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The impulse operator in quantum mechanics is given by

\begin{align} \hat{p} = \frac{\hbar}{i}\nabla \end{align}

As a Hermitian operator, the expected value of this operator $\langle{p}\rangle = \langle \psi|\hat{p}\psi\rangle$ should be real. However, for a real wave function $\psi(\vec {r})\in \mathbb{R}$ (a valid solution to the Schrödinger equation) the resulting integral is imaginary:

\begin{align} \langle{p}\rangle = \frac{\hbar}{i}\int d^3r \cdot \psi \nabla \psi \end{align}

Is there an error in my thinking or is it impossible to calculate the expected value that way? An alternative approach would be to use the Fourier transform.

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The possibility that is not accounted for in the question is that the integral may be zero. In fact, it can be shown that a wave function corresponding to a stationary state can always be chosen real, and the momentum of a stationary state is definitely zero.

Another insight may come from considering wave function $$\phi_+(x) = \psi_k(x) + \psi_{-k}(x) = e^{ikx} + e^{-ikx} = 2\cos(kx).$$ The average momentum in this state is zero, as it is a sum of two states with opposite momenta, $\pm\hbar k$.

To conclude: your formula for the average momentum is correct, since it is obtained from general rule. And, since it would give an unphysical imaginary value for a real wave function, it means that all such wave functions correspond to states with zero momentum.

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  • $\begingroup$ Do you have a citation for that a stationary state can always be chosen real? I have doubts about that. $\endgroup$
    – lalala
    Jul 17, 2020 at 16:18
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    $\begingroup$ @lalala I think it was somewhere in the beginning of Landau&Livshitz's quantum mechanics. And I probably should have said bound state - not sure that notion of stationarity is applicable in this context. $\endgroup$ Jul 17, 2020 at 17:14
  • $\begingroup$ I vaguely remember. Probably together with the higher then eigenvalue the more nodes in the wave function. Probably this was limited tonHamiltonian of the form H= T+V $\endgroup$
    – lalala
    Jul 17, 2020 at 17:27
  • $\begingroup$ @lalala this is a good point. $\endgroup$ Jul 17, 2020 at 17:28
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Just to add to Vadim's answer: The integral $$\int_{-\infty}^{\infty} \psi \partial_x \psi dx= \frac12 \int_{-\infty}^{\infty} \partial_x( \psi^2) dx = [\psi^2]_{-\infty}^{\infty}=0 $$ fo all wavefunctions that vanish at infinity.

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If your wavefunction $\psi$ is real, as is the case when you are dealing with a solution to the time-independent Schrodinger equation, then indeed the expectation value is automatically $0$ since the expectation value must be real and the integral $-i/\hbar\int dx \psi^* (\nabla)\psi$ is necessarily complex unless it is $0$.

If the wavefunction is complex, then one cannot say: the expectation can be $0$ or not. For instance, the combination of h.o. wavefunctions \begin{align} \psi(x)=\alpha \psi_n(x)+i\beta\psi_{n+1}(x)\, ,\qquad \alpha^2+\beta^2=1\, ,\quad \alpha,\beta\in\mathbb{R} \end{align} will have non-zero $\langle p\rangle$. However, \begin{align} \psi(x)=\alpha \psi_n(x)+i\beta\psi_{n+2}(x)\, ,\qquad \alpha^2+\beta^2=1\, ,\quad \alpha,\beta\in\mathbb{R} \end{align} will have $\langle p\rangle=0$ even if it is a complex combination.

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