I'm confused by my books treatment of the Schrödinger equation. In steado f listing my questions at the end of my post, I'll add them as questions in parentheses after the line in question.

For a free particle:

$$i \hbar \left| \dot{\Psi} \right \rangle = H\left| \Psi \right \rangle = \frac{P^2}{2m}\left| \Psi \right \rangle$$ (where did the potential go?)

The normal mode solutions are of the form: $\left| \Psi \right \rangle = \left| E \right \rangle e^{-iEt/ \hbar}$

Feeding this into the equation above (the schrödinger equation written above), we get the time-independent equation for $\left| E \right \rangle$ :

$$H \left| E \right \rangle = \frac{P^2}{2m}\left| E \right \rangle = E \left| E \right \rangle$$

(this follows from the eigen-equation, where the eigenvalue must be equal to $E$?)

This problem can be solved without going into any basis. First note that any eigenstateof $P$ is also an eigenstate of $P^2$. So we feed the trial solution for $\left| p \right \rangle$ into the equation (the equation directly above).

$$\frac{P^2}{2m}\left| p \right \rangle = E\left| p \right \rangle $$

(why are we all of a sudden talking about the ket $p$?

Which means that $\left| p \right \rangle = \pm \sqrt{2mE}$

This gives us two eigenkets of $E$, which span an eigenspace.

The next steps are to set up the propagator (which I have questions about, but understanding the previous steps may help me clear them up myself).

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    $\begingroup$ This is a bit confusing since all of your equations are taken out of context. There is probably a bunch of text in the book between all of the equations that explains what's going on. Does your book explain what system you're working with? I'm guessing a free particle; does it state this? If so, that may address the potential energy issue. $\endgroup$
    – BMS
    Sep 27, 2013 at 0:47
  • $\begingroup$ Yes, let me add the text in-between $\endgroup$
    – Astrum
    Sep 27, 2013 at 0:53
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    $\begingroup$ Free particles have no potential to bind it somewhere (hence free particle) $\endgroup$
    – Kyle Kanos
    Sep 27, 2013 at 1:14
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    $\begingroup$ That "clicked" after BMS asked about it being a free particle. For some reason, I didn't make the connection between "free" and being without a potential. $\endgroup$
    – Astrum
    Sep 27, 2013 at 1:16

3 Answers 3


Just adding some precisions to @AlfredCentauri 's answer.

When you use a ket notation $|p\rangle$, or $|E\rangle$, or $|x\rangle$, by definition, these kets are eigenvectors of their corresponding operators :

$P|p\rangle=p|p\rangle, \quad H|E\rangle = E |E\rangle, \quad X|x\rangle = x|x\rangle$

It is not a calculus, it it a definition.

So, if you try to find a solution of the Schrödinger equation by $\left| \Psi \right \rangle = \left| \Lambda \right \rangle e^{-iEt/ \hbar}$, where you don't know $|\Lambda \rangle$, the first lines of your question prove that $|\Lambda \rangle = |E\rangle$, thanks to the definition of $|E\rangle$

Finally, starting from the definition of the ket $|p\rangle$ : $P|p\rangle=p|p\rangle$, we have obviously $\large \frac{P^2}{2m}|p\rangle=\frac{p^2}{2m}|p\rangle$.

In the case of a free particle, you have $H = \frac{P^2}{2m}$, so finally $ H|p\rangle=\frac{p^2}{2m}|p\rangle = E_p|p\rangle$.

This shows that $|p\rangle$ is a eigenvector of $H$, with the eigenvalue $E_p=\frac{p^2}{2m}$

In a one-dimensional problem, for one value of $E$, you have to possibilites for $p$, such as $E=E_p$, these are $ p\pm = \pm \sqrt{2mE}$.


(where did the potential go?)

A potential implies a force and a free particle is not under the influence of a force (else it wouldn't be free from force).

this follows from the eigen-equation, where the eigenvalue must be equal to E?

It is not uncommon, when the context is appropriate, to denote a state by its eigenvalue, e.g. the context is states of definite energy.

(why are we all of a sudden talking about the ket p?)

Because, for a free particle, the Hamiltonian and momentum operators commute, i.e., states of definite energy are also states of definite momentum. Thus, in this context, an eigenstate can carry the energy eigenvalue label or the momentum eigenvalue label.


To answer your question: I assume your book is talking about the free particle and not the more general problem. The free particle itself (eg wavepackets spreading, or just simple time evolution of free partiles initial condition) is an interesting problem in itself. In this case, the potential is constant---and we can pick any constant, so V=0 works best and makes the reading and typesetting easiest and best to look at.

I also wanted to point out an error later in the post

$ \left| p \right \rangle = \pm \sqrt{2mE} $

This is not true, what you probably mean to say, is that the P operator has the effect of scaling any "eigen vector of P_op" or eigenstate |p> by $\pm \sqrt{2mE}$

  • $\begingroup$ Thanks bert, I too thought the highlighted line made no sense. $\endgroup$
    – Kenshin
    Sep 27, 2013 at 9:13

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