Let's say I have some state vector $|\Psi(t)\rangle$, and I express it as a linear combination of eigenstates of some operator, $\hat{Q}$, with a discrete spectrum, which we will call $|q_n\rangle$. Let's say that $\hat{Q}$ is not the Hamiltonian, so these are not energy eigenstates. Then, I have:

$$|\Psi(t)\rangle \ = \ \displaystyle\sum_{n} |q_n\rangle \langle q_n |\Psi(t)\rangle \ = \ \displaystyle\sum_{n} r_n(t) |q_n\rangle$$

Now, if I express this in the position basis, I get something like:

$$\Psi(x,\,t) \ = \ \displaystyle\sum_{n} r_n(t) \psi_n(x)$$

Now each term of this sum is orthogonal, so they obey the Schrödinger Equation. But isn't it true that the wavefunctions that are separable are equivalent to the set of energy eigenfunctions with time dependence? How can this be reconciled with the fact that $r_n(t) \psi_n(x)$ is a separable wavefunction that is not an eigenfunction of the Hamiltonian?

I'm probably making a stupid assumption/mistake somewhere in my reasoning, so any clarification is much appreciated.

  • $\begingroup$ But your wavefunction is $\Psi(x,t)$, not $r_n(t)\psi_n(x)$? $\endgroup$ Sep 28, 2019 at 21:26
  • $\begingroup$ Yeah, but each term of the sum should obey the Schrödinger equation right? $\endgroup$ Sep 28, 2019 at 21:27
  • $\begingroup$ If the $\psi_n(x)$ set is orthogonal $\endgroup$ Sep 28, 2019 at 21:27
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    $\begingroup$ Yes they should. Obeying the Schrodinger equation does not mean energy eigenstate though. A superposition of eigenstates of an operator is not necessarily an eigenstate itself. $\endgroup$ Sep 28, 2019 at 21:28
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    $\begingroup$ I would have to think about that. What I am saying is that your wavefunction is not $r_n(t)\psi_n(x)$, unless you are looking only at an eigenstate of $Q$. Also just because $Q\neq H$ does not mean the eigenstates of $Q$ are not eigenstates of $H$ $\endgroup$ Sep 28, 2019 at 21:38

3 Answers 3


Any separable solution to the Schrodinger equation is stationary. The mistaken step in your reasoning is "Each term of this sum is orthogonal, so they obey the Schrödinger Equation." That's not true.

If $Q$ commutes with the Hamiltonian (and there's no degeneracy), then $r_n(t) \psi_n(x)$ is a stationary state. If $Q$ does not commute with the Hamiltonian, then $r_n(t) \psi_n(x)$ does not solve the Schrodinger equation. In no situation is $r_n(t) \psi_n(x)$ a non-stationary separable solution to the Schrodinger equation.

  • $\begingroup$ Except the state vector $|\Psi(t)\rangle$ obeys the Schrödinger equation, I'm just expressing it as an expansion in the $\hat{Q}$ basis. Since $\hat{Q}$ is an observable, shouldn't it form a complete orthogonal basis? Thus, if $|\Psi(t)\rangle$ obeys the Schrödinger equation, then so should each term. $\endgroup$ Sep 30, 2019 at 11:52
  • $\begingroup$ @Jack Yes, $|\psi(t)\rangle$ solves the Schrodinger equation, and yes, the eigenvectors of Q form a complete orthogonal basis, but no, there's no reason why the individual terms in the expansion should each solve the Schrodinger equation. You keep repeating that claim without proving any reasoning for it, but it's not true. $\endgroup$
    – tparker
    Sep 30, 2019 at 12:01
  • $\begingroup$ @JackCeroni I apologize if I sound rude, but you appear to be fundamentally confused about what the term "orthogonal" means. $E \psi_n(x)$ is definitely not orthogonal to $\psi_n(x)$, nor does orthogonality have anything to do with solving the Schrodinger equation. Respectfully, I think that you may be better off speaking in person with someone knowledgeable about QM. $\endgroup$
    – tparker
    Sep 30, 2019 at 13:09
  • $\begingroup$ Sorry, that was a really stupid comment, I don’t know why I said that $\psi_n(x)$ is orthogonal to $E\psi_n(x)$, quite the opposite actually. $\endgroup$ Sep 30, 2019 at 13:32
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    $\begingroup$ @Jack If I'm correctly understanding what you're trying to say, then yes, that's correct. $\endgroup$
    – tparker
    Sep 30, 2019 at 13:59

The issue is either in the explicit time dependence you have assumed or in thinking your general $\Psi(x,t)$ is separable.

First, let's consider the case where your quantum state is an eigenstate of $Q$: $$|\Psi(t)\rangle=|q_n\rangle$$

The eigenstates $|q_n\rangle$ should be constant in time (I am assuming). But you are assuming an explicit time dependence on the left-hand side. Therefore, there is no contradiction for your separable state assumption in this case, because $r_n(t)=1$. The state is not an energy eigenstate, as assumed.

Moving to the more general case $$|\Psi(t)\rangle=\sum_n|q_n\rangle\langle q_n|\Psi(t)\rangle\to\Psi(x,t)=\sum_nr_n(t)\psi_n(x)$$

Your wavefunction is not separable. i.e. $$\Psi(x,t)\neq R(t)\psi(x)$$

So in this case there is still not a contradiction. Your state is not an energy eigenstate, as assumed.

I guess the root of my question is: Is $r_n(t)\psi_n(x)$ a separable solution to the Schrodinger equation that is not an energy eigenstate?

I am still thinking through this. Of course, if it was a separable solution then we would have a contradiction, as you have seen and as @tparker says in their answer. But I cannot think of a proof that does not involve a contradiction at the moment. I feel like it should be something simple that I am just not seeing for some reason.

Although I will say if two functions $A$ and $B$ each solve a linear differential equation $DA=0$ and $DB=0$ then it is true that the sum solves the differential equation: $D(A+B)=0$. However, the converse is not necessarily true. If we know that $D(A+B)=0$, then all we can say is $DA=-DB$, but it is not necessarily true that $DA=DB=0$. For even more terms in the sum we can say the same thing. i.e. if $D\left(\sum_n\Psi_n\right)=0$ it does not necessarily mean that $D\Psi_n=0$ for all $n$.

  • $\begingroup$ Ok, but even if $R(t)\psi(x)$ isn’t my general wave function, isn’t it still a solution to the Schrödinger equation? Consider the case where I express same state vector as a linear combo of energy eigenstates: $|\Psi(t)\rangle = \sum c_n(t) |E_n\rangle$. Each term in the sum should obey the Schrodinger equation, since the energy eigenstates form an orthogonal basis, which they do, and in the position basis the terms are separable solutions to the Schrodiger equation. How can this not be the case for the $\hat{Q}$ eigenstates? $\endgroup$ Sep 29, 2019 at 0:16
  • $\begingroup$ @JackCeroni Being a solution to the SE does not mean energy eigenstate. Or maybe I am still not understanding your point $\endgroup$ Sep 29, 2019 at 0:19
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    $\begingroup$ @JackCeroni I cannot think of a good explanation still, but I have undeleted my answer to be critiqued by others. $\endgroup$ Sep 29, 2019 at 20:10
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    $\begingroup$ @JackCeroni See my edit. I think it pertains to your most recent comment on tparker's answer. $\endgroup$ Sep 30, 2019 at 15:28
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    $\begingroup$ @JackCeroni Please upvote all useful answers, and select one answer as the accepted answer for future readers. $\endgroup$ Sep 30, 2019 at 15:49

If the wavefunction can be written as a product of a time-only and a time-independent(say position dependent) component, then every position will evolve the same way which is how energy eigenstates work.

You seem to prove for an arbitrary basis that it is an energy eigenstate basis, but only because of a mistake, I think.

You seem to have confused the abstract vector with the vector components. If you have abstract vectors $|\Psi\rangle $ , $|\Phi\rangle $ you can take the "abstract" inner product which is $\langle \Psi | \Phi \rangle$, but if you want to calculate with these vectors usefully it is useful to express them in a basis (e.g. the eigenvector basis of $\hat{Q}$). This is exactly analogous to regular vectors in 2D space. Now if you want to do any calculation on these vector you have to express them in the same basis!

You calculate the n vector components of abstract vector $|\Psi(t)\rangle \ $ which are $r_n(t)$ this is correct. In other words you expressed it in $\hat{Q}$ eigenvector basis. Now if you want to express $| \Psi(t) \rangle$ in position eigenstate basis (also known as position space), the rule is the same you take the inner product of the vector or its components with all the position eigenstates $|\psi^{[x]} \rangle$. You can do it with the abstract vector and you get $\displaystyle\sum_{x} |\psi^{[x]} \rangle \langle\psi^{[x]} | \Psi(t) \rangle$ or you can take the inner product using its known components in the $\hat{Q}$ eigenvector basis, but careful, in this case you have to express the position eigenfunctions in the $\hat{Q}$ basis too! While the position eigenvectors are simple Dirac delta functions in the position basis(any basis is just a bunch of deltas expressed in its on basis), in general the position basisvectors depend on space and time when expressed in any other basis. (as you probably know the position eigenvectors are Dirac deltas in their own basis and infinite waves in the momentum basis and vica-versa).

So you have to express your position eigenvectors(you have one fore each x) in in $\hat{Q}$ first: $$|\psi^{[x]}\rangle \ = \ \displaystyle\sum_{n} |q_n\rangle \langle q_n |\psi^{[x]}\rangle \ = \ \displaystyle\sum_{n} {\psi^{[x]}}_n|q_n\rangle$$

Then express in position basis (using Q-based components of the vectors) $$\Psi(x,t):=r^{[x]}(t) = \displaystyle\sum_{n} r_n(t){\psi^{[x]}}_n(t) $$

Now $r^{[x]}$(t) is just a complex number corresponding to every position eigenvector at every time t so we could rename it as just $\Psi(x,t)$ which is what we were looking for.


  1. When you calculate vectors you have to be in the same basis or no base at all.
  2. Position eigenvectors are time independent only in position eigenbase.

Update: Expressing a state vector in position basis

The wrong way to think about it is this: A state vector is like a little arrow and you can take the its inner product with the position basis vector and you get its x component.

But this is not what state vectors are. State vectors(or wavefunctions if expressed in position basis) can generally have a distinct value at every possible x position. A wave function $\Psi(x,t)$ of a single particle for example can be $1/\sqrt(2)$ at x=13 and x=19+$\pi$ and zero everywhere else. This is still just one state vector/wavefunction. Think of the state vector as a "vector" that has a unique value at every single x position. To describe it fully you need not just one x coordinate but one value(amplitude) for every possible x coordinates. (So just a particle in one dimension is an infinite dimensional state vector.).

To write the state vector in position basis you can't just ask what is the length in the x direction. You have to ask what is the amplitude at x=-1 , x=3, x=$\pi$ all of those. How do you do it? Simple! We have a basis vector for each of those $|\psi_{-1}\rangle $, $|\psi_3\rangle $, $|\psi_\pi \rangle $ etc. You take the value inner product with each of the infinite set of position basis vectors.


$$|\Psi \rangle = \displaystyle\sum_x |\psi_x\rangle \langle \psi_x | \Psi \rangle = \displaystyle\sum_x \Psi_x |\psi_x\rangle = \displaystyle\sum_x \Psi(x) |\psi_x\rangle $$

where $\Psi(x)$ is the amplitude(value) of $|\Psi\rangle $ statevectors at x.

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    $\begingroup$ But the OP in the question just does $$\langle x|\Psi(t)\rangle=\langle x|\left(\sum_n r_n(t)|q_b\rangle\right)=\sum_n r_n(t)\psi_n(x)$$ where $\langle x|q_n\rangle=\psi_n(x)$. I don't see the issue with that. $\endgroup$ Sep 29, 2019 at 12:21
  • $\begingroup$ Yes, I agree with Aaron’s comment, I still don’t really understand what is wrong with expanding in the position basis, by taking the inner product with the position bra $\endgroup$ Sep 29, 2019 at 17:43
  • $\begingroup$ Excellent, It just occurred to me that you might be thinking that way. I write a small update to address that. Thanks for the comments $\endgroup$ Sep 29, 2019 at 17:48
  • $\begingroup$ Ok, so then we have for out state vector $$|\Psi(t)\rangle=\int|x'\rangle\langle x'|\Psi(t)\rangle\,\text dx'$$ as well as $$|\Psi(t)\rangle=\sum_n|q_n\rangle\langle q_n|\Psi(t)\rangle$$ which means we also have $$\langle x|\Psi(t)\rangle=\langle x|\left(\int|x'\rangle\langle x'|\Psi(t)\rangle\,\text dx'\right)=\Psi(x,t)$$ as well as $$\langle x|\Psi(t)\rangle=\langle x|\left(\sum_n|q_n\rangle\langle q_n|\Psi(t)\rangle\right)=\sum_nr_n(t)\psi_n(x)$$ so then $$\Psi(x,t)=\sum_nr_n(t)\psi_n(x)$$ so there is still no issue. You keep looking for an error in the OP's work, but I don't think there is $\endgroup$ Sep 29, 2019 at 20:03
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    $\begingroup$ $|x\rangle$ is a position eigenstate. $\langle x|\Psi(t)\rangle$ is "picking out" the "position component" of $|\Psi(t)\rangle$. i.e. $\langle x|\Psi(t)\rangle=\Psi(x,t)$ You get a "component" for each $x$ value. $\endgroup$ Sep 29, 2019 at 20:30

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