rlm@0: #+TITLE: Bugs in quantum mechanics rlm@0: #+AUTHOR: Dylan Holmes rlm@0: #+DESCRIPTION: I explain hermitian operators and use them to solve a confusing problem in quantum mechanics. rlm@0: #+KEYWORDS: quantum mechanics, self-adjoint, hermitian, square well, momentum rlm@0: #+SETUPFILE: ../../aurellem/org/setup.org rlm@0: #+INCLUDE: ../../aurellem/org/level-0.org rlm@0: rlm@0: rlm@0: rlm@0: #Bugs in Quantum Mechanics rlm@0: #Bugs in the Quantum-Mechanical Momentum Operator rlm@0: rlm@0: rlm@4: rlm@0: I studied quantum mechanics the same way I study most subjects\mdash{} rlm@0: by collecting (and squashing) bugs in my understanding. One of these rlm@0: bugs persisted throughout two semesters of rlm@0: quantum mechanics coursework until I finally found rlm@0: the paper rlm@0: [[http://arxiv.org/abs/quant-ph/0103153][/Self-adjoint extensions of operators and the teaching of quantum rlm@0: mechanics/]], which helped me stamp out the bug entirely. I decided to rlm@0: write an article about the problem and its solution for a number of reasons: rlm@0: rlm@0: - Although the paper was not unreasonably dense, it was written for rlm@0: teachers. I wanted to write an article for students. rlm@0: - I wanted to popularize the problem and its solution because other rlm@0: explanations are currently too hard to find. (Even Shankar's rlm@0: excellent [[http://books.google.com/books/about/Principles_of_quantum_mechanics.html?id=2zypV5EbKuIC][textbook]] doesn't mention it, as far as I know.) rlm@0: - Attempting an explanation is my way of making rlm@0: sure that the bug really /is/ gone. rlm@0: # entirely eradicated. rlm@0: rlm@0: * COMMENT rlm@0: I recommend the rlm@0: paper not only for students who are learning rlm@0: quantum mechanics, but especially for teachers interested in debugging rlm@0: them. rlm@0: rlm@0: * COMMENT rlm@0: On my first exam in quantum mechanics, my professor asked us to rlm@0: describe how certain measurements would affect a particle in a rlm@0: box. Many of these measurement questions required routine application rlm@0: of skills we had recently learned\mdash{}first, you recall (or rlm@0: calculate) the eigenstates of the quantity rlm@0: to be measured; second, you write the given state as a linear rlm@0: sum of these eigenstates\mdash{} the coefficients on each term give rlm@0: the probability amplitude. rlm@0: rlm@0: rlm@0: * Two methods of calculation that give different results. rlm@0: rlm@0: In the infinitely deep well, there is a particle in the the rlm@0: normalized state rlm@0: rlm@0: \(\psi(x) = \begin{cases}A\; x(x-a)&\text{for }0\lt{}x\lt{}a;\\0,&\text{for }x\lt{}0\text{ or }x>a.\end{cases}\) rlm@0: rlm@0: This is apparently a perfectly respectable state: it is normalized ($A$ is a rlm@0: normalization constant), it is zero rlm@0: everywhere outside of the well, and it is moreover continuous. rlm@0: rlm@0: Even so, we will find a problem if we attempt to calculate the average rlm@0: energy-squared of this state (that is, the quantity \(\langle \psi | H^2 | \psi \rangle\)). rlm@0: rlm@0: ** First method rlm@0: rlm@0: For short, define a new variable[fn:1] \(|\bar \psi\rangle \equiv rlm@0: H|\psi\rangle\). We can express the state $|\bar\psi\rangle$ as a rlm@0: function of $x$ because we know how to express $H$ and $\psi$ in terms rlm@0: of $x$: $H$ is the differential operator $\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}$, and rlm@0: $\psi(x)$ is the function defined above. So, we get \(\quad\bar\psi(x) = \frac{-A\hbar^2}{m}\). For future reference, observe that $\bar\psi(x)$ rlm@0: is constant. rlm@0: rlm@0: Having introduced $|\bar\psi\rangle$, we can calculate the average energy-squared of $|\psi\rangle$ in the rlm@0: following way. rlm@0: rlm@0: \(\begin{eqnarray} rlm@0: \langle \psi | H^2 |\psi\rangle &=& \langle \psi H^\dagger | H rlm@0: \psi\rangle\\ rlm@0: &=& \langle \psi H | H\psi \rangle\\ rlm@0: &=& \langle \bar\psi | \bar\psi \rangle\\ rlm@0: &=& \int_0^a \left(\frac{A\hbar^2}{m}\right)^2\;dx\\ rlm@0: &=& \frac{A^2\hbar^4 a}{m^2}\\ rlm@0: \end{eqnarray}\) rlm@0: rlm@0: For future reference, observe that this value is nonzero rlm@0: (which makes sense). rlm@0: rlm@0: ** Second method rlm@0: We can also calculate the average energy-squared of $|\psi\rangle$ in the rlm@0: following way. rlm@0: rlm@0: \begin{eqnarray} rlm@0: \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ rlm@0: &=& \langle \psi |H \bar\psi \rangle\\ rlm@0: &=&\int_0^a Ax(x-a) rlm@0: \left(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\right)\frac{-A\hbar^2}{m}\;dx\\ rlm@0: &=& 0\quad (!)\\ rlm@0: \end{eqnarray} rlm@0: rlm@0: The second-to-last term must be zero because the second derivative rlm@0: of a constant \(\left(\frac{-A\hbar^2}{m}\right)\) is zero. rlm@0: rlm@0: * What is the problem? rlm@0: rlm@0: To recap: We used two different methods to calculate the average rlm@0: energy-squared of a state $|\psi\rangle$. For the first method, we rlm@0: used the fact that $H$ is a hermitian operator, replacing \(\langle rlm@0: \psi H^\dagger | H \psi\rangle\) with \(\langle \psi H | H rlm@0: \psi\rangle\). Using this substitution rule, we calculated the answer. rlm@0: rlm@0: For the second method, rlm@0: #we didn't use the fact that $H$ was hermitian; rlm@0: we instead used the fact that we know how to represent $H$ and $\psi$ rlm@0: as functions of $x$: $H$ is a differential operator rlm@0: \(\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}\) and $\psi$ is a quadratic rlm@0: function of $x$. By applying $H$ to $\psi$, we took several rlm@0: derivatives and arrived at our answer. rlm@0: rlm@0: These two methods gave different results. In the following sections, rlm@0: I'll describe and analyze the source of this difference. rlm@0: rlm@0: ** Physical operators only act on physical wavefunctions rlm@0: :PROPERTIES: rlm@0: :ORDERED: t rlm@0: :END: rlm@0: #In quantum mechanics, an operator is a function that takes in a rlm@0: #physical state and produces another physical state as ouput. Some rlm@0: #operators correspond to physical quantities such as energy, rlm@0: #momentum, or position; the mathematical properties of these operators correspond to rlm@0: #physical properties of the system. rlm@0: rlm@0: #Eigenstates are an example of this correspondence: an rlm@0: rlm@0: Physical states are represented as wavefunctions in quantum rlm@0: mechanics. Just as we disallow certain physically nonsensical states rlm@0: in classical mechanics (for example, we consider it to be nonphysical rlm@0: for an object to spontaneously disappear from one place and reappear rlm@0: in another), we also disallow certain wavefunctions in quantum rlm@0: mechanics. rlm@0: rlm@0: For example, since wavefunctions are supposed to correspond to rlm@0: probability amplitudes, we require wavefunctions to be normalized rlm@0: \((\langle \psi |\psi \rangle = 1)\). Generally, we disallow rlm@0: wavefunctions that do not satisfy this property (although there are rlm@0: some exceptions[fn:2]). rlm@0: rlm@0: As another example, we generally expect probability to vary smoothly\mdash{}if rlm@0: a particle is very likely or very unlikely to be found at a particular rlm@0: location, it should also be somewhat likely or somewhat unlikely to be rlm@0: found /near/ that location. In more precise terms, we expect that for rlm@0: physically meaningful wavefunctions, the probability rlm@0: \(\text{Pr}(x)=\int^x \psi^*\psi\) should be a continuous function of rlm@0: $x$ and, again, we disallow wavefunctions that do not satisfy this rlm@0: property because we consider them to be physically nonsensical. rlm@0: rlm@0: So, physical wavefunctions must satisfy certain properties rlm@0: like the two just described. Wavefunctions that do not satisfy these properties are rlm@0: rejected for being physically nonsensical: even though we can perform rlm@0: calculations with them, the mathematical results we obtain do not mean rlm@0: anything physically. rlm@0: rlm@0: Now, in quantum mechanics, an *operator* is a function that converts rlm@0: states into other states. Some operators correspond to rlm@0: physical quantities such as energy, momentum, or position, and as a rlm@0: result, the mathematical properties of these operators correspond to rlm@0: physical properties of the system. Such operators are called rlm@0: /hermitian operators/; one important property of hermitian operators rlm@0: is this rule: rlm@0: rlm@0: #+begin_quote rlm@0: *Hermitian operator rule:* A hermitian operator must only operate on rlm@0: the wavefunctions we have deemed physical, and must only produce rlm@0: physical wavefunctions[fn:: If you require a hermitian operator to rlm@0: have physical eigenstates (which is reasonable), you get a very strong result: you guarantee rlm@0: that the operator will convert every physical wavefunction into rlm@0: another physical wavefunction: rlm@0: rlm@0: For any linear operator $\Omega$, the eigenvalue equation is rlm@0: \(\Omega|\omega\rangle = \omega |\omega\rangle\). Notice that if an rlm@0: eigenstate $|\omega\rangle$ is a physical wavefunction, the rlm@0: eigenvalue equation forces $\Omega|\omega\rangle$ to be a rlm@0: physical wavefunction as well. To elaborate, if the eigenstates of rlm@0: $\Omega$ are physical functions, then $\Omega$ is guaranteed to rlm@0: convert them into other physical functions. Even more is true if the rlm@0: operator $\Omega$ is also hermitian: there is a theorem which states rlm@0: that \ldquo{}If \Omega is hermitian, then every physical wavefunction rlm@0: can be written as a weighted sum of eigenstates of \Omega.\rdquo{} This rlm@0: theorem implies that /if $\Omega$ is hermitian/, and /if the eigenstates rlm@0: of \Omega are physically allowed/, then \Omega is guaranteed to rlm@0: convert every physically allowed wavefunction into another physically rlm@0: allowed wavefunction.]. rlm@0: #+end_quote rlm@0: rlm@0: As you can see, this rule comes in two pieces. The first part is a rlm@0: constraint on *you*, the physicist: you must never feed a nonphysical rlm@0: state into a Hermitian operator, as it may produce nonsense. The rlm@0: second part is a constraint on the *operator*: the operator is rlm@0: guaranteed only to produce physical wavefunctions. rlm@0: rlm@0: In fact, this rule for hermitian operators is the source of our rlm@0: problem, as we unknowingly violated it when applying our second rlm@0: method! rlm@0: rlm@0: ** The Hamiltonian is nonphysical rlm@0: You'll remember that in the second method we had wavefunctions within rlm@0: the well rlm@0: rlm@0: \( rlm@0: \begin{eqnarray} rlm@0: \psi(x) &=& A\;x(x-a)\\ rlm@0: \bar\psi(x)&\equiv& H|\psi\rangle = \frac{-A\hbar^2}{m}\\ rlm@0: \end{eqnarray} rlm@0: \) rlm@0: rlm@0: Using this, we wrote rlm@0: rlm@0: rlm@0: \(\begin{eqnarray} rlm@0: \langle \psi | H^2 |\psi\rangle &=& \langle \psi | H\,H \psi \rangle\\ rlm@0: &=& \langle \psi |H \bar\psi \rangle\\ rlm@0: & \vdots&\\ rlm@0: &=& 0\\ rlm@0: \end{eqnarray}\) rlm@0: rlm@0: However, there are two issues here. First, we were not allowed to operate with $H$ on the wavefunction rlm@0: $|\bar \psi\rangle$, because $H$ is supposed to be a physical operator and $|\bar rlm@0: \psi\rangle$ is a nonphysical state. Indeed, the constant function $|\bar\psi\rangle$ does rlm@0: not approach zero at the edges of the well. By rlm@0: feeding $H$ a nonphysical wavefunction, we obtained nonsensical rlm@0: results. rlm@0: rlm@0: Second, and more importantly, we were wrong to claim that $H$ was a rlm@0: physical operator\mdash{}that $H$ was hermitian. According to the rlm@0: rule, a hermitian operator must convert physical states into other rlm@0: physical states. But $|\psi\rangle$ is a physical state, as we said rlm@0: when we first introduced it \mdash{}it is a normalized, continuous rlm@0: function which approaches zero at the edges of the well and doesn't rlm@0: exist outside it. On the other hand, $|\bar\psi\rangle$ is nonphysical rlm@0: because it does not go to zero at the edges of the well. It is rlm@0: therefore impermissible for $H$ to transform the physical state rlm@0: $|\psi\rangle$ into the nonphysical state $|\bar\psi\rangle$. Because rlm@0: $H$ converts some physical states into nonphysical states, it cannot rlm@0: be a hermitian operator as we assumed. rlm@0: rlm@0: # Boundary conditions affect hermiticity rlm@0: ** Boundary conditions alter hermiticity rlm@0: It may surprise you (and it certainly surprised me) to find that the rlm@0: Hamiltonian is not hermitian. One of the fundamental principles of rlm@0: quantum mechanics is that hermitian operators correspond to physically rlm@0: observable quantities; for this reason, surely the rlm@0: Hamiltonian\mdash{}the operator corresponding to energy\mdash{}ought to be hermitian? rlm@0: rlm@0: But we must understand the correspondence between physically rlm@0: observable quantities and hermitian operators: every hermitian rlm@0: operator corresponds to a physically observable quantity, but not rlm@0: every quantity that intuitively \ldquo{}ought\rdquo{} to be observable rlm@0: will correspond to a hermitian operator[fn::For a simple example, rlm@0: consider the differential operator \(D=\frac{d}{dx}\); although our rlm@0: intuitions might suggest that $D$ is observable which leads us to rlm@0: guess that $D$ is hermitian, it isn't. Still, the very closely related rlm@0: operator $P=i\hbar\frac{d}{dx}$ /is/ hermitian. The point is that we rlm@0: ought to validate our intuitions by checking the definitions.]. The rlm@0: true definition of a hermitian operator imply that the Hamiltonian rlm@0: stops being hermitian in the infinitely deep well. Here we arrive at a rlm@0: crucial point: rlm@0: rlm@0: Operators do not change /form/ between problems: the one-dimensional rlm@0: Hamiltonian is always $H=\frac{-\hbar^2}{2m}\frac{d^2}{dx^2}+V$, the rlm@0: one-dimensional canonical momentum is always $P=i\hbar\frac{d}{dx}$, rlm@0: and so on. rlm@0: rlm@0: However, operators do change in this respect: hermitian operators must rlm@0: only take in physical states, and must only produce physical states; because rlm@0: in different problems we /do/ change the requirements for being a rlm@0: physical state, we also change what it takes for rlm@0: an operator to be called hermitian. As a result, an operator that rlm@0: is hermitian in one setting may fail to be hermitian in another. rlm@0: rlm@0: Having seen how boundary conditions can affect hermiticity, we rlm@0: ought to be extra careful about which conditions we impose on our rlm@0: wavefunctions. rlm@0: rlm@0: ** Choosing the right constraints rlm@0: rlm@0: We have said already that physicists rlm@0: require wavefunctions to satisfy certain properties in order to be rlm@0: deemed \ldquo{}physical\rdquo{}. To be specific, wavefunctions in the rlm@0: infinitely deep well rlm@0: - Must be *normalizable*, because they correspond to rlm@0: probability amplitudes. rlm@0: - Must have *smoothly-varying probability*, because if a particle is very rlm@0: likely to be at a location, it ought to be likely to be /near/ rlm@0: it as well. rlm@0: - Must *not exist outside the well*, because it rlm@0: would take an infinite amount of energy to do so. rlm@0: rlm@0: These conditions are surely reasonable. However, physicists sometimes rlm@0: assert that in order to satisfy the second and third conditions, rlm@0: physical wavefunctions rlm@0: rlm@0: - (?) Must *smoothly approach zero* towards the edges of the well. rlm@0: rlm@0: This final constraint is our reason for rejecting $|\bar\psi\rangle$ rlm@0: as nonphysical and is consequently the reason why $H$ is not hermitian. If rlm@0: we can convince ourselves that the final constraint is unnecessary, rlm@0: $H$ may again be hermitian. This will satisfy our intuitions that the rlm@0: energy operator /ought/ to be hermitian. rlm@0: rlm@0: But in fact, we have the following mathematical observation to save rlm@0: us: a function $f$ does not need to be continuous in order for the rlm@0: integral \(\int^x f\) to be continuous. As a particularly relevant rlm@0: example, you may now notice that the function $\bar\psi(x)$ is not rlm@0: itself continuous, although the integral $\int_0^x \bar\psi$ /is/ rlm@0: continuous. Evidently, it doesn't matter that the wavefunction rlm@0: $\bar\psi$ itself is not continuous; the probability corresponding to rlm@0: $\bar\psi$ /does/ manage to vary continuously anyways. Because the rlm@0: probability corresponding to $\bar\psi$ is the only aspect of rlm@0: $\bar\psi$ which we can detect physically, we /can/ safely omit the rlm@0: final constraint while keeping the other three. rlm@0: rlm@0: ** Symmetric operators look like hermitian operators, but sometimes aren't. rlm@0: rlm@0: rlm@0: #+end_quote rlm@0: ** COMMENT Re-examining physical constraints rlm@0: rlm@0: We have now discovered a flaw: when applied to the state rlm@0: $|\psi\rangle$, the second method violates the rule that physical rlm@0: operators must only take in physical states and must only produce rlm@0: physical states. Let's examine the problem more closely. rlm@0: rlm@0: We have said already that physicists require wavefunctions to satisfy rlm@0: certain properties in order to be deemed \ldquo{}physical\rdquo{}. To rlm@0: be specific, wavefunctions in the infinitely deep well rlm@0: - Must be *normalizable*, because they correspond to rlm@0: probability amplitudes. rlm@0: - Must have *smoothly-varying probability*, because if a particle is very rlm@0: likely to be at a location, it ought to be likely to be /near/ rlm@0: it as well. rlm@0: - Must *not exist outside the well*, because it rlm@0: would take an infinite amount of energy to do so. rlm@0: rlm@0: We now have discovered an important flaw in the second method: when rlm@0: applied to the state $|\bar\psi\rangle$, the second method violates rlm@0: the rule that physical operators must only take in rlm@0: physical states and must only produce physical states. The problem is rlm@0: even more serious, however rlm@0: rlm@0: rlm@0: rlm@0: [fn:1] I'm defining a new variable just to make certain expressions rlm@0: look shorter; this cannot affect the content of the answer we'll rlm@0: get. rlm@0: rlm@0: [fn:2] For example, in vaccuum (i.e., when the potential of the rlm@0: physical system is $V(x)=0$ throughout all space), the momentum rlm@0: eigenstates are not normalizable\mdash{}the relevant integral blows rlm@0: up to infinity instead of converging to a number. Physicists modify rlm@0: the definition of normalization slightly so that rlm@0: \ldquo{}delta-normalizable \rdquo{} functions like these are included rlm@0: among the physical wavefunctions. rlm@0: rlm@0: rlm@0: rlm@0: * COMMENT: What I thought I knew rlm@0: rlm@0: The following is a list of things I thought were true of quantum rlm@0: mechanics; the catch is that the list contradicts itself. rlm@0: rlm@0: 1. For any hermitian operator: Eigenstates with different eigenvalues are orthogonal. rlm@0: 2. For any hermitian operator: Any physically allowed state can be rlm@0: written as a linear sum of eigenstates of the operator. rlm@0: 3. The momentum operator and energy operator are hermitian, because rlm@0: momentum and energy are measureable quantities. rlm@0: 4. In the vacuum potential, the momentum and energy operators have these eigenstates: rlm@0: - the momentum operator has an eigenstate rlm@0: \(p(x)=\exp{(ipx/\hbar)}\) for each value of $p$. rlm@0: - the energy operator has an eigenstate \(|E\rangle = rlm@0: \alpha|p\rangle + \beta|-p\rangle\) for any \(\alpha,\beta\) and rlm@0: the particular choice of momentum $p=\sqrt{2mE}$. rlm@0: 5. In the infinitely deep potential well, the momentum and energy rlm@0: operators have these eigenstates: rlm@0: - The momentum eigenstates and energy eigenstates have the same form rlm@0: as in the vacuum potential: $p(x) = rlm@0: \exp{(ipx/\hbar)}$ and $|E\rangle = \alpha|p\rangle + \beta|-p\rangle$. rlm@0: - Even so, because of the boundary conditions on the rlm@0: well, we must make the following modifications: rlm@0: + Physically realistic states must be impossible to find outside the well. (Only a state of infinite rlm@0: energy could exist outside the well, and infinite energy is not rlm@0: realistic.) This requirement means, for example, that momentum rlm@0: eigenstates in the infinitely deep well must be rlm@0: \(p(x) rlm@0: = \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a; rlm@0: \\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\) rlm@0: + Physically realistic states must vary smoothly throughout rlm@0: space. This means that if a particle in some state is very unlikely to be rlm@0: /at/ a particular location, it is also very unlikely be /near/ rlm@0: that location. Combining this requirement with the above rlm@0: requirement, we find that the momentum operator no longer has rlm@0: an eigenstate for each value of $p$; instead, only values of rlm@0: $p$ that are integer multiples of $\pi \hbar/a$ are physically rlm@0: realistic. Similarly, the energy operator no longer has an rlm@0: eigenstate for each value of $E$; instead, the only energy rlm@0: eigenstates in the infinitely deep well rlm@0: are $E_n(x)=\sin(n\pi x/ a)$ for positive integers $n$. rlm@0: rlm@0: * COMMENT: rlm@0: rlm@0: ** Eigenstates with different eigenvalues are orthogonal rlm@0: rlm@0: #+begin_quote rlm@0: *Theorem:* Eigenstates with different eigenvalues are orthogonal. rlm@0: #+end_quote rlm@0: rlm@0: ** COMMENT : rlm@0: I can prove this: if $\Lambda$ is any linear operator, suppose $|a\rangle$ rlm@0: and $|b\rangle$ are eigenstates of $\Lambda$. This means that rlm@0: rlm@0: rlm@0: \( rlm@0: \begin{eqnarray} rlm@0: \Lambda |a\rangle&=& a|a\rangle,\\ rlm@0: \Lambda|b\rangle&=& b|b\rangle.\\ rlm@0: \end{eqnarray} rlm@0: \) rlm@0: rlm@0: If we take the difference of these eigenstates, we find that rlm@0: rlm@0: \( rlm@0: \begin{eqnarray} rlm@0: \Lambda\;\left(|a\rangle-|b\rangle\right) &=& \Lambda |a\rangle - \Lambda |b\rangle rlm@0: \qquad \text{(because $\Lambda$ is linear.)}\\ rlm@0: &=& a|a\rangle - b|b\rangle\qquad\text{(because $|a\rangle$ and rlm@0: $|b\rangle$ are eigenstates of $\Lambda$)} rlm@0: \end{eqnarray}\) rlm@0: rlm@0: rlm@0: which means that $a\neq b$. rlm@0: rlm@0: ** Eigenvectors of hermitian operators span the space of solutions rlm@0: rlm@0: #+begin_quote rlm@0: *Theorem:* If $\Omega$ is a hermitian operator, then every physically rlm@0: allowed state can be written as a linear sum of eigenstates of rlm@0: $\Omega$. rlm@0: #+end_quote rlm@0: rlm@0: rlm@0: rlm@0: ** Momentum and energy are hermitian operators rlm@0: This ought to be true because hermitian operators correspond to rlm@0: observable quantities. Since we expect momentum and energy to be rlm@0: measureable quantities, we expect that there are hermitian operators rlm@0: to represent them. rlm@0: rlm@0: rlm@0: ** Momentum and energy eigenstates in vacuum rlm@0: An eigenstate of the momentum operator $P$ would be a state rlm@0: \(|p\rangle\) such that \(P|p\rangle=p|p\rangle\). rlm@0: rlm@0: ** Momentum and energy eigenstates in the infinitely deep well rlm@0: rlm@0: rlm@0: rlm@0: * COMMENT Can you measure momentum in the infinitely deep well? rlm@0: In summary, I thought I knew: rlm@0: 1. For any hermitian operator: eigenstates with different eigenvalues rlm@0: are orthogonal. rlm@0: 2. For any hermitian operator: any physically realistic state can be rlm@0: written as a linear sum of eigenstates of the operator. rlm@0: 3. The momentum operator and energy operator are hermitian, because rlm@0: momentum and energy are observable quantities. rlm@0: 4. (The form of the momentum and energy eigenstates in the vacuum potential) rlm@0: 5. (The form of the momentum and energy eigenstates in the infinitely deep well potential) rlm@0: rlm@0: Additionally, I understood that because the infinitely deep potential rlm@0: well is not realistic, states of such a system are not necessarily rlm@0: physically realistic. Instead, I understood rlm@0: \ldquo{}realistic states\rdquo{} to be those that satisfy the physically rlm@0: unrealistic Schr\ouml{}dinger equation and its boundary conditions. rlm@0: rlm@0: With that final caveat, here is the problem: rlm@0: rlm@0: According to (5), the momentum eigenstates in the well are rlm@0: rlm@0: \(p(x)= \begin{cases}\exp{(ipx/\hbar)},& \text{for }0\lt{}x\lt{}a;\\0, & \text{for }x<0\text{ or }x>a. \\ \end{cases}\) rlm@0: rlm@0: (Additionally, we require that $p$ be an integer multiple of $\pi\hbar/a$.) rlm@0: rlm@0: However, /these/ states are not orthogonal, which contradicts the rlm@0: assumption that (3) the momentum operator is hermitian and (2) rlm@0: eigenstates of a hermitian are orthogonal if they have different eigenvalues. rlm@0: rlm@0: #+begin_quote rlm@0: *Problem 1. The momentum eigenstates of the well are not orthogonal* rlm@0: rlm@0: /Proof./ If $p_1\neq p_2$, then rlm@0: rlm@0: \(\begin{eqnarray} rlm@0: \langle p_1 | p_2\rangle &=& \int_{\infty}^\infty p_1^*(x)p_2(x)\,dx\\ rlm@0: &=& \int_0^a p_1^*(x)p_2(x)\,dx\qquad\text{ Since }p_1(x)=p_2(x)=0\text{ rlm@0: outside the well.}\\ rlm@0: &=& \int_0^a \exp{(-ip_1x/\hbar)\exp{(ip_2x/\hbar)\,dx}}\\ rlm@0: &=& \int_0^a \exp{\left[i(p_1-p_2)x/\hbar\right]}\,dx\\ rlm@0: \end{eqnarray}\) rlm@0: $\square$ rlm@0: rlm@0: #+end_quote rlm@0: rlm@0: rlm@0: rlm@0: ** COMMENT Momentum eigenstates rlm@0: rlm@0: In free space, the Hamiltonian is \(H=\frac{1}{2m}P^2\) and the rlm@0: momentum operator $P$ has eigenstates \(p(x) = \exp{(-ipx/\hbar)}\). rlm@0: rlm@0: In the infinitely deep potential well, the Hamiltonian is the same but rlm@0: there is a new condition in order for states to qualify as physically rlm@0: allowed: the states must not exist anywhere outside of well, as it rlm@0: takes an infinite amount of energy to do so. rlm@0: rlm@0: Notice that the momentum eigenstates defined above do /not/ satisfy rlm@0: this condition. rlm@0: rlm@0: rlm@0: rlm@0: * COMMENT rlm@0: For each physical system, there is a Schr\ouml{}dinger equation that rlm@0: describes how a particle's state $|\psi\rangle$ will change over rlm@0: time. rlm@0: rlm@0: \(\begin{eqnarray} rlm@0: i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& rlm@0: H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) rlm@0: rlm@0: This is a differential equation; each solution to the rlm@0: Schr\ouml{}dinger equation is a state that is physically allowed for rlm@0: our particle. Here, physically allowed states are rlm@0: those that change in physically allowed ways. However, like any differential rlm@0: equation, the Schr\ouml{}dinger equation can be accompanied by rlm@0: /boundary conditions/\mdash{}conditions that further restrict which rlm@0: states qualify as physically allowed. rlm@0: rlm@0: rlm@0: rlm@0: rlm@0: ** Eigenstates of momentum rlm@0: rlm@0: rlm@0: rlm@0: rlm@0: #In the infinitely deep well potential $V(x)=0$, the Schr\ouml{}dinger rlm@0: rlm@0: #\(i\hbar\frac{\partial}{\partial t}|\psi\rangle = H|\psi\rangle\) rlm@0: rlm@0: rlm@0: rlm@0: rlm@0: rlm@0: rlm@0: rlm@0: * COMMENT rlm@0: rlm@0: #* The infinite square well potential rlm@0: rlm@0: A particle exists in a potential that is rlm@0: infinite everywhere except for a region of length \(a\), where the potential is zero. This means that the rlm@0: particle exists in a potential[fn:coords][fn:infinity] rlm@0: rlm@0: rlm@0: \(V(x)=\begin{cases}0,&\text{for }\;0< x< a;\\\infty,&\text{for rlm@0: }\;x<0\text{ or }x>a.\end{cases}\) rlm@0: rlm@0: The Schr\ouml{}dinger equation describes how the particle's state rlm@0: \(|\psi\rangle\) will change over time in this system. rlm@0: rlm@0: \(\begin{eqnarray} rlm@0: i\hbar \frac{\partial}{\partial t}|\psi\rangle &=& rlm@0: H |\psi\rangle \equiv\frac{P^2}{2m}|\psi\rangle+V|\psi\rangle \end{eqnarray}\) rlm@0: rlm@0: This is a differential equation; each solution to the rlm@0: Schr\ouml{}dinger equation is a state that is physically allowed for rlm@0: our particle. Here, physically allowed states are rlm@0: those that change in physically allowed ways. However, like any differential rlm@0: equation, the Schr\ouml{}dinger equation can be accompanied by rlm@0: /boundary conditions/\mdash{}conditions that further restrict which rlm@0: states qualify as physically allowed. rlm@0: rlm@0: rlm@0: Whenever possible, physicists impose these boundary conditions: rlm@0: - A physically allowed state ought to be a /smoothly-varying function of position./ This means rlm@0: that if a particle in the state is likely to be /at/ a particular location, rlm@0: it is also likely to be /near/ that location. rlm@0: rlm@0: These boundary conditions imply that for the square well potential in rlm@0: this problem, rlm@0: rlm@0: - Physically allowed states must be totally confined to the well, rlm@0: because it takes an infinite amount of energy to exist anywhere rlm@0: outside of the well (and physically allowed states ought to have rlm@0: only finite energy). rlm@0: - Physically allowed states must be increasingly unlikely to find very rlm@0: close to the walls of the well. This is because of two conditions: the above rlm@0: condition says that the particle is /impossible/ to find rlm@0: outside of the well, and the smoothly-varying condition says rlm@0: that if a particle is impossible to find at a particular location, rlm@0: it must be unlikely to be found nearby that location. rlm@0: rlm@0: #; physically allowed states are those that change in physically rlm@0: #allowed ways. rlm@0: rlm@0: rlm@0: #** Boundary conditions rlm@0: Because the potential is infinite everywhere except within the well, rlm@0: a realistic particle must be confined to exist only within the rlm@0: well\mdash{}its wavefunction must be zero everywhere beyond the walls rlm@0: of the well. rlm@0: rlm@0: rlm@0: [fn:coords] I chose my coordinate system so that the well extends from rlm@0: \(0