Infinite Square Well Eigenstates

[phrygian]

Homework Statement

The eigenstates of the infinite square well are not energy eigenstates and are not momentum eigenstates.

Homework Equations

The Attempt at a Solution

I don't understand how this can be? If the eigenstates of the infinite square well are energy eigenstates, and thus have a definite energy, how can that not have a definite momentum given that momentum in the well is just Sqrt[(2*m*KE)]?

 

[diazona]

What eigenstates are you talking about? i.e. the eigenstates of which operator? There's no such thing as "eigenstates of the infinite square well", there are only eigenstates of each individual operator in the infinite square well potential.

 

[phrygian]

I am talking about the energy eigenstates of the infinite square well, how are these not also momentum eigenstates since p = Sqrt[(2*m*KE)]

 

[diazona]

I am talking about the energy eigenstates of the infinite square well, how are these not also momentum eigenstates since p = Sqrt[(2*m*KE)]

It's not that simple in quantum mechanics. The formula you have really only works well for regular numbers. But when you're dealing with operators (as you are in QM), you run into problems with multiple operators squaring to the same thing. It's like how you can have two different numbers that have the same square, one positive and one negative, but more complex, because there can be a much larger number of operators with the same square, and only one of them can really be considered the "square root" of the original operator.

You can actually check this for yourself: take the square well energy eigenfunctions
$$\psi(x) = A \sin(k_n x)$$
and first apply the momentum operator to them:
$$\hat{p}\psi = -i\hbar\frac{d}{d x}\psi = \cdots$$
Notice that what you get is not a multiple of the original wavefunction, so it's not an eigenfunction. But if you apply the momentum operator again (try it!), you'll get something that is a multiple of the original eigenfunction. So the square well energy eigenstates are eigenstates of $$\hat{p}^2$$ (because that's what is in the Hamiltonian), but not of $$\hat{p}$$. This can happen because $$\hat{p}$$ is not the operator that you'd call the "square root" of $$\hat{p}^2$$:
$$\hat{p}\neq \sqrt{\hat{p}^2}$$
kind of like how $-2 \neq \sqrt{(-2)^2}$.

 

[paranormal]

There are a few key concepts that need to be clarified in order to understand why the eigenstates of the infinite square well are not energy or momentum eigenstates.

Firstly, the eigenstates of a system are the states in which the system's properties are well-defined and unchanging. In the case of the infinite square well, the eigenstates are the stationary states of the system, meaning they do not change with time.

Secondly, energy and momentum are both conserved quantities in a system. This means that they can only take on certain discrete values, rather than being continuous. In the case of the infinite square well, the energy levels are quantized, meaning they can only take on certain discrete values, and the same applies for momentum.

Now, coming to the question of energy and momentum eigenstates, these are states in which the energy or momentum of the system is well-defined and unchanging. In the case of the infinite square well, the eigenstates are not energy or momentum eigenstates because they do not have a definite energy or momentum. Instead, they are a superposition of different energy and momentum states.

To understand this, imagine a particle in the infinite square well. The particle can have different energy and momentum states, and the probability of finding the particle in each of these states is determined by the eigenstate of the system. However, the particle does not have a definite energy or momentum, as it is in a superposition of different states.

In conclusion, the eigenstates of the infinite square well are not energy or momentum eigenstates because they do not have a definite energy or momentum, but rather are a superposition of different states. This concept is a fundamental aspect of quantum mechanics and is necessary to understand the behavior of particles at the microscopic level.