Energy Eigenfunction Completeness

It's my understanding that eigenfunctions are complete (span the space). I don't know what the solution to the (Time-Dependent) Schrödinger Equation is, but whatever it is, any solution (no matter the potential $V$) can be expanded in terms of say position eigenfunctions or momentum eigenfunctions. I'd like to emphasize the phrase - no matter the potential $V$ - with doubt because this ties into my question. Energy eigenfunctions can also be used to represent a general solution. However, this is where my question begins:

Consider a set of energy eigenfunctions $\psi_n$ which satisfy by definition $\hat{H}\psi_n = E_n\psi_n$. It seems to me that the sum $\Psi = \sum c_n\psi_ne^{-iE_nt/\hbar}$ is a general solution to the Schrödinger equation only when the potential $V$ of the Schrödinger equation matches the potential $V$ in the Time-Independent Schrödinger Equation used to find the $\psi_n$'s. Is this correct? If it is, could one say that energy eigenfunctions are complete only with respect to the specific potential $V$ from which they are derived while (say) momentum eigenfunctions are complete with respect to any potential. If this is not true, if energy eigenfunctions of the Time-Independent Schrödinger equation $\hat{H}\psi_n = E_n\psi_n$ are complete with respect to any potential $V$ used in the (Time-Dependent) Schrödinger equation, why can't we use the $\psi_n$'s of say the infinite square well to construct general solutions $\Psi$ of the delta-function well, finite potential well, free particle, etc. Why are we always solving the time-independent Schrödinger equation when we can just use the energy eigenfunctions of the infinite square well?

• As long as the boundary conditions are the same (one must be careful with infinite potentials, since they are imposing boundary conditions), I don't see any problem. – Adam Sep 4 '17 at 19:09

Before I begin let me pause to observe that there is a slight lie in the following words (and, more or less, in the entire undegraduate curriculum) that one discovers when one gets very mathematically involved; right now it's present in this Wikipedia article as "subtleties of the unbounded case"... our "Hermitian" operators are not in general well-defined for all the states that we'd like. This becomes especially important as we look at position and momentum eigenstates -- very often states get unnormalizable and physics can get somewhat clunky in terms of these.

With that caveat, yes: the eigenfunctions of any given Hamiltonian are always a complete basis for the entire space. For example one can approach any 1D Hamiltonian with the eigenfunctions of the harmonic oscillator; those are valid wavefunctions which span the space.

Whether this is useful or not is a different story. Let's say you have a bunch of functions $\hat H_1 |\psi_n\rangle = E_n |\psi_n\rangle$ but then you bring them to a new Hamiltonian $\hat H_2$. In general $|\psi_n\rangle$ is no longer going to be an eigenvector for $\hat H_2,$ and therefore its evolution under that new Schrödinger equation is not going to be $|\psi_n(t)\rangle = e^{-iE_n t/\hbar} |\psi_n(0)\rangle,$ so these energies and wavefunctions are not obviously helpful in this new context.

Well, there is a way to make them useful, but of course it only really does a good job when $\hat H_1$ and $\hat H_2$ have some sort of nice relationship. A Schrödinger equation $i \hbar \partial_t |\Psi\rangle = \hat H |\Psi\rangle$ can be phrased purely as a unitary operator $|\Psi(t)\rangle = \hat U(t) |\Psi_0\rangle.$ The condition is that $i\hbar \partial_t \hat U = \hat H \hat U,$ but this is no problem in theory. This means that all of our expectation values in the second case take the form $$A(t) = \langle \Psi(t)|\hat A |\Psi(t)\rangle = \langle \Psi_0|\hat U_2^\dagger \hat A \hat U_2|\Psi_0\rangle.$$ Now since a unitary operator is defined by $U^\dagger U = 1$ we will insert strategic $\hat U_1^\dagger \hat U_1$ terms to rewrite this same expectation value as $$A(t) = \langle \Psi_0|\hat U_1^\dagger ~\Big(\hat U_1 \hat U_2^\dagger \hat A \hat U_2 \hat U_1^\dagger\Big)\hat U_1 |\Psi_0\rangle.$$Note that now there is some complicated time dependence for this parenthesized operator $\tilde A = \hat U_1 \hat U_2^\dagger \hat A \hat U_2 \hat U_1^\dagger$, but the outermost wavefunctions obey the Schrödinger equation for $\hat H_1$, not $\hat H_2$. The cost is that we have to shift our operators to have this complicated time dependence $\tilde A(t)$, which takes the form of a really big product rule,\begin{align} i\hbar\partial_t \tilde A = &H_1 U_1 U^\dagger_2 \hat A U_2 U_1^\dagger - U_1 U^\dagger_2 H_2 \hat A U_2 U_1^\dagger + i\hbar\tilde{\dot A} + \\ &U_1 U^\dagger_2 \hat A H_2 U_2 U_1^\dagger - U_1 U^\dagger_2 \hat A U_2 U_1^\dagger H_1. \end{align}(The minus sign for the daggers comes from taking the conjugate transpose of the earlier Schrodinger equation.)

Going "whole hog" with this requires replacing those $\hat A$ operators with $U_2 U_1^\dagger \tilde A U_1 U_2^\dagger$ which yields: \begin{align} i\hbar\partial_t \tilde A = &H_1 \tilde A - U_1 U^\dagger_2 H_2 U_2 U_1^\dagger \tilde A + i\hbar\tilde{\dot A} + \\ &\tilde A U_1 U_2^\dagger H_2 U_2 U_1^\dagger - \tilde A H_1. \end{align}We see that the only complicated thing that's left is that we also need $\tilde H_2$ to factor into these expressions, rather than the $t=0$ value of $H_2$. Once that's done we find just$$i\hbar\partial_t \tilde A = [H_1 - \tilde H_2, \tilde A] + i\hbar\tilde{\dot A}.$$ This is called an "interaction picture" because usually what we do is we use some easy-to-solve orthogonal states $\hat H_1 = H_0$ and then add some interaction term which couples them, $\hat H_2 = H_0 + V.$ The equation for a time-independent $\tilde H_2$ is just $$i\hbar\partial_t \tilde H_2 = [H_1 - \tilde H_2, \tilde H_2] = [H_1, \tilde H_2],$$and in many cases this just slaps some phase factors around the terms in $V$. Then, instead of calculating all-new basis states we can use the ones that are most familiar, instead preferring to find differential equations for the observables that we're interested in and solving them in some limits.

• So just to be specific, lets consider the Schrodinger equation (SE) for a free particle. The Hamiltonian $H$ of the free particle has a continuous eigenvalue spectrum of energy $E$, with associated eigenfunctions $\psi_E(x)$. So the general solution to this specific SE is $\int c(E) \psi_E(x) e^{-iEt/\hbar}dE$. However the general solution to this specific SE can also be expanded in terms of the eigenfunctions of the infinite square well $\psi_n(x)$ associated with discrete energy eigenvalues $E_n$. The general solution for the free particle SE (or any SE but specifically the free particle – DWade64 Sep 14 '17 at 22:18
• here) is $\sum_n c_n(t)\psi_n(x)$ ? It just seems weird since the free particle has a continuous energy spectrum but it's being represented by eigenfunctions corresponding to a discrete energy spectrum – DWade64 Sep 14 '17 at 22:20
• Well the case that you're talking about (free particle FP / particle in a box PB) is one of the very few cases where it actually would work out that way, in general you could not take eigenfunctions $\psi_n$ of Hamiltonian $H_1$ and express a solution for Hamiltonian $H_2$ as $\sum_n c_n(t) \psi_n(x)$ for any $c_n(t).$ But the FP and PB Hamiltonians are both identical on the only domain where PB is defined, so if you happen to consider a highly localized Gaussian wavepacket that happens to be inside the box, it evolves identically until it hits the sides of the box. – CR Drost Sep 15 '17 at 15:25
• Remember that the expression $\sum_n c_n(t) \psi_n(x)$ (or $\int dn$ for the continuous case) is actually an attempt to solve a PDE by separation of variables. So $i\hbar\partial_t\Psi=H_x[\Psi],$ we assume $\Psi=\int dn T_n(t)X_n(x),$ we find $i\hbar T_n'/T_n=H_x[X_n]/X_n=\text{const.}$ because the left hand side is a function of only $t$ and the right hand side is a function of only $x$. We find that we need $H_x[X_n]=\hbar\omega_n~X_n$ for this decomposition to work in the first place. Non-eigenfunctions will in general not remain the same under Schrodinger evolution. – CR Drost Sep 15 '17 at 15:32
• As for spectra, that's a property of the operator and not a property of the wavefunctions. But let me attempt to allay your fears: you are very worried about us in QM representing an arbitrary function $f(x)$ in terms of some countable basis states $g_n(x)$ as $\sum_n c_n~g_n(x).$ I claim that this is purely you not having the courage of your convictions because you are feeling very scared of this new quantum world you are in. "How could he possibly prove that to me?!" you might ask. Here's how: you have absolutely no problem with this if $g_n(x) = x^n$ and I call it a "Taylor series." QED. – CR Drost Sep 15 '17 at 15:39

You have to be a bit more careful. When you say that "momentum eigenfunctions are complete with respect to any potential" you're saying that any function can be decomposed as $$f(x) = \int \frac{dp}{2\pi} \, e^{ipx} \hat{f}(p)\,, \quad \hat{f}(p) = \int dx\, e^{-ipx} f(x)\,.$$ But this is formal nonsense: for a general function $f(x)$, the integral defining $\hat{f}(p)$ doesn't exist (it diverges). You need to impose some conditions on $f$, for instance that $$\int dx \, |f(x)|^2 < \infty\,.$$ By imposing such constraints (there are various choices you can make), you're defining a class of functions which is precisely a Hilbert space. This leads you down the mathematician's way to define the Fourier transform.

The same holds for the type of problem you're interested in, namely a complete basis of wavefunctions $\{\psi_n\}$ with respect to some potential $V(x)$. There you have a decomposition $$f(x) = \sum_n f_n \psi_n(x)\,, \quad f_n = \int dx\, w(x)\, \psi_n^*(x)\, f(x)$$ where this is possibly some weight function $w(x)$. Again the integral defining $f_n$ diverges if you plug in a general function $f(x)$. Once more, you have to work within some Hilbert space $\mathcal{H}$ to make sense of these manipulations. If you do the exercise, you will see that the boundary conditions at infinity (or at some boundary points, if you're working on a finite interval) are essentially encoded by the potential $V(x)$.

This whole business is called Sturm-Liouville theory, and if you want to understand QM at any serious level, you should spend some time studying it (it's not difficult).

If your potential is time-dependent, I don't see a way of justifying the use of the "time-independent" Schroedinger equation. Let us assume that that $H$ is an operator that only contains derivatives and multiplication operators relative to the variables $\mathbf x$. Forgetting numerical factors, one can write Schroedinger equation as

$$H\psi = \dot\psi.$$

By the Sturm-Liouville theory, let us look for solutions of the form $\psi(\mathbf x,t) = \phi(\mathbf x)\chi(t)$. We then get to the equality

$$\frac{H\phi}\phi(\mathbf x) = \frac{\dot\chi}\chi(t),\qquad\forall\ \mathbf x,t$$

which means that each term is constant, i.e. we get the "time-independent" equation

$$H\phi(\mathbf x) = E\phi(\mathbf x).$$

If $H$ has a dependency on $t$ through $V$ I don't see how to get to the same conclusions.

• By "time-dependent", the OP just means the $i\partial_t \psi = H\psi$, not that the potential is time-dependent, so the answer is off topic. – Adam Sep 4 '17 at 19:07
• then I don't quite get "...when the potential $V$ of the Schrödinger equation matches the potential $V$ in the Time-Independent Schrödinger Equation...". – Phoenix87 Sep 4 '17 at 19:10
• Solve the time independent SE for potential V1. Can you use the corresponding eigenfunctions to describe the time evolution under potential V2. – Adam Sep 4 '17 at 19:13

protected by Qmechanic♦Sep 4 '17 at 20:22

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