ladder operators in quantum

[physica]

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

 

[swansont]

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.

[studiot]

The substitution swansont mentions involves change of variable.

 

What does your Hamiltonian look like initially?

Edited September 10, 2014 by studiot

[physica]

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.

[elfmotat]

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, 2014 by elfmotat

[physica]

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.