Dear all

 

I’m studying quantum mechanics for my third year. I’m finding ladder operators a real struggle. Attached is a solution to an exercise. I understand that these operators cannot be swapped round, as they will have different results. However, I’m finding the solution hard to follow. I don’t see how or why the Hamiltonian operator comes into the right hand side and how or why it turns into a energy function. The other thing that’s confusing me is why the lowering operator is pulled out at the start of the solution then slotted back in. I would ask my lecturer but my year starts in October. I’ve nearly finished 2 textbooks back to back, would like the majority of my last year taken up on revision and project work. Many thanks

 

The Hamiltonian acting on the wave function gives you the energy, by definition. [math]H\Psi = E\Psi[/math]

 

It shows up because you can define H in terms of a^† and a. It's just a substitution.

 

For me (and this was years ago) it helped to know that the ladder operators were used in terms of photon creation and destruction; if you go up a level, it costs you a photon, and down a level generates one.

The substitution swansont mentions involves change of variable.

 

What does your Hamiltonian look like initially?

Edited September 10, 201411 yr by studiot

thanks swansont, the currency view really helped. As for the the initial Hamiltonian i've sent a picture via private message to studiot. For some reason after the first initial post I can never submit pictures.

Remember that [math]\widehat{H} = (\widehat{A}^\dagger \widehat{A} +\frac{1}{2}) \hbar \omega_0 [/math].

 

What they're doing is rearranging the equation so that [math]\widehat{H}[/math] acts on the wavefunction before [math]\widehat{A}[/math] does. The reason they do this is so that you can replace the Hamiltonian with its eigenvalue energy, [math]E[/math]. To do this they used the commutator relation and factored out a lowering operator:

[math]\widehat{H} [\widehat{A} \psi ]= (\widehat{A}^\dagger \widehat{A} +\frac{1}{2}) \hbar \omega_0 [\widehat{A} \psi ]= \hbar \omega_0( \widehat{A} \widehat{A}^\dagger -1+ \frac{1}{2}) [\widehat{A} \psi][/math]

 

[math]=\hbar \omega_0(\widehat{A} \widehat{A}^\dagger \widehat{A}-\widehat{A}+\frac{1}{2} \widehat{A})\psi = \hbar \omega_0 \widehat{A}( \widehat{A}^\dagger \widehat{A}+ \frac{1}{2} -1) \psi =\widehat{A}(\widehat{H}-\hbar \omega_0)\psi [/math]

 

Now, remember that the Hamiltonian acting on a state is its eigenvalue, which is the energy of the state: [math]\widehat{H}\psi =E \psi [/math]. So we have:

 

[math]\widehat{H} [\widehat{A} \psi ]=\widehat{A}(E-\hbar \omega_0)\psi =(E- \hbar \omega_0) [\widehat{A} \psi ][/math]

 

So we see that the lowering operator has the effect of lowering the energy by an amount [math]\hbar \omega_0[/math].

Edited September 10, 201411 yr by elfmotat

Amazing thank you. Had to give a 1+ as it clarified what was going on. God I love this forum, pity some people don't appreciate it.