Expectation value of momentum for bound states

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Homework Help Overview

The discussion revolves around proving that the expectation value of momentum for any bound state is zero. The context involves quantum mechanics, specifically the properties of wavefunctions in bound states and their implications on momentum calculations.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • The original poster attempts to establish a proof based on the parity of wavefunctions and their derivatives. Questions arise regarding the parity of bound states in symmetric versus asymmetric potentials.

Discussion Status

Participants explore the implications of symmetry in potentials on the parity of bound states and the expectation value of momentum. Some guidance is offered regarding the application of Ehrenfest's theorem, but questions remain about its applicability to free particles and the conditions under which momentum expectations are zero.

Contextual Notes

There is a discussion about the nature of potentials (symmetric vs. asymmetric) and how they affect the parity of wavefunctions. The potential implications of these characteristics on the expectation values are also considered.

blue_leaf77
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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:
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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:
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 x<0, and is a quadratic potential (a quantum spring) for x\geq 0. Here the eigenfunctions are definitely zero for x\leq 0, and nonzero for x> 0.
 
So is there a more formal way to prove that expectation value of momentum for bound states is zero?
 
Yes there is.
Ehrenfest's theorem.
 
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??
 
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.
 
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