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- Thread starter Silviu
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ChrisVer

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An example, let's say you have initiallly some state [itex]|0>[/itex]... you add a [itex]b_{sp}^\dagger[/itex] to get another state: [itex]b_{sp}^\dagger |0>[/itex]... you keep doing that thing and you end up with something that looks like this:

[itex] \Big( b_{sp}^\dagger\Big)^n|0>[/itex]...

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 [itex]n[/itex] 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

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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?

An example, let's say you have initiallly some state [itex]|0>[/itex]... you add a [itex]b_{sp}^\dagger[/itex] to get another state: [itex]b_{sp}^\dagger |0>[/itex]... you keep doing that thing and you end up with something that looks like this:

[itex] \Big( b_{sp}^\dagger\Big)^n|0>[/itex]...

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 [itex]n[/itex] 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).

- #4

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##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.

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