If <p> = 0 how can we know the energy of the system

[talabax]

So when doing the expectation value of <p> on a stationary state we get a 0 and I've seen the standard proof showing why as well as for <p^2> not being 0. But how can momentum be 0 and
yet have a energy defined for the system? Is the energy only due to Vψ = Eψ

 

[atyy]

The possible energies are not given by Vψ = Eψ. They are given by Hψ = Eψ, and H = p²/2m + V.

 

[talabax]

but if <p> is 0 how can we have a kinetic energy involved why even have it in the hamiltonian?

 

[atyy]

The Hamiltonian (together with the possible states of the system) is the fundamental description of the system. The Hamiltonian describes how any given state of the system evolves in time.

For any given state ψ, the average value of an observable A in that state is <ψ|A|ψ>.
If you are interested in the average momentum of the state then you choose A = p.
If you are interested in the average kinetic energy of the state then you choose A = p²/2m.
If you are interested in the average energy of the state then you choose A = H, because H is the energy operator, and also called the Hamiltonian.

So although some states have zero average momentum, it doesn't mean that all states have zero average momentum.

 

[Jilang]

If the particle spends half the time moving to the left and half the time moving to the right <p> will be zero. This doesn't mean that <|p|> is zero.

 

[Matterwave]

So when doing the expectation value of <p> on a stationary state we get a 0 and I've seen the standard proof showing why as well as for <p^2> not being 0. But how can momentum be 0 and
yet have a energy defined for the system? Is the energy only due to Vψ = Eψ

If you understand that <p^2> does not necessarily equal zero, then you should be able to understand that <KE> does not necessarily equal zero, since the two are related by a constant (1/2m).

 

[Demystifier]

So when doing the expectation value of <p> on a stationary state we get a 0 and I've seen the standard proof showing why as well as for <p^2> not being 0. But how can momentum be 0 and
yet have a energy defined for the system? Is the energy only due to Vψ = Eψ

Average kinetic energy is proportional to <p^2>, not to <p>^2.

In fact, you can understand it even with classical statistical physics. Suppose that you have a gas made of classical particles. The gas is in a box at rest. At any given moment of time, some particle move to the left inside the box, while other particles move to the right. In other words, some particles have positive momentum while other particles have negative momentum, so that the average momentum is zero. Nevertheless, each of these particles has a positive kinetic energy, so the average kinetic energy is positive too.