# I Dirac wrong Hamiltonian

1. Jul 6, 2017

### Silviu

Hello! I read that if we apply the exactly same procedure for Dirac theory as we did for Klein Gordon, in quantizing the field, we obtain this hamiltonian: $H=\int{\frac{d^3p}{(2\pi)^3}\sum(E_pa_p^{s\dagger}a_p^s-E_pb_p^{s\dagger}b_p^s)}$ and this is wrong as by applying the creation operator $b^\dagger$ you can lower the energy indefinitely. I am not sure I understand why, as at the same time, $a^\dagger$ should rise the energy by the same amount (in the end this hamiltonian wouldn't change the energy at all). So can someone explain to me how this is working? Thank you!

2. Jul 6, 2017

### ChrisVer

The thing is that this Hamiltonian is not bounded from below...You can have configurations in which you cannot be bounded from below: you can keep producing particles that are described by the $v$ spinor without wanting to create particles that are described by the $u$.
An example, let's say you have initiallly some state $|0>$... you add a $b_{sp}^\dagger$ to get another state: $b_{sp}^\dagger |0>$... you keep doing that thing and you end up with something that looks like this:
$\Big( b_{sp}^\dagger\Big)^n|0>$...
The |0> state had higher energy than the b|0> and the b|0> had higher energy than the bb|0> and so on... increasing the $n$ you keep going to lower energies.
It should be obvious that the KG procedure wouldn't work since it involves commutators instead of anticommutators, and lacks the Pauli exclusion principle (you can as I wrote have several particles in the same state without a problem).

3. Jul 6, 2017

### Silviu

Thank you for your answer! I understand physically why it doesn't work and I get your argument. But mathematically, when you apply H to $|0>$, aren't both $a^\dagger$ and $b^\dagger$ going to act on it? Like they appear in H, so how can you suppress one of them?

4. Jul 7, 2017

### Dr.AbeNikIanEdL

If you apply $H$ to $|0\rangle$ you get $0$ as it should be. The point is what happens when you take

$H \left(b^\dagger_{\mathrm{s}\mathrm{p}}\right)^n |0\rangle ~.$

The term with the $a$'s in $H$ still gives 0, but the second terms with the $b$'s will give something like $-nE_\mathrm{p} \left(b^\dagger_{\mathrm{s}\mathrm{p}}\right)^n |0\rangle$, so by applying enough $b^\dagger_{\mathrm{s}\mathrm{p}}$ to $|0\rangle$ you can reach an eigenstate of $H$ with arbitrary low eigenvalue.

5. Jul 8, 2017

### vanhees71

You must consider that Dirac particles are fermions. That's why there's an additional minus sign in the 2nd term when doing the normal ordering, and that makes $\hat{H}$ bounded from below as it must be. It's an example of the spin-statistics theorem: half-integer fields are necessarily to be quantized as fermions, i.e., with anti-commutator relations.