Constant particle motion due to zero-point energy....

[asimov42]

Hi all,

Just a clarification to ask about: if a have an electron (all by its lonesome) in its ground state, it will have non-zero kinetic energy (zero-point energy), even at absolute zero. This should mean the particle (oscillating field excitation in QFT) is always moving.

Now, to be clear, I may measure the momentum of the particle to be zero several times in a row, statistically speaking (to the best accuracy of my measurement device). The energy is the expectation of the Hamiltonian - so despite those measured values, the particle has a fixed zero-point kinetic energy and is always in motion.

Sorry, this is probably obvious - I'm basically asking about relationship between energy as the expectation, and the fact that you may randomly measure the particle momentum to be zero at certain times despite the fact it has non-zero energy always.

 

[asimov42]

Related question (purely hypothetical): let's say you could cool our electron down to absolute zero, and you ignored the zero-point energy. Then, the particle would sit motionless at the bottom of its potential well. In this case it's 3-momentum would be exactly zero in all frames of reference - but it's 4-momemtum would continue to be nonzero and it would have it's standard world line, just with the same spatial coordinates always. Correct? (again purely hypothetical non-quantum question here).

 

[PeterDonis]

if a have an electron (all by its lonesome) in its ground state

If you have a free electron all by its lonesome, the concept of "ground state" isn't well-defined. The electron's energy and momentum are frame-dependent, and there is no unique "ground state" of lowest energy.

let's say you could cool our electron down to absolute zero, and you ignored the zero-point energy. Then, the particle would sit motionless at the bottom of its potential well.

If the electron is free, there is no "potential well". The potential is zero everywhere (more precisely, the concept of "potential" has no meaning since there is nothing else for the electron to interact with, and "potential" requires some kind of interaction).

In this case it's 3-momentum would be exactly zero in all frames of reference

No, it wouldn't. There is no such state possible. 3-momentum is frame-dependent.

You need to rethink your entire scenario in the light of the above; as it stands it doesn't make sense.

 

[asimov42]

If you have a free electron all by its lonesome, the concept of "ground state" isn't well-defined. The electron's energy and momentum are frame-dependent, and there is no unique "ground state" of lowest energy.

If the electron is free, there is no "potential well". The potential is zero everywhere (more precisely, the concept of "potential" has no meaning since there is nothing else for the electron to interact with, and "potential" requires some kind of interaction).

Thanks @PeterDonis - yep, I realized afterwards that my questions didn't make sense as posed.

So here's a slightly better question (to which I believe I already know the answer, but to check). Let's take QFT and the following scenario:

I have a lonesome electron in an otherwise empty universe (empty of particles, but still filled with quantum fields of course). There is no well-defined potential for this electron - however, the Heisenberg uncertainty principle must still apply to it, correct? Even though there is nothing for the electron to interact with other than the vacuum.

 

[PeterDonis]

There is no well-defined potential for this electron

No well-defined potential? Or a well-defined potential of zero everywhere?

(The latter can be analyzed using QFT. The former can't.)

the Heisenberg uncertainty principle must still apply to it, correct?

Sure; there won't be any possible states of the electron, or more precisely the electron field, that have precise values for non-commuting observables.

However, you have to be careful to properly specify what kind of state of the electron field you are talking about. When you say "I have one electron", that doesn't mean there's a little billiard ball bouncing around in a vacuum filled with quantum fields. In fact the exact meaning of this statement is ambiguous: it could mean either of the following:

(1) The electron field is in an eigenstate of the electron number operator with eigenvalue 1.

(2) The electron field is in some state (such as a coherent state) whose expectation value for electron number is 1.

Choice #1 specifies a particular quantum field state and so it makes definite predictions for all possible measurement probabilities. Choice #2 does not specify a particular quantum field state; there are an infinite number of possible quantum field states that have an expectation value of 1 for electron number, and they will make different predictions for measurement probabilities.