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I'm trying to understand the relation between the time dependent and time independent Schrödinger equations. In particular, we know that the TDSE is $$H\Psi=i\hbar \frac{\partial \Psi}{\partial t}$$ whereas the independent equation is the eigenvalue problem $$H\psi=E\psi$$ My main question is this: if we allow $\Psi$ to be independent of time (which is my interpretation of a 'time independent equation') then why don't we just get $H\Psi=0$? I can see where the eigenvalue problem comes from: suppose we had a separable solution to the TDSE $\Psi(x,t)=\psi(x)T(t)$. Then, $$TH\psi=i\hbar \dot{T}\psi \implies i\hbar \frac{\dot{T}}{T}=\frac{H\psi}{\psi}=E$$ For some constant $E$, so we get $T(t)=Ae^{-iEt/\hbar}$ and $H\psi=E\psi$.

This is interesting, but doesn't quite answer my question: why does the argument that $H\psi=0$ not work, and what about solutions which are not separable?

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  • $\begingroup$ Why should $H\psi$ be equal to zero? Did you mean $H \Psi = 0$? $\endgroup$
    – Courage
    Nov 12, 2015 at 16:50
  • $\begingroup$ Well, yes, but what I mean by it is "why doesn't assuming $\Psi(x,t)=\psi(x)$ is independent of time give rise to the time-independent equation $H\psi=0$?" $\endgroup$
    – preferred_anon
    Nov 12, 2015 at 16:54
  • $\begingroup$ Why can't there be a non-zero eigenvalue associated with the spatially-dependent operator? $\endgroup$
    – Kyle Kanos
    Nov 12, 2015 at 17:03
  • $\begingroup$ Of course it can, but I don't see how any non-zero eigenvalues would arise in that situation. I.e. making the time-dependent SE independent of time doesn't seem to give the time independent SE (the eigenvalue problem). $\endgroup$
    – preferred_anon
    Nov 12, 2015 at 17:16
  • $\begingroup$ Think linear algebraically: $A\mathbf v=\lambda\mathbf v$ is the general eigenvalue problem, no? In QM, $A\equiv H$, $\psi\equiv\mathbf v$ and, thus, $E\equiv\lambda$ which would be zero for particular cases and not the general solution. $\endgroup$
    – Kyle Kanos
    Nov 12, 2015 at 17:43

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The "independent" in "time-independent Schrödinger equation" doesn't mean that the wavefunction $\psi(x,t)$ is independent of time, but that the quantum state it defines doesn't change with time.

Since $\psi(x)$ and $\mathrm{e}^{\mathrm{i}\phi}\psi(x)$ for any $\phi\in\mathbb{R}$ define the same quantum state, this does not imply $\partial_t\psi(x,t) = 0$. Indeed, as the solution shows, the time dependence $\mathrm{e}^{\mathrm{i}Et}$ is precisely the kind of dependence that is allowed.

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To understand what is going on you have you to distinguish a definition from an equation.

As an example you could consider the heat equation $\partial_{xx}u=k\partial_t u.$ Both sides have their own meaning, and the equation says that for the solution to the heat equation, the two things are equal. You have to first be able to compute the left hand side (second derivatives) and compute the right hand side (single derivative) and then anything that fails to have them be equal (and there are lots of such functions) are simply thrown out for not being the solutions you are looking for.

So in particular, the Hamiltonian (like $\partial_{xx}$) is its own thing and it is not defined by the Schrödinger equation it only provides the left hand side.

So when you write $$H\Psi=i\hbar \frac{\partial \Psi}{\partial t}$$ the left hand side has a meaning and the meaning isn't to take a single time derivative. Its meaning is to take multiple spatial derivatives and do some other stuff.

If you took a nonzero time independent solution to $H\psi=E\psi$ with nonzero energy $E$ then you'd notice right away that $H\psi=E\psi \neq0=i\hbar\partial_t \psi$ which means that function is simply not a solution to the time dependent equation.

Just like how most functions are not a solution to the heat equation.

if we allow $\Psi$ to be independent of time (which is my interpretation of a 'time independent equation') then why don't we just get $H\Psi=0$?

That is not what a time independent equation means. That is looking for an equilibrium or steady state. Again, go back to the heat equation. That is using the time dependent equation to look for particular solutions to the tine dependent equation that happen to be time independent. That is not what we are doing. We are making a different equation.

We are not requiring that $\psi$ be independent of time and be a solution to $H\psi=i\hbar\partial_t \psi.$ We require it be time independent and a solution to a completely new and different equation, $H\psi=E \psi.$

Why? Because then you can solve the separable equations as you describe by putting in a particularly simple time dependency.

what about solutions which are not separable?

If you take your initial conditions, then you can write is as a (linear) combination of solutions to the time independent equation. Then when you write the corresponding (linear) combination of separable solutions you get a solution to the time dependent equation that matches your initial conditions.

And often, that's all you really want. And you could do the same thing with the heat equation.

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  • $\begingroup$ I don't think that answers my question, I'll try to explain: I understand that we have two equations, the time independent SE and the time dependent one (I'll call them TISE and TDSE for brevity). My point is tht we should be able to recover the TISE by removing all the dependencies on time from the TDSE. But if we actully do that, we get the equation $H\psi=0$, rather than the TISE $H\psi=E\psi$ that we would expect. $\endgroup$
    – preferred_anon
    Nov 12, 2015 at 17:21
  • $\begingroup$ To continue your analogy with the heat equation, it would be like calling $\nabla^{2}u=k\dot{u}$ the "time depdendent heat equation". Then in steady-states we would get the time-independent equation $\nabla^{2}u=0$. We wouldn't get $\nabla^{2}u=\lambda u$, which is what seems to happen in this case. $\endgroup$
    – preferred_anon
    Nov 12, 2015 at 17:23
  • $\begingroup$ @DanielLittlewood If your question was about terminology all along, then ACuriousMind answered your question. But you should have used the tag terminology or the soft-question tag or said that. You don't get a time independent equation by trying to solve for time independent solutions to the original equation. It's a different equation. Its solutions can be used to find solutions to the more complex equation. And the more complex equation can be interpreted as telling you how waves evolve in time from given initial conditions, and the solutions to the TISE evolve the simplest way. $\endgroup$
    – Timaeus
    Nov 12, 2015 at 17:35
  • $\begingroup$ Fair point, sorry for not including the tag. $\endgroup$
    – preferred_anon
    Nov 12, 2015 at 18:14

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