# Momentum in time independent schrodinger equation

In the time independent Schrödinger equation I have read that the wave function is $$\psi(x)=\mathrm e^{-iEt/\hbar}$$ This means that $$\rho(x,t)=|\psi(x,t)|^2=|\psi(x)|^2=ρ(x)$$, that is, $\rho$ is not a function of $t$. That's why it is called stationary state.

Therefore, $$\left<x\right>=\int \psi^*x\,\psi \mathrm{d}x=\int \psi(x)^*x\,\psi(x) \mathrm{d}x=\text{constant}$$ and hence $$\left<p\right>=0$$ , the average value of momentum is zero.

What does this mean? Does this mean particle is in rest?

I think this is something like probability of going in one direction is equal to the probability of going in opposite direction, like $MV-MV =0$. Is this sense right? Or something else is happening?

• Please fix the math of this question. See MathJax basic tutorial and quick reference for a basic MathJax tutorial and quick reference. Thank you for your collaboration. – AccidentalFourierTransform Jan 4 '17 at 19:32
• I don't understand why are you giving me math tutorial, this is not math ,it is quantum mechanics I want my conceptions clear with proper logic that's why I want to clear myself why average of momentum is zero – user101134 Jan 4 '17 at 19:44
• @user101134 It's just ugly to read it this way. – Ali Jan 4 '17 at 19:46
• Just because $\langle x \rangle$ is a constant doesn't mean that $\langle p \rangle = 0$. Consider, for example, the wavefunction $\psi(x) = e^{ipx/\hbar}$, for which the probability distribution $\rho(x)$ is constant with respect to time. – Michael Seifert Jan 4 '17 at 19:46
• @user101134 In your first line you wrote ψ(x) e^(-iEt/h). If you change this into $\psi(x)=e^{-iEt/\hbar}$ it becomes $\psi(x)=e^{-iEt/\hbar}$, which is much more clear. Do that with the rest of math formulas in your post, and youll improve its quality substantially. – AccidentalFourierTransform Jan 4 '17 at 19:47

Your reasoning in the OP is correct, except when you say that the particle is at rest. Recall that in QM the exact position of a particle, and its state of motion, are not well defined. This means that it doesn't make sense to say that a particle is at rest. You could only say that a particle is at rest if $$\Delta p\equiv 0$$ but, because of the Heisenberg uncertainty principle, $\Delta p$ is always positive.

On the other hand, the correct statement is that it has an equal probability to be moving to the right and to be moving to the left, and thus, on average, its mean velocity is zero. You observed this in the OP, and this is the correct interpretation of $\langle p\rangle=0$.

• What if the potential is non-symmetric, or a scattering potential at 0<x<a, applied to a source at -b<0? Is <p>=0? OP specified no specific potential. – Bill N Jan 4 '17 at 20:31
• @BillN the potential is irrelevant. If $\langle x\rangle=\text{constant}$ then $\langle p\rangle=0$ because of Ehrenfest's theorem. – AccidentalFourierTransform Jan 4 '17 at 20:55

In the case of stationary states, the probability and hence the expectation values do not vary with time. As @AccidentalFourierTransform indicated, the reasoning is true that this leads to $\langle x\rangle=\text{constant}$. However, this doesn't mean that the particle is at rest.

If you draw the probability distribution versus the position, you see that the probability does not peak at a single point, instead there is a spread (remember that $\vert \psi(x)\vert^2 dx$ gives the probability to locate the particle at an "elemental length", not at the point $x$), which is as given by the uncertainty principle.

If the spread in the position space is less, then the particle is more localized, and the corresponding spread in the momentum space will be very large. So one couldn't expect $\langle p\rangle=0$. Ideally, if the probability in the position space peaks exactly at a particular value of $x$, the probability spreads to an infinite extend in the momentum space and this represents a plane wave state.

What the expectation value $\langle x\rangle=\text{constant}$ gives is the probability that the physical system represented by the state $\psi(x)$ about the point is a constant and hence the expectation value of being the state when it is at the point $x$ is a constant and the probability along some length about that point does not change with time.