# Commutation relations in relativistic quantum theory

• noospace
In summary, the Hamiltonian H = \vec{\alpha} \cdot \vec{p} c + \beta mc^2 is a Dirac Hamiltonian which can be written as a 4x4 matrix. The original Dirac equation is analogous to the Schrödinger equation but uses the Dirac Hamiltonian instead. By using special matrices, the Dirac equation can be written in a simplified form. The Hamiltonian is also a shortcut for [A,B] \Psi = C \Psi, where the commutator tells us the difference between AB \Psi and BA \Psi and how big this difference is. However, there is a paradox in this theory where the velocity of a relativistic electron always equals the
noospace
Given the Hamiltonian $H = \vec{\alpha} \cdot \vec{p} c + \beta mc^2$,

How should one interpret the commutator $[\vec{x}, H]$ which is supposedly related to the velocity of the Dirac particle? $\vec{x}$ is a 3-vector whereas H is a vector so how do we commute them. Is some sort of tensor product in order? ie

$[\vec{x}, H] = x_i H_{jk} - H_{jk}x_i$?

The Hamiltonian is not a vector but an operator on a Hilbert space which we can write as a matrix.

Commutators like $$[\vec{x}, H] = ...$$ are shortcuts for $$[x_i, H] = ..., \quad i \in {1,2,3}$$.

Note that $$x_i$$ is also an operator on a Hilbert space. Thus a commutator $$[A,B] = C$$ is also a shortcut for $$[A,B] \Psi = C \Psi$$ so the commutator tells you if there's a difference between $$AB \Psi$$ and $$BA \Psi$$ and how big this difference is.

I've never seen any of this before. Interesting. I guess x=x*I, where x is an orbital operator, and I is the 4x4 identity. The beta emm cee ^2 should therefore commute with x=x*I so that does not contribute to the velocity dx/dt. The momentum that the alpha is multiplying does not commute with x, but is given by [x,p]=ih. So in the end i got (if I didn't mess up):

$$\frac{dx^i}{dt}=c\alpha^i$$

So that the velocity is the 4-4 matrix alpha.

This is really weird. Maybe the canonical/Heisenberg formulation has been stretched too far?

The Hamiltonian almost looks like a Legendre transformation with the Lagrangian equal to mc^2.

RedX said:
The Hamiltonian almost looks like a Legendre transformation with the Lagrangian equal to mc^2.

$H_D = \vec{\alpha} \cdot \vec{p} c + \beta mc^2$ is the Dirac Hamiltonian. The original Dirac equation looks like the Schrödinger equation but it contains the Dirac Hamiltonian instead of the Schrödinger Hamiltonian:

$H_D \Psi(\vec{x},t) = i \hbar \partial_t \Psi(\vec{x},t)$

Using a special choice for the $\alpha^i$ and $\beta$ matrices, the Dirac $\gamma$ matrices, you can write it as

$(i \gamma^{\mu} \partial_{\mu} - m) \Psi(\vec{x},t) = 0$

Things kinda get confusing sometimes, and I think it has to do with the fact that we are interpreting the Dirac equation as an equation for a particle, instead of an equation for a field. We're treating the Hamiltonian as 4x4. In fact, if you sandwich this 4x4 Hamiltonian between two spinors $$\psi^{\dagger} H \psi$$, then you get the 1x1 Hamiltonian density for a field. Of course to get the Hamiltonian for a field, you just integrate the Hamiltonian density, so the Hamiltonian is 1x1 for a field, and 4x4 for a particle. Or at least that's what I think is going on.

RedX said:
I've never seen any of this before. Interesting. [...]
So in the end i got (if I didn't mess up):

$$\frac{dx^i}{dt}=c\alpha^i$$

So that the velocity is the 4-4 matrix alpha.
This is really weird. Maybe the canonical/Heisenberg formulation has been stretched too far?
This is an old "puzzle". You might want to read up on the Foldy-Wouthuysen transformation
(but not from Wiki - it's F-W page is not very good). Try a good RQM textbook like that of
Paul Strange. Also take a look at the Newton-Wigner position operator in parallel.

$$\frac{dx^i}{dt}=c\alpha^i$$

This is really weird. Maybe the canonical/Heisenberg formulation has been stretched too far?

Indeed you are right! Since the eigenvalues of $\alpha_{i}$ are $\pm 1$, we are faced with a paradox. That is the absolute value of the velocity of a relativistic electron always equals the velocity of light! Also, since the $\alpha_{i}$ do not commute with each other, the components of the velocity would not be simultaneously measurable! Clearly this nonsense (in cotradiction with the Ehrenfest theorem) would not give the classical result for the mean values.

To have a "consistent" relativistic QM, i.e., relativistic one-electron theory, we must necessarily define all physical quantities by EVEN operators. That is

$$\hat{O} \rightarrow [\hat{O}] = \frac{1}{2} (\hat{O} + \Omega \hat{O} \Omega )$$

where the Hermitian and unitary sign operator is defined ( in the momentum representation) by

$$\Omega = \frac{\alpha_{i}p_{i} + m \beta}{E_{p}}$$

Note that $[H_{D}] = H_{D}$ and $[\hat{p}_{i}] = \hat{p}_{i}$, but

$$[\alpha_{i}] = \hat{p}_{i} \frac{\Omega}{E_{p}}$$

Thus

$$[\frac{d \hat{x}_{i}}{dt}] = \frac{\hat{p}_{i}}{E_{p}} \Omega$$

and the velocity is $p_{i}/E_{p}$ for positive and $-p_{i}/E_{p}$ for negative free particle solutions. Notice that the classical picture corresponds only to the positive solutions. One could carry on to bring about the connection with Zitterbewegung motion arround the classical trajectory by solving the EOM in the Hiesenberg representation.

regards

sam

Thanks. Awhile ago I was reading a relativistic QM book by Bjorken and Drell, but I quit after the first couple of chapters because I wanted to jump to QFT (and also the Bjorken and Drell book seemed really old). So I think I missed this chapter on Foldy-Wouthuysen and Zitterbewegung. I'll be sure to read it when I have some spare time though as it seems interesting.

## 1. What are commutation relations in relativistic quantum theory?

Commutation relations in relativistic quantum theory describe the relationship between different observables, such as position and momentum, in a relativistic quantum system. These relations are important for understanding the behavior and properties of particles at the quantum level.

## 2. Why are commutation relations important in relativistic quantum theory?

Commutation relations are important because they allow us to determine the uncertainties in the measurement of different observables. They also provide a framework for understanding the fundamental principles of quantum mechanics and how they apply in a relativistic context.

## 3. How are commutation relations derived in relativistic quantum theory?

Commutation relations are derived using the principles of quantum mechanics, which involve the use of operators to describe physical quantities and their relationships. In relativistic quantum theory, these operators are modified to account for the effects of special relativity.

## 4. Can commutation relations be applied to all particles in relativistic quantum theory?

Yes, commutation relations can be applied to all particles in relativistic quantum theory, including both fundamental and composite particles. However, the specific form of the relations may vary depending on the characteristics of the particles being studied.

## 5. How do commutation relations impact our understanding of the quantum world?

Commutation relations play a crucial role in our understanding of the quantum world as they provide a mathematical framework for describing the behavior and properties of particles at the subatomic level. They also help us make predictions and calculations about the behavior of particles in different situations and conditions.

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