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$. + - the energy operator has an eigenstate \(|E\rangle = + \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and + the particular choice of momentum $p=\sqrt{2mE}$. +5. In the infinitely deep potential well, the momentum and energy + operators have these eigenstates: + - The momentum eigenstates and energy eigenstates have the same form + as in the vacuum potential: $p(x) = + \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$. + - Even so, because of the boundary conditions on the + well, we must make the following modifications: + + Physically realistic states must be impossible to find outside the well. (Only a state of infinite + energy could exist outside the well, and infinite energy is not + realistic.) This requirement means, for example, that momentum + eigenstates in the infinitely deep well must be + \(p(x) + = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; + \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\) + + Physically realistic states must vary smoothly throughout + space. This means that if a particle in some state is very unlikely to be + /at/ a particular location, it is also very unlikely be /near/ + that location. Combining this requirement with the above + requirement, we find that the momentum operator no longer has + an eigenstate for each value of $p$; instead, only values of + $p$ that are integer multiples of $\pi \hbar/a$ are physically + realistic. Similarly, the energy operator no longer has an + eigenstate for each value of $E$; instead, the only energy + eigenstates in the infinitely deep well + are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$. + +* COMMENT: + +** Eigenstates with different eigenvalues are orthogonal + +#+begin_quote +*Theorem:* Eigenstates with different eigenvalues are orthogonal. +#+end_quote + +** COMMENT : +I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$ +and $|b\rangle$ are eigenstates of $\Lambda$. This means that + + +\( +\begin{eqnarray} +\Lambda |a\rangle&=& a|a\rangle,\\ +\Lambda|b\rangle&=& b|b\rangle.\\ +\end{eqnarray} +\) + +If we take the difference of these eigenstates, we find that + +\( +\begin{eqnarray} +\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle +\qquad \text{(because $\Lambda$ is linear.)}\\ +&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and +$|b\rangle$ are eigenstates of $\Lambda$)} +\end{eqnarray}\) + + +which means that $a\neq b$. + +** Eigenvectors of hermitian operators span the space of solutions + +#+begin_quote +*Theorem:* If $\Omega$ is a hermitian operator, then every physically + allowed state can be written as a linear sum of eigenstates of + $\Omega$. +#+end_quote + + + +** Momentum and energy are hermitian operators +This ought to be true because hermitian operators correspond to +observable quantities. Since we expect momentum and energy to be +measureable quantities, we expect that there are hermitian operators +to represent them. + + +** Momentum and energy eigenstates in vacuum +An eigenstate of the momentum operator $P$ would be a state +\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\). + +** Momentum and energy eigenstates in the infinitely deep well + + + +* COMMENT Can you measure momentum in the infinitely deep well? +In summary, I thought I knew: +1. For any hermitian operator: eigenstates with different eigenvalues + are orthogonal. +2. For any hermitian operator: any physically realistic state can be + written as a linear sum of eigenstates of the operator. +3. The momentum operator and energy operator are hermitian, because + momentum and energy are observable quantities. +4. (The form of the momentum and energy eigenstates in the vacuum potential) +5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) + +Additionally, I understood that because the infinitely deep potential +well is not realistic, states of such a system are not necessarily +physically realistic. Instead, I understood +\ldquo{}realistic states\rdquo{} to be those that satisfy the physically +unrealistic Schr\ouml{}dinger equation and its boundary conditions. + +With that final caveat, here is the problem: + +According to (5), the momentum eigenstates in the well are + +\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\) + +(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.) + +However, /these/ states are not orthogonal, which contradicts the +assumption that (3) the momentum operator is hermitian and (2) +eigenstates of a hermitian are orthogonal if they have different eigenvalues. + +#+begin_quote +*Problem 1. The momentum eigenstates of the well are not orthogonal* + +/Proof./ If $p_1\neq p_2$, then + +\(\begin{eqnarray} +\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\ +&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ +outside the well.}\\ +&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\ +&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\ +\end{eqnarray}\) +$\square$ + +#+end_quote + + + +** COMMENT Momentum eigenstates + +In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the +momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\). + +In the infinitely deep potential well, the Hamiltonian is the same but +there is a new condition in order for states to qualify as physically +allowed: the states must not exist anywhere outside of well, as it +takes an infinite amount of energy to do so. + +Notice that the momentum eigenstates defined above do /not/ satisfy +this condition. + + + +* COMMENT +For each physical system, there is a Schr\ouml{}dinger equation that +describes how a particle's state $|\psi\rangle$ will change over +time. + +\(\begin{eqnarray} +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) + +This is a differential equation; each solution to the +Schr\ouml{}dinger equation is a state that is physically allowed for +our particle. Here, physically allowed states are +those that change in physically allowed ways. However, like any differential +equation, the Schr\ouml{}dinger equation can be accompanied by +/boundary conditions/\mdash{}conditions that further restrict which +states qualify as physically allowed. + + + + +** Eigenstates of momentum + + + + +#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger + +#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\) + + + + + + + +* COMMENT + +#* The infinite square well potential + +A particle exists in a potential that is +infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the +particle exists in a potential[fn:coords][fn:infinity] + + +\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for +}\;x<0\text{ or }x>a.\end{cases}\) + +The Schr\ouml{}dinger equation describes how the particle's state +\(|\psi\rangle\) will change over time in this system. + +\(\begin{eqnarray} +i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& +H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) + +This is a differential equation; each solution to the +Schr\ouml{}dinger equation is a state that is physically allowed for +our particle. Here, physically allowed states are +those that change in physically allowed ways. However, like any differential +equation, the Schr\ouml{}dinger equation can be accompanied by +/boundary conditions/\mdash{}conditions that further restrict which +states qualify as physically allowed. + + +Whenever possible, physicists impose these boundary conditions: +- A physically allowed state ought to be a /smoothly-varying function of position./ This means + that if a particle in the state is likely to be /at/ a particular location, + it is also likely to be /near/ that location. + +These boundary conditions imply that for the square well potential in +this problem, + +- Physically allowed states must be totally confined to the well, + because it takes an infinite amount of energy to exist anywhere + outside of the well (and physically allowed states ought to have + only finite energy). +- Physically allowed states must be increasingly unlikely to find very + close to the walls of the well. This is because of two conditions: the above + condition says that the particle is /impossible/ to find + outside of the well, and the smoothly-varying condition says + that if a particle is impossible to find at a particular location, + it must be unlikely to be found nearby that location. + +#; physically allowed states are those that change in physically +#allowed ways. + + +#** Boundary conditions +Because the potential is infinite everywhere except within the well, +a realistic particle must be confined to exist only within the +well\mdash{}its wavefunction must be zero everywhere beyond the walls +of the well. + + +[fn:coords] I chose my coordinate system so that the well extends from +\(0a.\end{cases}\) - -The Schr\ouml{}dinger equation describes how the particle's state -\(|\psi\rangle\) will change over time in this system. - -\(\begin{eqnarray} -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) - -This is a differential equation whose solutions are the physically -allowed states for the particle in this system. Like any differential -equation, - - -Like any differential equation, the Schr\ouml{}dinger equation -#; physically allowed states are those that change in physically -#allowed ways. - - -** Boundary conditions -Because the potential is infinite everywhere except within the well, -a realistic particle must be confined to exist only within the -well\mdash{}its wavefunction must be zero everywhere beyond the walls -of the well. - - -[fn:coords] I chose my coordinate system so that the well extends from -\(0a.\end{cases}\) - -The Schr\ouml{}dinger equation describes how the particle's state -\(|\psi\rangle\) will change over time in this system. - -\(\begin{eqnarray} -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) - -This is a differential equation whose solutions are the physically -allowed states for the particle in this system. Physically allowed -states are those that change in physically allowed ways. Like any -differential equation, the Schr\ouml{}dinger equation can be -accompanied by /boundary conditions/\mdash{}conditions that -further restrict which states qualify as physically allowed. - -Whenever possible, physicists impose these boundary conditions: -- The state should be a /continuous function of/ \(x\). This means - that if a particle is very likely to be /at/ a particular location, - it is also very likely to be /near/ that location. -- - -#; physically allowed states are those that change in physically -#allowed ways. - - -** Boundary conditions -Because the potential is infinite everywhere except within the well, -a realistic particle must be confined to exist only within the -well\mdash{}its wavefunction must be zero everywhere beyond the walls -of the well. - - -[fn:coords] I chose my coordinate system so that the well extends from -\(0a \\ \end{cases}\) -#0,&\text{for }x\lt{}0\text{ or }x\gt{}a\end{cases}\) - - no longer has an eigenstate for each value - of $p$. Instead, only values of $p$ that are integer multiples of - $\pi a/\hbar$ are physically realistic. - - - -* COMMENT: - -** Eigenstates with different eigenvalues are orthogonal - -#+begin_quote -*Theorem:* Eigenstates with different eigenvalues are orthogonal. -#+end_quote - -** COMMENT : -I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$ -and $|b\rangle$ are eigenstates of $\Lambda$. This means that - - -\( -\begin{eqnarray} -\Lambda |a\rangle&=& a|a\rangle,\\ -\Lambda|b\rangle&=& b|b\rangle.\\ -\end{eqnarray} -\) - -If we take the difference of these eigenstates, we find that - -\( -\begin{eqnarray} -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle -\qquad \text{(because $\Lambda$ is linear.)}\\ -&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and -$|b\rangle$ are eigenstates of $\Lambda$)} -\end{eqnarray}\) - - -which means that $a\neq b$. - -** Eigenvectors of hermitian operators span the space of solutions - -#+begin_quote -*Theorem:* If $\Omega$ is a hermitian operator, then every physically - allowed state can be written as a linear sum of eigenstates of - $\Omega$. -#+end_quote - - - -** Momentum and energy are hermitian operators -This ought to be true because hermitian operators correspond to -observable quantities. Since we expect momentum and energy to be -measureable quantities, we expect that there are hermitian operators -to represent them. - - -** Momentum and energy eigenstates in vacuum -An eigenstate of the momentum operator $P$ would be a state -\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\). - -** Momentum and energy eigenstates in the infinitely deep well - - - -* Can you measure momentum in the infinite square well? - - - -** COMMENT Momentum eigenstates - -In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the -momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\). - -In the infinitely deep potential well, the Hamiltonian is the same but -there is a new condition in order for states to qualify as physically -allowed: the states must not exist anywhere outside of well, as it -takes an infinite amount of energy to do so. - -Notice that the momentum eigenstates defined above do /not/ satisfy -this condition. - - - -* COMMENT -For each physical system, there is a Schr\ouml{}dinger equation that -describes how a particle's state $|\psi\rangle$ will change over -time. - -\(\begin{eqnarray} -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) - -This is a differential equation; each solution to the -Schr\ouml{}dinger equation is a state that is physically allowed for -our particle. Here, physically allowed states are -those that change in physically allowed ways. However, like any differential -equation, the Schr\ouml{}dinger equation can be accompanied by -/boundary conditions/\mdash{}conditions that further restrict which -states qualify as physically allowed. - - - - -** Eigenstates of momentum - - - - -#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger - -#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\) - - - - - - - -* COMMENT - -#* The infinite square well potential - -A particle exists in a potential that is -infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the -particle exists in a potential[fn:coords][fn:infinity] - - -\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for -}\;x<0\text{ or }x>a.\end{cases}\) - -The Schr\ouml{}dinger equation describes how the particle's state -\(|\psi\rangle\) will change over time in this system. - -\(\begin{eqnarray} -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) - -This is a differential equation; each solution to the -Schr\ouml{}dinger equation is a state that is physically allowed for -our particle. Here, physically allowed states are -those that change in physically allowed ways. However, like any differential -equation, the Schr\ouml{}dinger equation can be accompanied by -/boundary conditions/\mdash{}conditions that further restrict which -states qualify as physically allowed. - - -Whenever possible, physicists impose these boundary conditions: -- A physically allowed state ought to be a /smoothly-varying function of position./ This means - that if a particle in the state is likely to be /at/ a particular location, - it is also likely to be /near/ that location. - -These boundary conditions imply that for the square well potential in -this problem, - -- Physically allowed states must be totally confined to the well, - because it takes an infinite amount of energy to exist anywhere - outside of the well (and physically allowed states ought to have - only finite energy). -- Physically allowed states must be increasingly unlikely to find very - close to the walls of the well. This is because of two conditions: the above - condition says that the particle is /impossible/ to find - outside of the well, and the smoothly-varying condition says - that if a particle is impossible to find at a particular location, - it must be unlikely to be found nearby that location. - -#; physically allowed states are those that change in physically -#allowed ways. - - -#** Boundary conditions -Because the potential is infinite everywhere except within the well, -a realistic particle must be confined to exist only within the -well\mdash{}its wavefunction must be zero everywhere beyond the walls -of the well. - - -[fn:coords] I chose my coordinate system so that the well extends from -\(0a. \\ \end{cases}\) - + Physically realistic states must vary smoothly throughout - space. This means that if a particle in some state is very unlikely to be - /at/ a particular location, it is also very unlikely be /near/ - that location. Combining this requirement with the above - requirement, we find that the momentum operator no longer has - an eigenstate for each value of $p$; instead, only values of - $p$ that are integer multiples of $\pi a/\hbar$ are physically - realistic. Similarly, the energy operator no longer has an - eigenstate for each value of $E$; instead, the only energy - eigenstates in the infinitely deep well - are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$. - -* COMMENT: - -** Eigenstates with different eigenvalues are orthogonal - -#+begin_quote -*Theorem:* Eigenstates with different eigenvalues are orthogonal. -#+end_quote - -** COMMENT : -I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$ -and $|b\rangle$ are eigenstates of $\Lambda$. This means that - - -\( -\begin{eqnarray} -\Lambda |a\rangle&=& a|a\rangle,\\ -\Lambda|b\rangle&=& b|b\rangle.\\ -\end{eqnarray} -\) - -If we take the difference of these eigenstates, we find that - -\( -\begin{eqnarray} -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle -\qquad \text{(because $\Lambda$ is linear.)}\\ -&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and -$|b\rangle$ are eigenstates of $\Lambda$)} -\end{eqnarray}\) - - -which means that $a\neq b$. - -** Eigenvectors of hermitian operators span the space of solutions - -#+begin_quote -*Theorem:* If $\Omega$ is a hermitian operator, then every physically - allowed state can be written as a linear sum of eigenstates of - $\Omega$. -#+end_quote - - - -** Momentum and energy are hermitian operators -This ought to be true because hermitian operators correspond to -observable quantities. Since we expect momentum and energy to be -measureable quantities, we expect that there are hermitian operators -to represent them. - - -** Momentum and energy eigenstates in vacuum -An eigenstate of the momentum operator $P$ would be a state -\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\). - -** Momentum and energy eigenstates in the infinitely deep well - - - -* Can you measure momentum in the infinitely deep well? -In summary, I thought I knew: -1. For any hermitian operator: eigenstates with different eigenvalues - are orthogonal. -2. For any hermitian operator: any physically realistic state can be - written as a linear sum of eigenstates of the operator. -3. The momentum operator and energy operator are hermitian, because - momentum and energy are observable quantities. -4. (The form of the momentum and energy eigenstates in the vacuum potential) -5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) - -Additionally, I understood that because the infinitely deep potential -well is not realistic, states of such a system are not necessarily -physically realistic. Instead, I understood -\ldquo{}realistic states\rdquo{} to be those that satisfy the physically -unrealistic Schr\ouml{}dinger equation and its boundary conditions. - -With that final caveat, here is the problem: - -According to (5), the momentum eigenstates in the well are - -\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\) - -However, /these/ states are not orthogonal, which contradicts the -assumption that (3) the momentum operator is hermitian and (2) -eigenstates of a hermitian are orthogonal if they have different eigenvalues. - -#+begin_quote -*Problem 1. The momentum eigenstates of the well are not orthogonal* - -/Proof./ If $p_1\neq p_2$, then - -\(\begin{eqnarray} -\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)dx\\ -&=& \int_0^a p_1^*(x)p_2(x)dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ -outside the well.}\\ -&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)dx}} -\end{eqnarray}\) -$\square$ - -#+end_quote - - - -** COMMENT Momentum eigenstates - -In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the -momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\). - -In the infinitely deep potential well, the Hamiltonian is the same but -there is a new condition in order for states to qualify as physically -allowed: the states must not exist anywhere outside of well, as it -takes an infinite amount of energy to do so. - -Notice that the momentum eigenstates defined above do /not/ satisfy -this condition. - - - -* COMMENT -For each physical system, there is a Schr\ouml{}dinger equation that -describes how a particle's state $|\psi\rangle$ will change over -time. - -\(\begin{eqnarray} -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) - -This is a differential equation; each solution to the -Schr\ouml{}dinger equation is a state that is physically allowed for -our particle. Here, physically allowed states are -those that change in physically allowed ways. However, like any differential -equation, the Schr\ouml{}dinger equation can be accompanied by -/boundary conditions/\mdash{}conditions that further restrict which -states qualify as physically allowed. - - - - -** Eigenstates of momentum - - - - -#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger - -#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\) - - - - - - - -* COMMENT - -#* The infinite square well potential - -A particle exists in a potential that is -infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the -particle exists in a potential[fn:coords][fn:infinity] - - -\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for -}\;x<0\text{ or }x>a.\end{cases}\) - -The Schr\ouml{}dinger equation describes how the particle's state -\(|\psi\rangle\) will change over time in this system. - -\(\begin{eqnarray} -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) - -This is a differential equation; each solution to the -Schr\ouml{}dinger equation is a state that is physically allowed for -our particle. Here, physically allowed states are -those that change in physically allowed ways. However, like any differential -equation, the Schr\ouml{}dinger equation can be accompanied by -/boundary conditions/\mdash{}conditions that further restrict which -states qualify as physically allowed. - - -Whenever possible, physicists impose these boundary conditions: -- A physically allowed state ought to be a /smoothly-varying function of position./ This means - that if a particle in the state is likely to be /at/ a particular location, - it is also likely to be /near/ that location. - -These boundary conditions imply that for the square well potential in -this problem, - -- Physically allowed states must be totally confined to the well, - because it takes an infinite amount of energy to exist anywhere - outside of the well (and physically allowed states ought to have - only finite energy). -- Physically allowed states must be increasingly unlikely to find very - close to the walls of the well. This is because of two conditions: the above - condition says that the particle is /impossible/ to find - outside of the well, and the smoothly-varying condition says - that if a particle is impossible to find at a particular location, - it must be unlikely to be found nearby that location. - -#; physically allowed states are those that change in physically -#allowed ways. - - -#** Boundary conditions -Because the potential is infinite everywhere except within the well, -a realistic particle must be confined to exist only within the -well\mdash{}its wavefunction must be zero everywhere beyond the walls -of the well. - - -[fn:coords] I chose my coordinate system so that the well extends from -\(0a.\end{cases}\) - -This is apparently a perfectly respectable state: it is normalized ($A$ is a -normalization constant), it is zero -everywhere outside of the well, and it is moreover continuous. - -Even so, we will find a problem if we attempt to calculate the average -energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)). - -** First method - -For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv -H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a -function of $x$ because we know how to express $H$ and $\psi$ in terms -of $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and -$\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$ -is constant. - -Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the -following way. - -\(\begin{eqnarray} -\langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H -\psi\rangle\\ -&=& \langle \psi H | H\psi \rangle\\ -&=& \langle \bar\psi | \bar\psi \rangle\\ -&=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\ -&=& \frac{A^2\hbar^4 a}{m^2}\\ -\end{eqnarray}\) - -For future reference, observe that this value is nonzero -(which makes sense). - -** Second method -We can also calculate the average energy-squared of $|\psi\rangle$ in the -following way. - -\begin{eqnarray} -\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ -&=& \langle \psi |H \bar\psi \rangle\\ -&=&\int_0^a Ax(x-a) -\left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\ -&=& 0\quad (!)\\ -\end{eqnarray} - -The second-to-last term must be zero because the second derivative -of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero. - -* What is the problem? - -To recap: We used two different methods to calculate the average -energy-squared of a state $|\psi\rangle$. For the first method, we -used the fact that $H$ is a hermitian operator, replacing \(\langle -\psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H -\psi\rangle\). Using this substitution rule, we calculated the answer. - -For the second method, we didn't use the fact that $H$ was hermitian; -instead, we used the fact that we know how to represent $H$ and $\psi$ -as functions of $x$: $H$ is a differential operator -\(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic -function of $x$. By applying $H$ to $\psi$, we took several -derivatives and arrived at our answer. - -These two methods gave different results. In the following sections, -I'll describe and analyze the source of this difference. - -** Physical operators only act on physical wavefunctions - :PROPERTIES: - :ORDERED: t - :END: -#In quantum mechanics, an operator is a function that takes in a -#physical state and produces another physical state as ouput. Some -#operators correspond to physical quantities such as energy, -#momentum, or position; the mathematical properties of these operators correspond to -#physical properties of the system. - -#Eigenstates are an example of this correspondence: an - -Physical states are represented as wavefunctions in quantum -mechanics. Just as we disallow certain physically nonsensical states -in classical mechanics (for example, we consider it to be nonphysical -for an object to spontaneously disappear from one place and reappear -in another), we also disallow certain wavefunctions in quantum -mechanics. - -For example, since wavefunctions are supposed to correspond to -probability amplitudes, we require wavefunctions to be normalized -\((\langle \psi |\psi \rangle = 1)\). Generally, we disallow -wavefunctions that do not satisfy this property (although there are -some exceptions[fn:2]). - -As another example, we generally expect probability to vary smoothly\mdash{}if -a particle is very likely or very unlikely to be found at a particular -location, it should also be somewhat likely or somewhat unlikely to be -found /near/ that location. In more precise terms, we expect that for -physically meaningful wavefunctions, the probability -\(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of -$x$ and, again, we disallow wavefunctions that do not satisfy this -property because we consider them to be physically nonsensical. - -So, physical wavefunctions must satisfy certain properties -like the two just described. Wavefunctions that do not satisfy these properties are -rejected for being physically nonsensical: even though we can perform -calculations with them, the mathematical results we obtain do not mean -anything physically. - -Now, in quantum mechanics, an *operator* is a function that converts -states into other states. Some operators correspond to -physical quantities such as energy, momentum, or position, and as a -result, the mathematical properties of these operators correspond to -physical properties of the system. Physical operators are furthermore -subject to the following rule: they are only allowed to operate on -#physical wavefunctions, and they are only allowed to produce -#physical wavefunctions[fn:why]. -the wavefunctions we have deemed physical, and they only produce physical wavefunctions[fn:: - - If you require a hermitian operator to have physical - eigenstates, you get a very strong result: you guarantee that the - operator will convert /every/ physical wavefunction into another - physical wavefunction: - - For any linear operator $\Omega$, the eigenvalue equation is -\(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an -eigenstate $|\omega\rangle$ is a physical wavefunction, the -eigenvalue equation equation forces $\Omega|\omega\rangle$ to be a -physical wavefunction as well. To elaborate, if the eigenstates of -$\Omega$ are physical functions, then $\Omega$ is guaranteed to -convert them into other physical functions. Even more is true if the -operator $\Omega$ is also hermitian: there is a theorem which states -that \ldquo{}If \Omega is hermitian, then every physical wavefunction -can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This -theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates -of \Omega are physically allowed/, then \Omega is guaranteed to -convert every physically allowed wavefunction into another physically -allowed wavefunction.]. - -In fact, this rule for physical operators is the source of our -problem, as we unknowingly violated it when applying our second -method! - -** The violation - -I'll start explaining this violation by being more specific about the -infinitely deep well potential. We have said already that physicists -require wavefunctions to satisfy certain properties in order to be -deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the -infinitely deep well -- Must be *normalizable*, because they correspond to - probability amplitudes. -- Must have *smoothly-varying probability*, because if a particle is very - likely to be at a location, it ought to be likely to be /near/ - it as well. -- Must *not exist outside the well*, because it - would take an infinite amount of energy to do so. - -Additionally, by combining the second and third conditions, some -physicists reason that wavefunctions in the infinitely deep well - -- Must *become zero* towards the edges of the well. - - - - -You'll remember we had - -\( -\begin{eqnarray} -\psi(x) &=& A\;x(x-a)\\ -\bar\psi(x)&=& \frac{-A\hbar^2}{m}\\ -&&\text{for }0\lt{}x\lt{}a\\ -\end{eqnarray} -\) - -In our second method, we wrote - - -\(\begin{eqnarray} -\langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ -&=& \langle \psi |H \bar\psi \rangle\\ -& \vdots&\\ -&=& 0\\ -\end{eqnarray}\) - -However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction -$|\bar \psi\rangle$, because $H$ is a physical operator and $|\bar -\psi\rangle$ is a nonphysical state: in the infinite square well, -physical wavefunctions must approach zero at the edges of the well, -which the constant function $|\bar\psi\rangle$ does not do. By -feeding $H$ a nonphysical wavefunction, we obtained nonsensical -results. - -Second, we claimed that $H$ was a physical operator\mdash{}that $H$ -was hermitian. According to the rule, this means $H$ must convert physical states into other -physical states. But $H$ converts the physical state $|\psi\rangle$ -into the nonphysical state $|\bar\psi\rangle = H|\psi\rangle$ is not. Because $H$ converts some -physical states into nonphysical states, it cannot be a hermitian operator. - -** Boundary conditions affect hermiticity -We have now discovered a flaw: when applied to the state -$|\psi\rangle$, the second method violates the rule that physical -operators must only take in physical states and must only produce -physical states. This suggests that the problem was with the state -$|\bar\psi\rangle$ that we wrongly fed into $H$. However, the problem -is more serious still: the state $|\psi\rangle - -** COMMENT Re-examining physical constraints - -We have now discovered a flaw: when applied to the state -$|\psi\rangle$, the second method violates the rule that physical -operators must only take in physical states and must only produce -physical states. Let's examine the problem more closely. - -We have said already that physicists require wavefunctions to satisfy -certain properties in order to be deemed \ldquo{}physical\rdquo{}. To -be specific, wavefunctions in the infinitely deep well -- Must be *normalizable*, because they correspond to - probability amplitudes. -- Must have *smoothly-varying probability*, because if a particle is very - likely to be at a location, it ought to be likely to be /near/ - it as well. -- Must *not exist outside the well*, because it - would take an infinite amount of energy to do so. - -We now have discovered an important flaw in the second method: when -applied to the state $|\bar\psi\rangle$, the second method violates -the rule that physical operators must only take in -physical states and must only produce physical states. The problem is -even more serious, however - - - -[fn:1] I'm defining a new variable just to make certain expressions - look shorter; this cannot affect the content of the answer we'll - get. - -[fn:2] For example, in vaccuum (i.e., when the potential of the - physical system is $V(x)=0$ throughout all space), the momentum - eigenstates are not normalizable\mdash{}the relevant integral blows - up to infinity instead of converging to a number. Physicists modify - the definition of normalization slightly so that - \ldquo{}delta-normalizable \rdquo{} functions like these are included - among the physical wavefunctions. - - - -* COMMENT: What I thought I knew - -The following is a list of things I thought were true of quantum -mechanics; the catch is that the list contradicts itself. - -1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. -2. For any hermitian operator: Any physically allowed state can be - written as a linear sum of eigenstates of the operator. -3. The momentum operator and energy operator are hermitian, because - momentum and energy are measureable quantities. -4. In the vacuum potential, the momentum and energy operators have these eigenstates: - - the momentum operator has an eigenstate - \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$. - - the energy operator has an eigenstate \(|E\rangle = - \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and - the particular choice of momentum $p=\sqrt{2mE}$. -5. In the infinitely deep potential well, the momentum and energy - operators have these eigenstates: - - The momentum eigenstates and energy eigenstates have the same form - as in the vacuum potential: $p(x) = - \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$. - - Even so, because of the boundary conditions on the - well, we must make the following modifications: - + Physically realistic states must be impossible to find outside the well. (Only a state of infinite - energy could exist outside the well, and infinite energy is not - realistic.) This requirement means, for example, that momentum - eigenstates in the infinitely deep well must be - \(p(x) - = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; - \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\) - + Physically realistic states must vary smoothly throughout - space. This means that if a particle in some state is very unlikely to be - /at/ a particular location, it is also very unlikely be /near/ - that location. Combining this requirement with the above - requirement, we find that the momentum operator no longer has - an eigenstate for each value of $p$; instead, only values of - $p$ that are integer multiples of $\pi \hbar/a$ are physically - realistic. Similarly, the energy operator no longer has an - eigenstate for each value of $E$; instead, the only energy - eigenstates in the infinitely deep well - are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$. - -* COMMENT: - -** Eigenstates with different eigenvalues are orthogonal - -#+begin_quote -*Theorem:* Eigenstates with different eigenvalues are orthogonal. -#+end_quote - -** COMMENT : -I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$ -and $|b\rangle$ are eigenstates of $\Lambda$. This means that - - -\( -\begin{eqnarray} -\Lambda |a\rangle&=& a|a\rangle,\\ -\Lambda|b\rangle&=& b|b\rangle.\\ -\end{eqnarray} -\) - -If we take the difference of these eigenstates, we find that - -\( -\begin{eqnarray} -\Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle -\qquad \text{(because $\Lambda$ is linear.)}\\ -&=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and -$|b\rangle$ are eigenstates of $\Lambda$)} -\end{eqnarray}\) - - -which means that $a\neq b$. - -** Eigenvectors of hermitian operators span the space of solutions - -#+begin_quote -*Theorem:* If $\Omega$ is a hermitian operator, then every physically - allowed state can be written as a linear sum of eigenstates of - $\Omega$. -#+end_quote - - - -** Momentum and energy are hermitian operators -This ought to be true because hermitian operators correspond to -observable quantities. Since we expect momentum and energy to be -measureable quantities, we expect that there are hermitian operators -to represent them. - - -** Momentum and energy eigenstates in vacuum -An eigenstate of the momentum operator $P$ would be a state -\(|p\rangle\) such that \(P|p\rangle=p|p\rangle\). - -** Momentum and energy eigenstates in the infinitely deep well - - - -* COMMENT Can you measure momentum in the infinitely deep well? -In summary, I thought I knew: -1. For any hermitian operator: eigenstates with different eigenvalues - are orthogonal. -2. For any hermitian operator: any physically realistic state can be - written as a linear sum of eigenstates of the operator. -3. The momentum operator and energy operator are hermitian, because - momentum and energy are observable quantities. -4. (The form of the momentum and energy eigenstates in the vacuum potential) -5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) - -Additionally, I understood that because the infinitely deep potential -well is not realistic, states of such a system are not necessarily -physically realistic. Instead, I understood -\ldquo{}realistic states\rdquo{} to be those that satisfy the physically -unrealistic Schr\ouml{}dinger equation and its boundary conditions. - -With that final caveat, here is the problem: - -According to (5), the momentum eigenstates in the well are - -\(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\) - -(Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.) - -However, /these/ states are not orthogonal, which contradicts the -assumption that (3) the momentum operator is hermitian and (2) -eigenstates of a hermitian are orthogonal if they have different eigenvalues. - -#+begin_quote -*Problem 1. The momentum eigenstates of the well are not orthogonal* - -/Proof./ If $p_1\neq p_2$, then - -\(\begin{eqnarray} -\langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\ -&=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ -outside the well.}\\ -&=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\ -&=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\ -\end{eqnarray}\) -$\square$ - -#+end_quote - - - -** COMMENT Momentum eigenstates - -In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the -momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\). - -In the infinitely deep potential well, the Hamiltonian is the same but -there is a new condition in order for states to qualify as physically -allowed: the states must not exist anywhere outside of well, as it -takes an infinite amount of energy to do so. - -Notice that the momentum eigenstates defined above do /not/ satisfy -this condition. - - - -* COMMENT -For each physical system, there is a Schr\ouml{}dinger equation that -describes how a particle's state $|\psi\rangle$ will change over -time. - -\(\begin{eqnarray} -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) - -This is a differential equation; each solution to the -Schr\ouml{}dinger equation is a state that is physically allowed for -our particle. Here, physically allowed states are -those that change in physically allowed ways. However, like any differential -equation, the Schr\ouml{}dinger equation can be accompanied by -/boundary conditions/\mdash{}conditions that further restrict which -states qualify as physically allowed. - - - - -** Eigenstates of momentum - - - - -#In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger - -#\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\) - - - - - - - -* COMMENT - -#* The infinite square well potential - -A particle exists in a potential that is -infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the -particle exists in a potential[fn:coords][fn:infinity] - - -\(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for -}\;x<0\text{ or }x>a.\end{cases}\) - -The Schr\ouml{}dinger equation describes how the particle's state -\(|\psi\rangle\) will change over time in this system. - -\(\begin{eqnarray} -i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& -H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) - -This is a differential equation; each solution to the -Schr\ouml{}dinger equation is a state that is physically allowed for -our particle. Here, physically allowed states are -those that change in physically allowed ways. However, like any differential -equation, the Schr\ouml{}dinger equation can be accompanied by -/boundary conditions/\mdash{}conditions that further restrict which -states qualify as physically allowed. - - -Whenever possible, physicists impose these boundary conditions: -- A physically allowed state ought to be a /smoothly-varying function of position./ This means - that if a particle in the state is likely to be /at/ a particular location, - it is also likely to be /near/ that location. - -These boundary conditions imply that for the square well potential in -this problem, - -- Physically allowed states must be totally confined to the well, - because it takes an infinite amount of energy to exist anywhere - outside of the well (and physically allowed states ought to have - only finite energy). -- Physically allowed states must be increasingly unlikely to find very - close to the walls of the well. This is because of two conditions: the above - condition says that the particle is /impossible/ to find - outside of the well, and the smoothly-varying condition says - that if a particle is impossible to find at a particular location, - it must be unlikely to be found nearby that location. - -#; physically allowed states are those that change in physically -#allowed ways. - - -#** Boundary conditions -Because the potential is infinite everywhere except within the well, -a realistic particle must be confined to exist only within the -well\mdash{}its wavefunction must be zero everywhere beyond the walls -of the well. - - -[fn:coords] I chose my coordinate system so that the well extends from -\(0