# Heisenberg equation of motion for the Dirac field?

1. Jun 12, 2014

### pellman

I would expect that the Heisenberg equation of motion for the Dirac field would yield the Dirac equation. Indeed, these lecture notes claim it as a fact in eq 7.7 but without proof. My trouble is that I know the anti-commutation rules for the Dirac field but I don't know how to calculate the commutator with Hamiltonian, even if the Hamiltonian is expressed in terms of the field itself.

Can someone here provide some hints or tips?

Here is an old thread I started years ago on the same question.

2. Jun 13, 2014

### king vitamin

The trick is to decompose the commutator in terms of anti-commutators. Since the Hamiltonian for the Dirac field is quadratic in the fermion fields, you use identities such as

$$[A,BC] = \{A,B\}C - B\{C,A\}$$

Since you know these anti-commutators, the rest of the calculation is straight-forward.

3. Jun 18, 2014

### pellman

Using the lagrangian (in natural units)

$$\mathcal{L}=\bar{\psi}(i\gamma^\mu \partial_\mu -m)\psi$$

where $\psi(x)$ is the 4-spinor field and $\bar{\psi}={\psi}^\dagger \gamma^0$ we get the Hamiltonian

$$H=\int{\bar{\psi}(x')(-i\vec{\gamma}\cdot\nabla + m)\psi(x')d^3x'}$$

where $\vec{\gamma}$ are three space gamma matrices. The commutation rule is

$$\{\psi_j(\vec{x},t),\psi^{\dagger}_k (\vec{x}',t)\}=\delta_{jk}\delta^3(\vec{x}-\vec{x}')$$

Using $[H,\psi_j]=2H\psi_j-\{H,\psi_j\}$ we get for the Heisenberg equation of motion

$$-i\partial_t \psi_j=[H,\psi_j]=2H\psi_j-\{H,\psi_j\}$$

$$\{H,\psi_j\}=-i{\gamma^0}_{jk}\vec{\gamma}_{kl}\cdot\nabla\psi_l + m{\gamma^0}_{jk}\psi_k$$

Multiplying by $\gamma^0$ and rearranging, we get

$$2\gamma^0 H\psi+(i\gamma^\mu \partial_\mu -m)\psi=0$$

So for the Heisenberg equation of motion to be equivalent to the Dirac equation would require that $\gamma^0 H\psi =0$. But I don't know how to show that is the case. Or maybe I went wrong somewhere?

4. Jun 18, 2014

### Avodyne

Your evaluation of {H,ψ} is not correct. If you write it out, and keep track of the ordering of the ψ's, you will find that you don't get the anticommutator of ψ and ψ, but rather the commutator (and this doesn't simplify). You need to follow the previous suggestion to get this to work.

5. Jun 18, 2014

### stevendaryl

Staff Emeritus
I tried looking in "Peskin & Schroeder" and they skirt the issue. In the case of boson fields, they prove that $[H, \phi] = i \frac{d}{dt} \phi$, but then they don't prove that for fermion fields.

On the one hand, it's got to be true, since they claim that $\psi(\vec{x}, t) = e^{-iHt/\hbar} \psi(x,0) e^{+i Ht/\hbar}$, but I'm with you, I don't see how it reproduces the Dirac equation.

6. Jun 18, 2014

### pellman

I don't think so. You can see my steps in this image. http://s23.postimg.org/c8bzm0yej/dirac_commutator.jpg

7. Jun 18, 2014

### stevendaryl

Staff Emeritus
Okay, we're trying to compute the commutator:

$[H, \psi(x)]$

$H$ has the form $\int d^3y\ \psi^\dagger(y)_\alpha \ O_{\alpha\beta}\ \psi(y)_\beta$

where $O$ involves gamma matrices, and the gradient and the mass, and so forth. Let's just look at $[H, \psi(x)_\tau]$, one component. For simplicity, I'm going to leave off the integral sign, because it's easy enough to restore them later. So we have

$\psi^\dagger(y)_\alpha \ O_{\alpha\beta}\ \psi(y)_\beta\ \psi(x)_\tau - \psi(x)_\tau\ \psi^\dagger(y)_\alpha \ O_{\alpha\beta}\ \psi(y)_\beta$

We can write $\psi(x)_\tau\ \psi^\dagger(y)_\alpha = \delta_{\tau \alpha} \delta^3(x-y) - \psi^\dagger(y)_\alpha\ \psi(x)_\tau$. So substituting this into the above expression gives:

$\psi^\dagger(y)_\alpha \ O_{\alpha\beta}\ \psi(y)_\beta\ \psi(x)_\tau - \delta_{\tau \alpha} \delta^3(x-y)\ O_{\alpha\beta}\ \psi(y)_\beta + \psi^\dagger(y)_\alpha\ \psi(x)_\tau\ O_{\alpha\beta}\ \psi(y)_\beta$

Now, we can rewrite $\psi(x)_\tau\ O_{\alpha\beta} = O_{\alpha\beta}\ \psi(x)_\tau$.

The reason why is because $O_{\alpha \beta}$ is not a matrix, but a single cell of a matrix. It's a scalar whose only operator part is $\nabla$, but since $\nabla$ is a derivative with respect to $y$, not $x$, so it doesn't affect $\psi(x)$. So they commute. So rearranging the three terms gives:

$- \delta_{\tau \alpha} \delta^3(x-y)\ O_{\alpha\beta}\ \psi(y)_\beta + \psi^\dagger(y)_\alpha \ O_{\alpha\beta}\ \psi(y)_\beta\ \psi(x)_\tau + \psi^\dagger(y)_\alpha\ O_{\alpha\beta}\ \psi(x)_\tau\ \psi(y)_\beta$

The last two terms can be combined using an anti-commutator, giving:

$- \delta_{\tau \alpha} \delta^3(x-y)\ O_{\alpha\beta}\ \psi(y)_\beta + \psi^\dagger(y)_\alpha \ O_{\alpha\beta}\ \{ \psi(y)_\beta, \psi(x)_\tau \}$

Since the anti-commutator is zero, we just get

$- \delta_{\tau \alpha} \delta^3(x-y)\ O_{\alpha\beta}\ \psi(y)_\beta$

After restoring the integral, and summing over the matrix indices, we have:

$- O_{\tau \beta}\ \psi(x)_\beta = - (O\ \psi(x))_\tau$

I think the operator $O$ is just $-i \alpha \cdot \nabla + \beta m$. So I think it works out. Since there are two field operators in $H$, I think you get the same answer regardless of whether you use commutators or anti-commutators.

8. Jun 19, 2014

### stevendaryl

Staff Emeritus
I think I see the problem in your derivation: You write:

$m \psi_k^\dagger(x',t) \gamma^0_{kl} \psi_l(x',t) \psi_j(x, t)$
$= + m \psi_k^\dagger(x',t) \psi_j(x, t) \gamma^0_{kl} \psi_l(x',t)$

I think the right-hand side should be a minus sign, since the field operators anti-commute. Similarly for other terms.

9. Jun 19, 2014

### pellman

Right. Thanks! I will look over your other post closely and get back asap.

10. Jun 19, 2014

### pellman

That the field operators anti-commute was the key! It all fell right into place. Big thanks!