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Having trouble understanding something here, hoping someone can help...when dealing with a SHO, we can define two ladder operators a and a-dagger. The way I understand it is, applying a-dagger to an eigenstate of H (and that has, for instance, eigenvalue E) will give us a new eigenstate that has eigenvalue E+hw (h=h-bar). Similarly, applying a would have given eigenstate with eigenvalue E-hw, right?

The problem is, if you apply a and then a-dagger one after the other on some eigenstate, there should be no effect (one lowers the energy, one raises it), right?... so:

a a-dagger|phi> = |phi>

and surely:

a-dagger a|phi> = |phi>

so shouldn't the commutator between a and a-dagger be zero? (but I know that it is actually 1, so my reasoning is wrong, but why?)

Thanks

James

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# Ladder Operators for a SHO

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