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How do we prove that the expectation value of momentum operator in any bound state is zero ?

$$ \langle \hat{P} \rangle_{\text{bound state}}=0 $$

and what about the position expectation value ?

How do we prove it for a general case. How do we get a physical intuition from the corresponding wavefunctions ?


marked as duplicate by ACuriousMind, Gert, Kyle Kanos, Bill N, John Rennie Nov 6 '15 at 7:28

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    $\begingroup$ Possible duplicate [physics.stackexchange.com/q/100277/] $\endgroup$ – Oswald Nov 5 '15 at 5:39
  • $\begingroup$ @Vishwaas thanx for the comment. but is there any general proof for this ?. I expect a much more detailed and complete answer. $\endgroup$ – ss1729 Nov 5 '15 at 5:46
  • $\begingroup$ Dear @soorajs - you obtain the answer (the general proof) if you move the "cursor" of the "mouse" above "physics.stack.exchange..." above and press the left button on the "mouse". And then press "page down" twice to get from a copy of your question to the answer. $\endgroup$ – Luboš Motl Nov 5 '15 at 6:37
  • $\begingroup$ @LubošMotl i'm srry .... didnt get it? $\endgroup$ – ss1729 Nov 5 '15 at 9:44
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    $\begingroup$ Intuitively, if a bound state would have an average nonzero momentum along the confined dimension, it would no longer be "bound" because it would just propagate away. You could maybe start by proving that any real wave function has $\left< p^n \right>=0$ for $n$ odd. Just use the fact that $p$ is hermitian so that its expectation value must be real for normalizable wave functions. $\endgroup$ – Praan Nov 5 '15 at 10:55

I think for 1-dimensional bound states, this is the proof :

The expectation value of the momentum operator,

$\langle \hat{P} \rangle$=$\langle \psi|\hat{P} \psi\rangle$=$\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{\partial}{\partial x}\psi(x)dx$=$\frac{\hbar}{i}\int_{-\infty}^{+\infty}\psi^{*}(x)\frac{\partial}{\partial x}\psi(x)dx$=$\frac{\hbar}{2i}\int_{-\infty}^{+\infty}\frac{\partial}{\partial x}({\psi}^2(x))dx$ =$\frac{\hbar}{2i}{\psi}^2(x) \Big|_{-\infty}^{+\infty}$= 0

Since $\psi(x)$=0 at x=$\pm \infty$ for bound states.

and similarly for position expectation value in the bound state,

$\langle \hat{x} \rangle$ = 0

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    $\begingroup$ The fourth equality in your proof is wrong unless the wave function is real. And the second statement is only generally true for nondegenerate bound states in symmetric potentials. By noting that bound states can always be chosen real, this rounds up the proof. See the accepted answer of physics.stackexchange.com/questions/77894/… $\endgroup$ – Praan Nov 5 '15 at 18:14
  • $\begingroup$ @Praan it'd be nice if u could kindly edit the answer or post a more detailed one. $\endgroup$ – ss1729 Nov 6 '15 at 5:38
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    $\begingroup$ This method of proof is already explained in the answer of user gonenc to the linked duplicate question. For the last statement, just think of a displaced harmonic oscillator to see that is false except in special cases. If a potential is symmetric $V(-x)=V(x)$ then the wave function of any nondegenerate eigenstate $\left| n \right>$ is either odd or even so that $\left< n \right| x \left| n \right> = 0$ since the integrand is always odd. $\endgroup$ – Praan Nov 6 '15 at 11:57
  • $\begingroup$ @Praan soo both the statements ($\langle \hat{P} \rangle=0$,$\langle \hat{X} \rangle=0$) are only true for non-degenerate bound states in symmetric potentials and of cause for non-relativistic case, right ? $\endgroup$ – ss1729 Nov 6 '15 at 13:14
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    $\begingroup$ No, the first statement $\langle P \rangle = 0$ is true for all bound stationary states. The second statement $\langle x \rangle = 0$ is true for nondegenerate eigenstates of a Hamiltonian with a symmetric potential. $\endgroup$ – Praan Nov 6 '15 at 14:02

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