# HG changeset patch # User Robert McIntyre # Date 1318918675 25200 # Node ID f743fd0f4d8b14a7b9d891aa97c063b89c8cabd7 initial commit of dylan's stuff diff -r 000000000000 -r f743fd0f4d8b org/bk.org --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/org/bk.org Mon Oct 17 23:17:55 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 +\(0a.\end{cases}\) + +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; +we instead 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. Such operators are called +/hermitian operators/; one important property of hermitian operators +is this rule: + +#+begin_quote +*Hermitian operator rule:* A hermitian operator must only operate on +the wavefunctions we have deemed physical, and must only produce +physical wavefunctions[fn:: If you require a hermitian operator to +have physical eigenstates (which is reasonable), 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 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.]. +#+end_quote + +As you can see, this rule comes in two pieces. The first part is a +constraint on *you*, the physicist: you must never feed a nonphysical +state into a Hermitian operator, as it may produce nonsense. The +second part is a constraint on the *operator*: the operator is +guaranteed only to produce physical wavefunctions. + +In fact, this rule for hermitian operators is the source of our +problem, as we unknowingly violated it when applying our second +method! + +** The Hamiltonian is nonphysical +You'll remember that in the second method we had wavefunctions within +the well + +\( +\begin{eqnarray} +\psi(x) &=& A\;x(x-a)\\ +\bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\ +\end{eqnarray} +\) + + Using this, 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 supposed to be a physical operator and $|\bar +\psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does +not approach zero at the edges of the well. By +feeding $H$ a nonphysical wavefunction, we obtained nonsensical +results. + +Second, and more importantly, we were wrong to claim that $H$ was a +physical operator\mdash{}that $H$ was hermitian. According to the +rule, a hermitian operator must convert physical states into other +physical states. But $|\psi\rangle$ is a physical state, as we said +when we first introduced it \mdash{}it is a normalized, continuous +function which approaches zero at the edges of the well and doesn't +exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical +because it does not go to zero at the edges of the well. It is +therefore impermissible for $H$ to transform the physical state +$|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$. Because +$H$ converts some physical states into nonphysical states, it cannot +be a hermitian operator as we assumed. + +# Boundary conditions affect hermiticity +** Boundary conditions alter hermiticity +It may surprise you (and it certainly surprised me) to find that the +Hamiltonian is not hermitian. One of the fundamental principles of +quantum mechanics is that hermitian operators correspond to physically +observable quantities; for this reason, surely the +Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian? + +But we must understand the correspondence between physically +observable quantities and hermitian operators: every hermitian +operator corresponds to a physically observable quantity, but not +every quantity that intuitively \ldquo{}ought\rdquo{} to be observable +will correspond to a hermitian operator[fn::For a simple example, +consider the differential operator \(D=\frac{d}{dx}\); although our +intuitions might suggest that $D$ is observable which leads us to +guess that $D$ is hermitian, it isn't. Still, the very closely related +operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we +ought to validate our intuitions by checking the definitions.]. The +true definition of a hermitian operator imply that the Hamiltonian +stops being hermitian in the infinitely deep well. Here we arrive at a +crucial point: + +Operators do not change /form/ between problems: the one-dimensional +Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the +one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$, +and so on. + +However, operators do change in this respect: hermitian operators must +only take in physical states, and must only produce physical states; because +in different problems we /do/ change the requirements for being a +physical state, we also change what it takes for +an operator to be called hermitian. As a result, an operator that +is hermitian in one setting may fail to be hermitian in another. + +Having seen how boundary conditions can affect hermiticity, we +ought to be extra careful about which conditions we impose on our +wavefunctions. + +** Choosing the right constraints + + 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. + +These conditions are surely reasonable. However, physicists sometimes +assert that in order to satisfy the second and third conditions, +physical wavefunctions + +- (?) Must *smoothly approach zero* towards the edges of the well. + +This final constraint is our reason for rejecting $|\bar\psi\rangle$ +as nonphysical and is consequently the reason why $H$ is not hermitian. If +we can convince ourselves that the final constraint is unnecessary, +$H$ may again be hermitian. This will satisfy our intuitions that the +energy operator /ought/ to be hermitian. + +But in fact, we have the following mathematical observation to save +us: a function $f$ does not need to be continuous in order for the +integral \(\int^x f\) to be continuous. As a particularly relevant +example, you may now notice that the function $\bar\psi(x)$ is not +itself continuous, although the integral $\int_0^x \bar\psi$ /is/ +continuous. Evidently, it doesn't matter that the wavefunction +$\bar\psi$ itself is not continuous; the probability corresponding to +$\bar\psi$ /does/ manage to vary continuously anyways. Because the +probability corresponding to $\bar\psi$ is the only aspect of +$\bar\psi$ which we can detect physically, we /can/ safely omit the +final constraint while keeping the other three. + +** Symmetric operators look like hermitian operators, but sometimes aren't. + + +#+end_quote +** 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