Foldy-Wouthusien velocity operator

[noospace]

If one takes the derivative of the position operator in the Dirac Hamiltonian, the result is $\dot{\vec{x}} = c \vec{\alpha}$. This, however, disagrees with the classical limit in which $\dot{\vec{x}}\sim \dot{\vec{p}}/m$.

I'm trying to show that the time derivative of the position operator $\vec{X} = U^\dag \vec{x} U$ is $\frac{c^2\vec{p}}{E^2}H$, and thus satisfies the non-relativistic approximation.

$U = \frac{\beta \vec{\alpha}\cdot \vec{p} c + mc^2 + E}{\sqrt{2E(mc^2+E)}}$ is the FW transformation (Relativistic quantum mechanics by Greiner p. 277 or Advanced Quantum Mechanics by Sakurai p. 176) .

I tried using the Heisenberg equation of motion

$\dot{\vec{X}} = \frac{1}{i\hbar}[X,H]$

but the expression for X is long and complicated so I wasn't able to get to the result using this method.

I note that $\frac{1}{i\hbar}[\vec{x},E] = (\vec{p}c^2)/E$ which looks a bit like the answer.

I'm also aware that $U^\dag H U = \beta E$. So I want to show that

$\frac{1}{i\hbar}[U^\dag \vec{x} U, H ] = \frac{1}{i\hbar}[\vec{x}, \beta U^\dag H U ]$ (since $\beta^2=1$) but I can't figure out how to show this.

Any help would be greatly appreciated!

 

[olgranpappy]

If one takes the derivative of the position operator in the Dirac Hamiltonian, the result is $\dot{\vec{x}} = c \vec{\alpha}$. This, however, disagrees with the classical limit in which $\dot{\vec{x}}\sim \dot{\vec{p}}/m$.

you mean
$$\frac{p}{m}$$
not
$$\frac{\dot p}{m}$$

I'm trying to show that the time derivative of the position operator $\vec{X} = U^\dag \vec{x} U$ is $\frac{c^2\vec{p}}{E^2}H$, and thus satisfies the non-relativistic approximation.

$U = \frac{\beta \vec{\alpha}\cdot \vec{p} c + mc^2 + E}{\sqrt{2E(mc^2+E)}}$ is the FW transformation (Relativistic quantum mechanics by Greiner p. 277 or Advanced Quantum Mechanics by Sakurai p. 176) .

I tried using the Heisenberg equation of motion

$\dot{\vec{X}} = \frac{1}{i\hbar}[X,H]$

but the expression for X is long and complicated so I wasn't able to get to the result using this method.

no, one finds (with H being the dirac hamiltonian)
$$\dot \vec x = -i[\vec x,H]/\hbar=-i[\vec x, \vec c\alpha\cdot\vec p]/\hbar=c\vec \alpha$$

 

[StatusX]

This also addresses your other thread (where my comment wasn't correct), but I thought it was more relevant here.

You can always just change the basis of your Hilbert space by applying an operator U to all states and operators (ie, an operator A becomes UAU^t, while a state $\Psi$ becomes U$\Psi$). The same operator should be applied at all times, whether you're in the Schrödinger or Heisenberg picture. But then all expectation values are unchanged, and, eg, for the Foldy-Wouthuysen transformation, you're not going to get a better behaved position operator.

What's done in the FW transformation is a unitary transformation is found that puts the Hamiltonian:

$$H = \beta m + \alpha \cdot p$$

into the form:

$$H' =UHU^{-1} = \beta \sqrt{m^2 + p^2}$$

Note this is written in terms of the old variables. Of course, if we wrote it in terms of p', we'd get something of the form of the first equation above, since all U does is affect a change of basis. Similarly, you'll trivially find that $\dot x = i [H,x] = \alpha$ is equal to $\dot x'$, if we define this as $i[H',x']$.

However, what we really want is a new position operator. So, we stay in the old basis, but define:

$$X = U^{-1}xU$$

(note this is not the same as x') This will turn out to be a sort of mean position operator. Then we have:

$$\dot X = i [H,X]$$

This is not equal to $\alpha$, but is simply related to the more easily computed $i[H',x]$, which is just the same equation written in terms of the alternate basis (and explains why we use X and not x'), and which you've already found.

Alternatively, you could work in the new basis, and just borrow the old position operator x, which will now describe a mean position operator. This is the way FW do it in their original paper, but it's slightly confusing since you're using the same position operator you started with to describe something completely different.

 

[noospace]

Thanks for replying.

How do you define $x',p'$ etc.? You also say that $i[H,X]$ is simply related to $i[H',x]$ which is indeed easy to compute. In fact

$i[H',x] = \frac{c^2\vec{p}}{E}$

I'm afraid I don't say what the simply relationship is? Could you please expand upon that?

 

[StatusX]

All operators in the new basis are related to the old ones by $A'=UAU^{-1}$ In particular:

$$i[H,X] = i[U^{-1}H'U,U^{-1}xU] = U^{-1} i[H',x] U$$

This isn't that hard to compute directly in terms of p, giving something pretty simple, but not quite what you're meant to show. However, if you define $P=U^{-1}pU$, a "mean momentum operator", you get:

$$\dot X = i[H,X] = \frac{c^2 P}{E(P)}$$

where $E(P)=\sqrt{P^2+m^2}$. In the non-relativistic limit it is these mean operators that we interpret as the position and momentum operators. If you have access to it, the http://prola.aps.org/abstract/PR/v78/i1/p29_1" is very readable.

 

[Hans de Vries]

If one takes the derivative of the position operator in the Dirac Hamiltonian, the result is $\dot{\vec{x}} = c \vec{\alpha}$. This, however, disagrees with the classical limit in which $\dot{\vec{x}}\sim \dot{\vec{p}}/m$.

Actually this is correct, The Dirac equation produces the right velocity operator.It is a matrix operator and it must operate on the four spinor in the correct way.
For instance if $-i\hbar\partial/\partial t$ is the energy operator then we must apply it like:

$$\frac{1}{2E}\left( \begin{array}{c} \psi_L^* \\ \psi_R^* \end{array} \right)\ \left(-i\hbar\ \mbox{\Large I}\ \frac{\partial}{\partial t}\ \right)\ \left( \begin{array}{c} \psi_L \\ \psi_R \end{array} \right)\ \ =\ \ E$$

Where 1/2E is a normalization constant. If one now uses the velocity operator
instead then we get:

$$\frac{1}{2E}\left( \begin{array}{c} \psi_L^* \\ \psi_R^* \end{array} \right)\ c \ \vec{\alpha}\ \left( \begin{array}{c} \psi_L \\ \psi_R \end{array} \right)\ \ =\ \ v$$This is most easily shown in the 2d Chiral Dirac equation where we have:

$$\frac{1}{2E}\left( \begin{array}{c} \psi_L^* \\ \psi_R^* \end{array} \right)\ \left( \begin{array}{cc} -c & 0 \\ \ \ 0 & c \ \ \end{array} \right)\ \left( \begin{array}{c} \psi_L \\ \psi_R \end{array} \right)\ \ =\ \ \ v$$Which for a plane wave solution comes down to:

$$\frac{1}{2E}\left( \begin{array}{c} \sqrt{E-cp} \\ \sqrt{E+cp} \end{array} \right)\ \left( \begin{array}{cc} -c & 0 \\ \ \ 0 & c \ \ \end{array} \right)\ \left( \begin{array}{c} \sqrt{E-cp} \\ \sqrt{E+cp} \end{array} \right)\ \ =\ \ \ \frac{2pc^2}{2E}\ \ =\ \ \frac{2\ \gamma mv\ c^2}{2\ \gamma mc^2}\ \ =\ \ v$$
See my book on the Dirac equation here, section 11.4
http://chip-architect.com/physics/Book_Chapter_Dirac.pdf"

The velocity operator for the full Chiral Dirac equation is:

$$c \left( \begin{array}{cc} -\sigma^i & 0 \\ \ \ 0 & \sigma^i \ \ \end{array} \right)$$Regards, Hans