There is probably a very simple answer to this question, but I have not been able to find it in my textbook or online. Please bear in mind that I am in my first week of quantum mechanics.

First, my textbook tells me that $$\langle p \rangle = m \frac{d}{dt} \langle x \rangle = - i \hbar \int_{-\infty}^{\infty} \left( \psi^* \frac{\partial \psi}{\partial x} \right) dx.$$

To explore this a little, I made up the normalized wave function $$ \psi(x) = \begin{cases} -\sqrt{5}ix^2 : & 0 \leq x \leq 1 \\ 0 : & \mbox{elsewhere}, \end{cases} $$ which gives us $$\langle x \rangle = \int_{0}^{1} x|\psi|^2 dx = \frac56.$$ From this, it follows that $$\langle p \rangle = m \frac{d}{dt} \langle x \rangle = 0.$$ However, \begin{eqnarray*} \langle p \rangle &=& - i \hbar \int_{-\infty}^{\infty} \left( \psi^* \frac{\partial \psi}{\partial x} \right) dx \\ &=& - 10 i \hbar \int_{0}^{1} x^3 dx \\ &=& - \frac52 i \hbar. \end{eqnarray*}

Why am I getting different answers for $\langle p \rangle$?

Edit: Changed $\psi(x,0)$ to $\psi(x)$.

  • 1
    $\begingroup$ You have to compute the expectation value for the solution $\psi(x,t)$ of the Schrödinger equation with that $\psi(x,0)$ as the initial condition. How is $\psi(x,0)$ supposed to know about the time dependence otherwise? $\endgroup$
    – ACuriousMind
    Sep 6, 2016 at 1:03
  • $\begingroup$ Assume there is no time dependence (I will edit the question to reflect this) I was given homework where the wave functions were of the form $\psi(x)$. I computed $\langle p \rangle$ with either of the two methods above, thinking they were interchangeable. Now, I am not so sure. $\endgroup$
    – theelder3
    Sep 6, 2016 at 1:19
  • $\begingroup$ @ACuriousMind is still correct. The relationship: $$m \frac{\operatorname{d}}{\operatorname{d} t} \langle x\rangle = \langle p \rangle$$ is known as a canonical equation of motion, and it only applies after the time evolution for $\psi$ has been correctly calculated. $\endgroup$ Sep 6, 2016 at 1:23

1 Answer 1


You're running into two problems. First, you're failing to have the wave function correctly evolve with time. You have to get the time evolution correct before applying these relationships. Second, you're using a discontinuous wave function, and taking the derivative of a discontinuous function requires more care than with continuous ones. When you take the derivative of a discontinuous function it generates Dirac delta functions. When you try to calculate the amount of energy associated with such a wave function you'll get $\infty$. It is because of this that it is generally taken as a requirement that all wave functions be at least continuous.

Here are a pair of better behaved wave functions for you to play with: $$\begin{align}\psi_{\mathrm{Gauss}}(x) & = \frac{1}{\sqrt{\sigma \sqrt{2\pi}}} \operatorname{e}^{-\frac{(x-x_0)^2}{4\sigma^2} + i\frac{p_0 x}{\hbar}}, \ \mathrm{and}\\ \psi_{\mathrm{sinc}}(x) &= \frac{2\hbar \sin\left(\frac{p_w [x-x_0]}{2\hbar}\right)}{p_w [x-x_0]} \operatorname{e}^{i\frac{p_0 [x-x_0]}{\hbar}} \sqrt{\frac{p_w}{2\pi \hbar}} .\end{align}$$

$\langle p\rangle$ can be calculated without first calculating the time evolution of $\psi$, but you'll need to watch those delta functions. The wave function in the question can be written:$$\psi(x) = -\Theta(x) \Theta(1-x) \sqrt{5} i x^2,$$ where $\Theta(x)$ is the Heaviside step function that is $0$ for $x < 0$ and $1$ for $x>0$. The first derivative of $\psi$ is: $$\psi'(x) = -\Theta(x) \Theta(1-x) 2\sqrt{5}i x + \Theta(x) \delta(1-x) \sqrt{5} i x^2.$$

  • $\begingroup$ Okay, that makes sense. I should have realized that the discontinuity would cause an issue with differentiation. Now I have two follow-up questions. Does the second method (i.e. the integral method) work for my discontinuous example? That is, is my second solution for $\langle p \rangle$ correct? Second, in general, can I count on $\langle p \rangle = m \frac{d \langle x \rangle}{dt}$ giving me the correct answer if $\langle x \rangle$ is continuous? $\endgroup$
    – theelder3
    Sep 6, 2016 at 1:27
  • $\begingroup$ If $\psi$ is continuous and the time evolution has been correctly applied, then the relationship will hold. Your integral calculation of $\langle p \rangle$, as shown, has an error: you failed to include the delta function. Calculate $\langle p\rangle$ for the functions I gave above. Notice what variables are in the answer? Also, $\langle p \rangle$ must be a real number. $\endgroup$ Sep 6, 2016 at 1:36

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