Expectation Value of Momentum in H-Atom

In summary, a discussion about calculating the expectation value of momentum in the H-atom took place. The conversation included attempts to evaluate the integral and discussions about the ground state wavefunction and the properties of the momentum operator. Ultimately, it was concluded that the expectation value of momentum for any stationary bound state must be 0. Different methods were suggested, such as using the Klein-Gordon momentum density operator, to get a real result without the imaginary part. The conversation also touched on the electron speed and the value of alpha.
  • #1
quantumdude
Staff Emeritus
Science Advisor
Gold Member
5,584
24
Here's a silly question. I'm sure I should know the answer, but alas my most recent QM course was 9 years ago.

I sat down to calculate the expectation value of momentum in the H-atom today, because some kid on another forum wanted to know how fast an electron in an atom is. I was going to calculate [itex]<p>[/itex] and divide it by [itex]m[/itex] for him.

So I wrote down [itex]<p>=\int_V\psi_{100}^*(-i\hbar\nabla)\psi_{100} d^3x[/itex] ([itex]\psi_{100}[/itex] being the ground state H-atom wavefunction). Upon noticing that I've got a solitary [itex]i[/itex] sandwiched between two purely real functions, I didn't even bother to continue. It's going to come out imaginary.

What am I doing wrong? :confused:
 
Physics news on Phys.org
  • #2
Not evaluating the integral. Just compute it and see.

Then, calculate <p^2> and take its root mean square and tell the kid that.
 
  • #3
The vector momentum =0.
 
  • #4
Tom Mattson said:
So I wrote down [itex]<p>=\int_V\psi_{100}^*(-i\hbar\nabla)\psi_{100} d^3x[/itex] ([itex]\psi_{100}[/itex] being the ground state H-atom wavefunction). Upon noticing that I've got a solitary [itex]i[/itex] sandwiched between two purely real functions, I didn't even bother to continue. It's going to come out imaginary.

What am I doing wrong? :confused:

It should come out 0, and in fact you just proved this:
the p operator being hermitean, its expectation value must be real for any vector. You showed that it must be a purely imaginary number. Hence it can only be 0.

And that's normal too: the expectation of momentum for any stationary bound state must be 0: given that a stationary state will not have the momentum change in time, and given that the fact that it is bound will confine (most of) the wavefunction to a bounded region, the C.O.G. cannot leave that bounded region, which it should, if there was a non-zero expectation value of momentum, after some time.
 
  • #5
The integral is not zero. The ground state wavefunction is:

[tex]\psi_{100}(r)=\frac{1}{\sqrt{\pi}}\frac{1}{a_0^{3/2}}\exp(-r/a_0)[/tex]

If we integrate [itex]\psi^*_{100}(-i\hbar\nabla)\psi_{100}[/itex] over all space we get:

[tex]\frac{-i\hbar}{\pi a_0^3}\int_0^{2\pi}\int_{-1}^1\int_0^{\infty}\exp(-r/a_0)\frac{d}{dr}\exp(-r/a_0)r^2drd(\cos(\theta))d\phi[/tex]

Since the wavefunction doesn't depend on the angular coordinates, this integral is proportional to the following:

[tex]i\int_0^{\infty}r^2\exp(-2r/a_0)dr[/tex]

The integrand is not identically zero, and is nonnegative everywhere. So it's integral cannot vanish. So, it seems this result is purely imaginary and not zero.

There's got to be a mistake somewhere, but I don't see where. :confused:
 
Last edited:
  • #6
The nabla is a vector operator. It produces something proportional to r_vector. Integrating that over spherically symmetrical space gives zero vector.
 
  • #7
Tom Mattson said:
The integral is not zero. The ground state wavefunction is:

[tex]\psi_{100}(r)=\frac{1}{\sqrt{\pi}}\frac{1}{a_0^{3/2}}\exp(-r/a_0)[/tex]

If we integrate [itex]\psi^*_{100}(-i\hbar\nabla)\psi_{100}[/itex] over all space we get:

[tex]\frac{-i\hbar}{\pi a_0^3}\int_0^{2\pi}\int_{-1}^1\int_0^{\infty}\exp(-r/a_0)\frac{d}{dr}\exp(-r/a_0)r^2drd(\cos(\theta))d\phi[/tex]

Since the wavefunction doesn't depend on the angular coordinates, this integral is proportional to the following:

[tex]i\int_0^{\infty}r^2\exp(-2r/a_0)dr[/tex]

The integrand is not identically zero, and is nonnegative everywhere. So it's integral cannot vanish. So, it seems this result is purely imaginary and not zero.

There's got to be a mistake somewhere, but I don't see where. :confused:

I agree with Smallphi.
You forgot to include [itex] \hat{r}[/itex]. Integrating this over all angles gives zero.
 
  • #8
One simple way to get an appropriate estimate of the velocity is to divide 13.6eV by the electron mass, multiply by 2 and then take the square-root thus finding that the electron travels at a bit less than one hundredth of the speed of light. This is correct because the expectation value of the kinetic energy is equal to the negative of the expectation value of the Hamiltonian for a Coulomb potential. I.e. -(-13.6eV).

Also, you can see that <\vec p> must be zero for any real wave-function (btw, eigenfunctions of this Hamiltonian can always be *chosen* as real) by just integrating by parts... you end up with the equation <\vec p>=-<\vec p> which implies that <\vec p>=0
 
  • #9
To get rid of the " i " you might want to use following momentum
density operator:

[tex]{ i\hbar \over 2} \left( \frac{\partial \psi^*}{\partial x_i}\psi\ -\ \psi^*\frac{\partial \psi}{\partial x_i }\right)[/tex]

Where the psi is the relativistic wave function related to the Schrödinger
wave function as follows:

[tex]\psi\ =\ \Psi \ e^{-imc^2t/\hbar}, \qquad \mbox{$\Psi$ is the Schroedinger wave function}[/tex]

This gives the same result without the imaginary part. It's always real.
This is the Klein-Gordon momentum density operator, as well as the Dirac
momentum density operator, (after the spin contributions are split off
via the Gordon decomposition).Regards, Hans
 
Last edited:
  • #10
olgranpappy said:
divide 13.6eV by the electron mass, multiply by 2 and then take the square-root thus finding that the electron travels at a bit less than one hundredth of the speed of light.

A little bit less than one hundredth of the speed of light is 1/137.036 to be
a bit more exact, a value also known as alpha. :wink:


Regards, Hans
 

Related to Expectation Value of Momentum in H-Atom

What is the expectation value of momentum in a hydrogen atom?

The expectation value of momentum in a hydrogen atom is the average value of the momentum of the electron in the atom. It is calculated by taking the product of the momentum operator and the wave function of the electron, and integrating over all space.

How is the expectation value of momentum related to the uncertainty principle?

The expectation value of momentum is related to the uncertainty principle in that it is always greater than or equal to the uncertainty in the momentum. This means that the more precisely we know the momentum of the electron, the less certain we are about its position, and vice versa.

Why is the expectation value of momentum important in quantum mechanics?

The expectation value of momentum is an important concept in quantum mechanics because it allows us to make predictions about the behavior of particles on a microscopic level. It also helps us understand the inherent uncertainty in the behavior of quantum particles.

How does the expectation value of momentum change for different energy levels in the hydrogen atom?

The expectation value of momentum increases as the energy level of the electron increases in the hydrogen atom. This is because electrons in higher energy levels have a greater range of possible momenta, leading to a larger uncertainty in the momentum and a higher expectation value.

Can the expectation value of momentum be experimentally measured?

Yes, the expectation value of momentum can be experimentally measured through techniques such as electron diffraction or particle scattering. These experiments involve measuring the momentum of a large number of particles and calculating the average value, which should closely match the theoretical expectation value.

Similar threads

Replies
2
Views
897
  • Quantum Physics
2
Replies
56
Views
3K
  • Quantum Physics
Replies
20
Views
3K
Replies
2
Views
601
Replies
9
Views
2K
  • Quantum Physics
Replies
8
Views
5K
Replies
1
Views
823
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
663
Back
Top