I How to find common eigenbases of momentum and energy in infinite well?

[LightPhoton]

Consider an infinite well between $0$ and $a$, the energy eigen functions are:
$$\phi(x)=\sqrt{\frac 2 a}\sin{\frac{n\pi x}a}$$

Since the Hamiltonian of this system is only a function of momentum operator $(\hat H=\hat p^2/2m)$, we should be able to find a common energy and momentum eigenbasis, since $[\hat H,\hat p]=0$.

But applying $\hat p$ to $\phi(x)$ we get:
$$\hat p \phi(x)=-i\hbar \partial_x \phi(x)=-i\hbar \frac{n \pi}a \sqrt{\frac 2 a}\cos{\frac{n\pi x}a}\neq k\phi(x)$$

Where $k$ is some constant.

So how do we find a common eigenbasis?

 

[PeterDonis]

the Hamiltonian of this system is only a function of momentum operator

No, it isn't. There is a potential energy term in the Hamiltonian that does not depend on the momentum operator, it depends on position.

 

[LightPhoton]

No, it isn't. There is a potential energy term in the Hamiltonian that does not depend on the momentum operator, it depends on position.

Thanks, makes much more sense now

 

[pines-demon]

Another way of seeing that momentum cannot be a good operator here is that the infinite-well is not translationally invariant, thus momentum is not a conserved quantity.

 

[LightPhoton]

Another way of seeing that momentum cannot be a good operator here is that the infinite-well is not translationally invariant, thus momentum is not a conserved quantity.

Can you expand on how does a "good operator" not being conserved quantity is an issue?

 

[pines-demon]

Can you expand on how does a "good operator" not being conserved quantity is an issue?

Do you know Noether's theorem? It also applies to quantum mechanics. Symmetries are related to conserved quantities (that is operators that commute with the Hamiltonian)

 

[LightPhoton]

Do you know Noether's theorem? It also applies to quantum mechanics. Symmetries are related to conserved quantities (that is operators that commute with the Hamiltonian)

Yeah, I get that, but I am not able to make the connection between conserved quantities and simultaneous eigen basis. Your comment seems to imply that only conserved quantities can share same eigen bases.

 

[PeroK]

Yeah, I get that, but I am not able to make the connection between conserved quantities and simultaneous eigen basis. Your comment seems to imply that only conserved quantities can share same eigen bases.

The momentum operator for the infinite square well is not well behaved. See this discussion:

https://www.physicsforums.com/threads/momentum-eigenfunctions-in-an-infinite-well.1051061/

 

[pines-demon]

Yeah, I get that, but I am not able to make the connection between conserved quantities and simultaneous eigen basis. Your comment seems to imply that only conserved quantities can share same eigen bases.

Well Noether's theorem in Hamiltonian formalism is roughly what you wrote
$$\frac{\mathrm d Q}{\mathrm d t}=\{Q,H\}+\frac{\partial Q}{\partial t}$$
where $Q$ is the quantity to check, $\{\cdots\}$ is the Poisson brackets, which in quantum mechanics reads
$$\frac{\mathrm d \hat{Q}}{\mathrm d t}=\frac{1}{i\hbar}[\hat{Q},\hat{H}]+\frac{\partial \hat{Q}}{\partial t}$$
where now we have commutators and operators.

How to relate $\hat{Q}$ to the symmetry? Well the operation $e^{\frac{i}{\hbar}\hat{Q} r}$ is the action on the wavefunction ($r$ is the conjugate variable to $\hat{Q}$). For $\hat{p}$ it leads to the translation operator $e^{\frac{i}{\hbar}Px_0 }\psi(x)=\psi(x+x_0)$.

 

[PeterDonis]

Your comment seems to imply that only conserved quantities can share same eigen bases.

I don't think that's generally true, but I think what @pines-demon might have been getting at is that two operators have to commute in order to have a set of simultaneous eigenstates, and, from the equations he gave in post #9, one can see that momentum will only commute with the Hamiltonian if it is a conserved quantity, i.e., if its total time derivative as shown in the Noether's theorem equation is zero.