Expectation value of momentum for bound states

In summary, the conversation discusses proving that the expectation value of momentum for any bound state is zero. It is suggested that this is true because bound states of symmetric potentials have a definite parity, making the Hamiltonian invariant under coordinate swap. It is also mentioned that for asymmetric potentials, the wavefunctions may not have definite parity but the expectation of momentum may still be zero. The use of Ehrenfest's theorem is suggested as a formal way to prove this. Finally, it is clarified that for a free particle, the expectation of momentum need not be zero unless the expectation of position is a constant value.
  • #1
blue_leaf77
Science Advisor
Homework Helper
2,637
786

Homework Statement


I'm curious in proving that expectation value of momentum for any bound state is zero. So the problem is how to prove this.

Homework Equations


$$ \langle \mathbf{p_n} \rangle \propto \int \psi^*(\mathbf{r_1}, \dots ,\mathbf{r_N}) \nabla_n \psi(\mathbf{r_1}, \dots ,\mathbf{r_N}) d^3\mathbf{r_1} \dots d^3\mathbf{r_N} $$
where ## n = 1,2, \dots N ##.

The Attempt at a Solution


If I assume that ## \psi(\mathbf{r_1}, \dots ,\mathbf{r_N}) ## is either even or odd then its first derivative will be its counterpart, so ## \nabla_n \psi(\mathbf{r_1}, \dots ,\mathbf{r_N}) ## has opposite parity in ## \mathbf{r_n} ##. Thus the integral will result zero.
My question is that is this true that all bound states can only be even or odd? If yes why?
 
Last edited:
Physics news on Phys.org
  • #2
All bound states of symmetric potentials have a definite parity.

This can be shown because in that case the entire Hamiltonian will have even parity, and will be invariant under a coordinate swap.

Whether this is true for bound states of asymmetric potentials (say three randomly placed square wells), I couldn't say. The wavefunctions would no longer have definite parity, but it may yet be the case that the expectation of the momentum is still zero.
 
Last edited:
  • #3
Okay, the eigenfunctions would definitely not have a definite parity if the potential is asymmetric. As an extreme example, consider the potential, which is an infinite square well for [itex]x<0[/itex], and is a quadratic potential (a quantum spring) for [itex]x\geq 0[/itex]. Here the eigenfunctions are definitely zero for [itex]x\leq 0[/itex], and nonzero for [itex]x> 0[/itex].
 
  • #4
So is there a more formal way to prove that expectation value of momentum for bound states is zero?
 
  • #5
Yes there is.
Ehrenfest's theorem.
 
  • #6
Ok thank you very much, now it's pretty clear why it should be zero for bound states. But first my question is does this theorem also applicable to free particle? If yes then the momentum of free particle is zero??
 
  • #7
For the free particle, the expectation of the position need not be a constant value like it would need to be for a bound state. Because of this, the expectation for the momentum need not be zero.

Whenever the expectation of the position is constant, the expectation of the momentum will be zero.
 
  • Like
Likes blue_leaf77

Related to Expectation value of momentum for bound states

1. What is the expectation value of momentum for bound states?

The expectation value of momentum for bound states is a measure of the average momentum of a particle in a bound state. It is calculated by taking the integral of the momentum operator over the wave function of the bound state.

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

According to the uncertainty principle, the product of the uncertainty in position and momentum is always greater than or equal to a constant. This means that as the expectation value of momentum increases, the uncertainty in momentum decreases, and vice versa.

3. Can the expectation value of momentum be negative?

No, the expectation value of momentum cannot be negative. This is because momentum is a vector quantity and the expectation value is a scalar. The expectation value represents the average direction and magnitude of the momentum, which is always positive.

4. How does the expectation value of momentum change for different bound states?

The expectation value of momentum can vary for different bound states, depending on the specific properties of the bound state. For example, the expectation value may be higher for a bound state with a larger energy, or for a bound state in a larger potential well.

5. What is the significance of the expectation value of momentum in quantum mechanics?

The expectation value of momentum is an important concept in quantum mechanics as it allows us to make predictions about the behavior of particles in bound states. It also helps us understand the relationship between position and momentum and how they can affect the properties of a bound state.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
360
Replies
1
Views
708
  • Calculus and Beyond Homework Help
Replies
3
Views
573
  • Quantum Interpretations and Foundations
Replies
5
Views
2K
Replies
21
Views
2K
  • Atomic and Condensed Matter
Replies
1
Views
2K
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Quantum Physics
Replies
3
Views
961
  • Advanced Physics Homework Help
Replies
9
Views
2K
Back
Top