Is it possible to know the exact values of momentum and velocity of a particle simultaneously?

Question

I know that by Heisenberg's Uncertainty Principle that it is not possible to know the exact values of position and momentum of a particle simultaneously, but can we know the exact values of momentum and velocity of a particle simultaneously? I would think the answer would be no because even if we were 100% certain of the particle's position, we would be completely unsure of the particle's momentum, thus making us also completely unsure of the particle's velocity. Does anyone have any insight into this?

Answer 1

It is quite common to discuss the two extremes of the uncertainty principle, sinusoid and delta function. One has a perfectly defined wavelength but no position, the other has a perfectly defined position but no wavelength.

However, neither one of these shapes is terribly physical for a particle's position wavefunction. A true sinusoidal wavefunction would extend through all space, which is absurd for several reasons (including the presence of other matter). A true delta function would be equally likely to have any momentum, which would probably violate conservation of energy. So, these two extreme limits are mathematically interesting, but not physically relevant.

Given the question "Does the uncertainty principle put some bound on momentum and velocity being simultaneously well-defined?", the answer is no.

Given the question "Does the uncertainty principle forbid me from measuring any single variable with infinite precision?", the answer is no.

Given the question "Does anything forbid me from measuring with infinite precision?", the answer is yes.

So, your question mentions 'exact values', which is a very interesting, thorny subject. (Is it ever possible to measure an exact value? How would we tell the difference?) Are you really curious about 'exact values'? Are you more curious about where the Heisenberg uncertainty principle does and does not apply? Or are you curious if there are other bounds on our ability to measure, in addition to the uncertainty principle?

Answer 2

If in your theory the momentum operator and velocity operator are proportional to each other, then yes. Knowing one's eigenvalue means knowing the other's. It is always the case with any function of a "known" operator.

Answer 3

The velocity eigenvalues of the Dirac equation are $\pm c$. This is well known since the equation was found; see Dirac's book, "The Principles of Quantum Mechanics, 4th Ed.,", Oxford University Press, Oxford 1958, Chapter XI "Relativistic Theory of the Electron", Section 69, "The motion of a free electron", page 262. It used to be a commonly taught fact of quantum mechanics, but I understand the down votes, it's now possible to get a PhD in physics without knowing the slightest thing about the following quite elementary calculation. Partly since this isn't taught much anymore, the derivation has been reappearing recently in the literature, for example see: Eur.Phys.J.C50:673-678,2007 Chiral oscillations in terms of the zitterbewegung effect / hep-th/0701091, around equation (11).

We begin by noting that velocity is the time rate of change of position, and that you can define the time rate of change of position by using the commutator:
$$\hat{v}_x = \dot{x} = -(i/\hbar)[\hat{x},H]$$
If the above appears to be magic to you, read the wikipedia entry on Ehrenfest's theorem which states the principle and gives the identical situation for non-relativistic quantum mechanics: $$\frac{d}{dt}\langle x\rangle = -(i/\hbar)\langle [\hat{x},H]\rangle = \langle p_x\rangle /m$$ and so $\;m v_x = m\dot{x} = p_x$ (for the non-relativistic case). Thus, for the non-relativistic electron model, it is possible to simultaneously measure velocity and momentum; their proportionality constant is the mass. But with relativity the proportionality does not happen so the situation is different.

For a state to be an eigenstate of velocity requires that:
$$\hat{v}_x\;\psi(x) = -(i/\hbar)[\hat{x},H]\;\psi(x) = \lambda\psi(x)$$
Dirac defined the free-particle Hamiltonian as $H=c\vec{\alpha}\cdot \vec{p} + \beta mc^2$. In modern notation, $\beta=\gamma^0$ and $\alpha^k = \gamma^0\gamma^k$, while $p$ is the usual momentum operator.

Note that the only thing that doesn't commute with $\hat{x}$ is the x-component of the momentum operator, which gives $[\hat{x},\hat{p}_x]=i\hbar$. Thus the above reduces to:
$$-(i/\hbar)[\hat{x},c\gamma^0\gamma^1p_x]\psi(x) = \lambda\psi(x) $$ $$-(ic/\hbar)\gamma^0\gamma^1[\hat{x},p_x]\psi(x) = \lambda\psi(x) $$ $$-(ic/\hbar)(i\hbar)\gamma^0\gamma^1\psi(x) = \lambda\psi(x) $$ $$c\gamma^0\gamma^1\psi(x) = \lambda\psi(x) $$

Using the wikipedia's choice of gamma matrix representation, we have: $$c\gamma^0\gamma^1 = c\left(\begin{array}{cccc} 1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{array}\right) \left(\begin{array}{cccc} 0&0&0&1\\0&0&1&0\\0&-1&0&0\\-1&0&0&0\end{array}\right) =c\left(\begin{array}{cccc} 0&0&0&1\\0&0&1&0\\0&1&0&0\\1&0&0&0\end{array}\right)$$ The eigenvalues are obtained by solving the characteristic polynomial. That is, compute the matrix determinant and set it to zero: $$\left[\begin{array}{cccc} -\lambda&0&0&c\\0&-\lambda&c&0\\0&c&-\lambda&0\\c&0&0&-\lambda\end{array}\right] = \lambda^4-2\lambda^2c^2 + c^4=0$$ I leave it as an exercise for the reader to show that there are two real roots, $\pm c$ each with order two.

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The four solutions to the velocity eigenvalue problem for the Dirac equation correspond to the the right and left handed electron and positron. That is, the velocity eigenstates of the Dirac equation are precisely the left and right-handed states used to represent fermions in the standard model.