# Heisenberg equation of motion for fermions?

1. May 14, 2008

### pellman

In the Heisenberg picture, we move the time dependence away from the states and incorporate them in the operators. That is, if we write the time dependent state in the Schrodinger picture as $$|\Psi(t)\rangle=e^{-iHt}|\Psi\rangle$$, then an expectation value for an operator Q at time t, which we would write in the Schrodinger picture as

$$\langle\Psi(t)|Q|\Psi(t)\rangle=(\langle\Psi|e^{+iHt})Q(e^{-iHt}|\Psi\rangle)$$

we think of it in the Heisenberg picture as

$$\langle\Psi|Q(t)|\Psi\rangle=\langle\Psi|(e^{+iHt}Qe^{-iHt})|\Psi\rangle$$

Same answer of course. But instead of having a state which satisfies the Schrodinger equation

$$H|\Psi(t)\rangle=i\hbar\frac{\partial}{\partial t}|\Psi(t)\rangle$$

we have an operator obeying the Heisenberg equation of motion

$$[H,Q]=-i\hbar \frac{\partial Q}{\partial t}$$

at least this is the case for bosons. (I write all this preface material just in case my understanding thus far is mistaken. Now the question.)

But what about fermions? Do operators related to fermions obey a Heisenberg equation with anti-commutators instead of commutators?

2. May 14, 2008

### malawi_glenn

When one derives HEQ (heisenberg equation of motion), one imposes (almost) nothing on the state $$|\Psi(t)\rangle$$.

It is an arbitrary state, arbitrary operator Q and hamiltonian H.

The only things you impose is that the hamiltonian commute with the time evolution operator and that the operator Q does not depend on time in Schrödinger picture.

The operator Q, can example be the Spin operator $S$ for spin-½ particles.

3. May 16, 2008

### pellman

Thanks, malawi_glenn

4. Oct 27, 2010

### pellman

For the free scalar field, one can express the Hamiltonian in terms of the field operator $$\phi(x,t)$$. Then given the commutator

$$[\phi(x,t),\dot{\phi}^\dag (x',t)]=im\delta^3(x-x')$$

one can directly show that

$$[H,\phi(x,t)]=-i\frac{\partial\phi}{\partial t}$$

Similarly, we can express the Hamiltonian for the Dirac field in terms of the field operators $$\psi(x,t)$$. But in this case we have the anti-commutation rule

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

So do I use this to find the anti-commutator with the Hamiltonian? (The result should be equivalent to the Dirac equation, I should think.) Or if the time evolution of a fermionic operator is governed by the commutator with the Hamiltonian just like with bosons, how can one calculate it since we are only given the anti-commutation rules?

5. Oct 27, 2010

### xepma

The Heisenberg equation of motion for any operator (fermionic or bosonic) is given by:

$$[H,\mathcal{O}(x,t)]=-i\frac{\partial\mathcal{O}}{\partial t}$$

This relation is nothing but the differential version of the transformation rule:

$$\mathcal{O}(t) = e^{iHt} \mathcal{O}(t=0)e^{-iHt}$$

This unitary transformation is defined to be the time evolution of the operator. Not that when you take the derivative on both side you obtain the Heisenberg equation of motion:

$$\frac{d}{dt} \mathcal{O}(t) = iH e^{iHt} \mathcal{O}(t=0)e^{-iHt} + e^{iHt} \mathcal{O}(t=0)e^{-iHt} (-iH) = iH \mathcal{O}(t) + \mathcal{O}(t) (-iH) = i(H \mathcal{O}(t) - \mathcal{O}(t)H) = i[H, \mathcal{O}(t)]$$

Fermionic / bosonic: it all stays the same.

To actually evaluate this commutator is ofcourse a different matter (and not always possible in fact). For a free theory you can do it -- for an interacting theory you can only do it perturbatively. This goes for both bosons and fermions. Best thing to do is to write the whole thing in terms of Fourier modes and move on from there.

6. Oct 27, 2010

### pellman

Thanks, xepma.

This is rather interesting. Because using the anti-commutator above and expression for the Hamiltonian in terms of $$\psi(x,t)$$, I find

$$\{H,\psi(x,t)\}=2i\frac{\partial\psi}{\partial t}$$

The similarity with the Heisenberg equation of motion makes one wonder. I'm fairly confident that I didn't make a mistake and that the factor of 2 is correct. Is there any significance to this relation?

7. Oct 28, 2010

### xepma

I take it you used an explicit expression for the Hamiltonian? Because what you wrote down will not be true in general (i.e. more general Hamiltonians).

The anti-commutation relations of the fermionic fields arises due to the statistics of the particle (which, in turn, is related to the spin of the particle through the spin-statistics theorem). However, it has nothing to do with the time evolution properties of the field.

8. Oct 28, 2010

### pellman

Yes.

Where I am coming from is this: If we start with a Lagrangian, the Euler-Lagrange equation yields the differential equation defining the evolution of the field. From that Lagrangian we can define a Hamiltonian. From that Hamiltonian, the Heisenberg equation of motion also determines the evolution of the field. So the results of the two should be identical.

For the Klein-Gordon case, the result $$[H,\phi]=-i\dot{\phi}$$ is something of an identity and does not imply the Klein-Gordon equation. This is because $$\dot{\phi}$$ itself appears in the commutator.

For the Dirac case, we should have $$[H,\psi]=-i\dot{\psi}$$ equivalent to the Dirac equation. I was trying to show it explicitly, but I didn't have a commutation rule to apply. Hence my question.

Thank you for the info.

Last edited: Oct 28, 2010