# B Help with Dirac’s QM book

1. Feb 23, 2017

### exmarine

In Dirac’s QM book, Revised Fourth Edition (~1967), Chapter XI Relativistic Theory of the Electron, page 262, he reaches the conclusion that “…a measurement of a component of the velocity of a free electron is certain to lead to the result ±c.”

Is this still the current thinking? (He is talking about the instantaneous velocity.)

He then goes on to “verify” that by using the uncertainty principle later on the same page. “The great accuracy with which the position of the electron is known (my edit: in order to measure instantaneous velocity) during the time interval must give rise, according to the principle of uncertainty, to an almost complete indeterminacy in its momentum. This means that almost all values of the momentum are equally probable, so that the momentum is almost certain to be infinite. An infinite value for a component of momentum corresponds to the value ±c for the corresponding component of velocity.” (my italics)

Equally probable values of momentum mean it is almost certain to be infinite? I don’t understand that. Any help appreciated.

Well I see it didn't hold my italics, but you can still understand my question.

2. Feb 23, 2017

### stevendaryl

Staff Emeritus
It's sort of true, and sort of meaningless. In the Heisenberg picture, you can define the time-dependence of a dynamic variable via:

$\frac{dV}{dt} = i [H, V]$

If you let $V=\vec{x}$, the position operator, then you find for the Dirac equation:

$\frac{d\vec{x}}{dt} = \vec{\alpha} c$

where $\vec{\alpha}$ are the 3 spatial Dirac matrices.

It's easy enough to see that for each $\alpha^j$, the eigenvalues are $\pm 1$, so leading to the conclusion that the eigenvalues for each velocity component are $\pm c$.

People don't really take that too seriously, these days. Relativistic particles don't have well-defined positions, so velocity is not really an observable. There are alternative definitions of the velocity operator that give more reasonable values, but I don't remember what they are.

3. Feb 23, 2017

### Jilang

Instaneous values are going to to different to expectation values measured over some period of time.

4. Feb 23, 2017

### Staff: Mentor

Its part of the re-normalization programme that bedeviled Dirac. He is heuristically correct and that is why you get infinity in calculations.

We now understand re-normalization much better thanks to the Nobel prize winning work of Wilson (that's right he got a Nobel for sorting it out) with the realization physics depends on your energy scale. If you push a theory like EM developed from our everyday energy scale too far you end up with absurdities like the Landau pole:
https://en.wikipedia.org/wiki/Landau_pole

But not to worry - Wilson to the rescue - long before then the electroweak theory takes over. So what you do is cut off interactions about that scale, express all your parameters in terms of measurable quantities and low and behold the cutoff disappears. In fact in re-normalizeable theories it doesn't matter what cutoff you chose, even ones that make no sense in light of knowing at what energy other theories take over, in makes no difference. That makes them special.

But we have other theories like gravity where that is not the case - one must choose an actual cutoff which is taken about the Plank scale:
https://arxiv.org/abs/1209.3511

Best to read more modern texts before delving into Dirac. I didn't and paid the price of taking longer to understand things.

Thanks
Bill

5. Feb 24, 2017

### stevendaryl

Staff Emeritus
Hmm. I hadn't heard a connection between the infinities of QFT and the fact that the velocity operator in Dirac's equations has eigenvalues $\pm c$.

6. Feb 24, 2017

### Demystifier

In nonrelativistic QM the velocity operator is essentially a derivative operator
$${\bf v}=\frac{-i\hbar{\bf \nabla}}{m}$$
Relativistic velocity operator must reduce to this in the non-relativistic limit, so it cannot be $c{\bf\alpha}$. Indeed, for relativistic Klein-Gordon equation a natural velocity operator is also the derivative operator above. Using the fact that Dirac equation implies also Klein-Gordon equation, it appears natural to define velocity operator as the derivative operator even in the Dirac case.

7. Feb 24, 2017

### stevendaryl

Staff Emeritus
Not really. For one thing, relativistically, $v = \frac{pc^2}{E} = \frac{p}{\sqrt{m^2 + \frac{p^2}{c^2}}}$, not $\frac{p}{m}$.

8. Feb 24, 2017

### Demystifier

By "velocity" I meant the spatial part $v^i$ of the 4-velocity $v^{\mu}$. Note that (in classical relativistic mechanics)
$$v^i=\frac{dx^i}{d\tau}\neq \frac{dx^i}{dx^0}$$