How to prove that for any bound electronic state, < p > = 0

[ani4physics]

Hey all. So, I understand that every bound electronic state will have zero average electronic momentum, because otherwise the electron will fly off the atom. But how do I show mathematically that < p > = 0 for any bound state. Any help or reference greatly appreciated. Thanks.

 

[Sando]

Try relating $$\langle p \rangle$$ to $$\frac{d}{dt} \langle x \rangle$$ (Ehrenfest's Theorem). If "bound state" means "stationary state in a time-independent potential", then use the fact that $$\frac{d}{dt} \langle x \rangle = 0$$.

 

[ani4physics]

Try relating $$\langle p \rangle$$ to $$\frac{d}{dt} \langle x \rangle$$ (Ehrenfest's Theorem). If "bound state" means "stationary state in a time-independent potential", then $$\frac{d}{dt} \langle x \rangle = 0$$.

If d<x>/dt = 0 then that means <p> is a constant. but how I prove that <p> = 0?

 

[Petr Mugver]

Hey all. So, I understand that every bound electronic state will have zero average electronic momentum, because otherwise the electron will fly off the atom. But how do I show mathematically that < p > = 0 for any bound state. Any help or reference greatly appreciated. Thanks.

The atom's hamiltonian commutes with the parity operator P. Hence any energy eigenstate $$|\psi>}$$ must have definite parity:

$$P|\psi>=\pm|\psi>$$

Now, the momentum operator changes sign under parity

$$PpP^\dagger=-p$$

So, using $$P^\dagger P = I$$, we find

$$<\psi|p|\psi>=(<\psi|P^\dagger)( P p P^\dagger) (P|\psi>)=-<\psi|p|\psi>$$

hence

$$<\psi|p|\psi>=0$$

 

[jostpuur]

The atom's hamiltonian commutes with the parity operator P. Hence any energy eigenstate $$|\psi>}$$ must have definite parity:

$$P|\psi>=\pm|\psi>$$

This is not properly justified. You did not use the knowledge that the state is "bounded" (whatever it means rigorously...?)

Now, the momentum operator changes sign under parity

$$PpP^\dagger=-p$$

So, using $$P^\dagger P = I$$, we find

$$<\psi|p|\psi>=(<\psi|P^\dagger)( P p P^\dagger) (P|\psi>)=-<\psi|p|\psi>$$

hence

$$<\psi|p|\psi>=0$$

For example set $V=0$. The Hamiltonian of a free particle commutes with the parity operator too, but still its eigenstates $\psi(x)=e^{ikx}$ don't satisfy $P|\psi\rangle = \pm|\psi\rangle$.

How do you prove, that bounded states are non-degenerate? That would help. In the previous counter example the eigenstates $\sin(kx)$ and $\cos(kx)$ would be eigenstates of parity operator, but their linear combinations are not.

 

[Sando]

For a time independent potential:

$$\langle p \rangle = m \frac{d}{dt} \langle x \rangle$$

which is zero for a bound state.

 

[ani4physics]

ok someone please tell me if this is right:

d<p>/dt = <F> : this is always valid. so if I want to calculate the average force on a particle in any stationary state / non stationary state, I can just calculate the average momentum and take the time derivative.

For example, if I have an atom in presence of a time-dependent potential, I can calculate the perturbed wave function using time-dependent perturbation theory, then take the average of p over the perturbed wave function and then take its time derivative to get the average force. Is that right?

 

[dextercioby]

You can easily show that it's time independent, but showing it equal to 0 is a much stronger statement. Compute the integral in the case of the infinite square potential in 1 dimension. Is the integral = 0 ?

 

[ani4physics]

You can easily show that it's time independent, but showing it equal to 0 is a much stronger statement. Compute the integral in the case of the infinite square potential in 1 dimension. Is the integral = 0 ?

yeah I can show that it's time independent. For example, consider the ground electronic state of an atom in absence of any external field. then in ground state,

<p> = <psi * Exp[i * w * t] ] p [psi * Exp [ -i * w *t]>

= <psi] p [psi>

= time independent.

but, is <p> = 0? someone told me it's = 0 but I can't understand why.

Also, someone please tell me if my previous post is right or wrong.