I haven't formally learned the separation of variables technique to solve partial differential equations, but my lecturer knew this when he wrote his notes, so hopefully a recourse into this is not needed. However, that is hardly the confusion I have. My confusion concerns the following snippet from my notes:

We isolate the $x$ & $t$ dependence by dividing both sides of the equation by our trial solution,

$$ \frac{1}{\psi_n(x)}\left\{ -\frac{\hbar^2}{2m} \frac{d^2\psi_n}{dx^2} + V(x) \psi_n(x) \right\} = \frac{1}{\theta_n(t)}\left\{ i\hbar \frac{\partial \theta_n}{\partial t} \right\} $$

The next step is crucial, so be sure that you follow the argument completely. It's not that it's so very complicated; it's actually rather obvious once you perceive its truth, but it can be subtle if you are seeing it for the first time. Read through it again if you do not understand, and be sure to ask your tutor (or myself) if you still cannot follow.

The left-hand side of this equation depends only on the variable $x$, and the right-hand side is a function only of the variable $t$, and these two variables are independent of one another. In other words, $x$ is not a function of time, because $x$ refers to a particular location in space relative to the origin, and that doesn't change with time if the coordinate axes are fixed. You may be used to seeing $x(t)$ used to represent the location of a particle as a function of time, but it is not being used that way here. That is something strictly for classical physics, because quantum particles do not follow localized trajectories through space. Note also that $t\neq t(x)$; the coordinate time does not depend on position.

Here's the key conclusion: because the left- and right- hand sides are functions of two different, independent variables, the only way the equality can hold for all values of $x$ & $t$ is if they are each separately equal to the same constant.

Imagine fixing our attention on one location in space, so that $x$ is held constant. The left-hand side cannot change with time, because it is only a function of position, and so the right-hand side also cannot change with time, otherwise the equality would not hold. The exact same argument can be made by holding $t$ fixed and allowing $x$ to vary. We must therefore have:

$$ -\frac{\hbar^2}{2m} \frac{d^2\psi_n}{dx^2} + V(x) \psi_n(x) = E_n\psi_n(x) \qquad\qquad i\hbar \frac{\partial \theta_n}{\partial t} = E_n\theta_n(t) $$

Here's what I'm getting from this:

We have an equality, despite having two functions of two distinct independent variables. For this to be true:

  1. The two variables have to be expressed in the same function, i.e $\sin(x) = \sin(t)$

  2. The LHS and RHS have to map $x$ and $t$ separately to the same constant for all $x$ and $t$.

I think (2) is what my lecturer is saying hopefully, but in that case, I would expect:

$$-\frac{\hbar}{2m} \frac{d^2 \Psi_n}{dx^2} + V(x) \Psi_n = E_n = i\hbar \frac{d\theta_n}{dt}$$

As opposed to what my lecturer wrote:

$$-\frac{\hbar}{2m} \frac{d^2 \Psi_n}{dx^2} + V(x) \Psi_n = E_n \psi_n(x)$$

$$i\hbar \frac{d\theta_n}{dt} = E_n \theta_n(t)$$

What am I missing here? I suspect I'm misinterpreting what's going on here.

  • 1
    $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$ Sep 18, 2018 at 20:28
  • 1
    $\begingroup$ Don't forget the denominator before the bracket=) $\endgroup$
    – OON
    Sep 18, 2018 at 20:47
  • $\begingroup$ @OON Where are you referring? $\endgroup$
    – sangstar
    Sep 18, 2018 at 20:54
  • $\begingroup$ @EmilioPisanty Sorry about that! It's just a lot to write. $\endgroup$
    – sangstar
    Sep 18, 2018 at 20:54
  • $\begingroup$ @sangstar There's this thing called copy-paste that lets you post text without needing to type it out. The legibility of your quote takes an enormous hit here for no good reason - there's frankly no excuse here. $\endgroup$ Sep 18, 2018 at 21:12

2 Answers 2


First of all, the option 1) you're mentioning isn't valid, $sin(x)\neq sin(t)$ in general. Consider for example the case with $x=0$ and $t=1$ to see why.

That said, what your lecturer wrote is right. You have $$ f(x) = \frac{1}{\psi(x)}\left(-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2}(x) + V(x) \psi(x)\right) \qquad\text{and}\qquad g(t)= \frac{1}{\theta(t)} \left(i\hbar\frac{d\theta}{dt}(t)\right) $$ and the first equation reads $$f(x)=g(t).$$ So it has to be $f(x)=E=g(t)$, that is $$ \frac{1}{\psi(x)}\left(-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2}(x) + V(x) \psi(x)\right) = E $$ or, multiplying both sides by $\psi(x)$, $$ -\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2}(x) + V(x) \psi(x) = E \psi(x) $$ and similar for $g(t)=E$.

You can see also that your reasoning is wrong by inserting your equations $$\tag{1}-\frac{\hbar}{2m} \frac{d^2 \Psi}{dx^2} + V(x) \Psi = E \qquad \text{and}\qquad i\hbar \frac{d\theta}{dt}=E$$ in the first one. You get $$ \frac{1}{\psi(x)} E = \frac{1}{\theta(t)} E \implies \psi(x) = \theta(t) $$ which by the same reasoning as before implies $\psi(x) = \theta(t) = \text{constant}$. If you insert this in (1) you can verify that it means $\psi(x)=\theta(t)=0$, which can't be since they appear in the denominator of the original equation.

  • $\begingroup$ I see that multiplying by $psi(x)$ above, you created the time-independent Schrodinger equation. Is this a coincidence? Multiplying by $\theta(t)$ then -- does this reveal a noteworthy equation as well? $\endgroup$
    – sangstar
    Sep 19, 2018 at 20:49
  • $\begingroup$ Not a coincidence. I'll leave that for you to think :) (it's the very reason of why the time-independent Schrodinger equation exists in the first place!) $\endgroup$
    – pp.ch.te
    Sep 19, 2018 at 22:17

If $f(x)=g(t)$ for all $x$ and $t$ and if $f(x)$ were not a constant then there would be $x_1$ and $x_2$ such that $f(x_1)\ne f(x_2)$. Similarly if $g(t)$ were not constant there would be $t_1$ and $t_2$ such that $g(t_1)\ne g(t_2)$. Then, from $f(x)=g(t)$, for all $x$ and $t$ we have that $f(x_1) = g(t_1)$ and $f(x_2)=g(t_1)$, but $f(x_1)\ne f(x_2)$ so this is a contradiction. We conclude that $f(x)$ is a constant. Similarly $g(t)$ is a constant. Finally we see that the two constants must equal each other.


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