# Impulse operator on real wave function

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.

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.

• Do you have a citation for that a stationary state can always be chosen real? I have doubts about that. Jul 17, 2020 at 16:18
• @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. Jul 17, 2020 at 17:14
• 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 Jul 17, 2020 at 17:27
• @lalala this is a good point. Jul 17, 2020 at 17:28

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.

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.