# Hamiltonian in terms of ladders, question.

1. Jun 21, 2008

### ballzac

I'm preparing for an exam at the moment and in one of the past exams the is a question asking to prove that the hamiltonian operator can be expressed in terms of the ladder operators.

The solution is this

(The minus sign didn't come out in the last line, and obviously there is one more step that I left of the end, but you get the picture.)

Following this from the second to the third line seems to imply

But I would expect

because the derivative of x wrt x is 1. My maths isn't terribly good, and there is obviously something simple that I am missing, but I've been staring at this for days and can't seem to get it. Could someone please explain what I'm missing?

I could answer this question if it comes up on the exam, because I can remember how it goes, but I want to actually be able to understand it.

2. Jun 21, 2008

### Fredrik

Staff Emeritus
These are operators, acting on functions. If you want to prove an operator identity A=B, you can do it by proving that Af=Bf for all f.

In this case, writing D instead of d/dx because I'm lazy...

Dx should be interpreted as the operator that takes f to D(xf). So

(Dx)(f)=D(xf)=f+xDf=(1+xD)(f)

This implies that

Dx=(1+xD)

3. Jun 21, 2008

### ballzac

Thank you so much. I've been agonising over that for days, and you've just made it so simple. :)

4. Jun 21, 2008

### ismaili

BTW, an interesting issue by writing the Hamiltonian in terms of creation and annihilation operators is due to "cluster decomposition principle."

By counting the number of adjustable parameters, we can show that all operators can be written as a sum of products of creation and annihilation operators. Those creation/annihilation operators would satisfy commutation or anticommutation relation, $$\[a(q'),a^\dagger(q)] = \delta(q'-q)$$ or $$\{a(q'),a^\dagger(q')\}=\delta(q'-q)$$ according to the field being bosonic or fermionic.

Then you write down the scattering matrix element, and using the (anti-)commutation relations to move all the annihilation operators to the right, you would generate lots of delta functions. Now, assigning some graphical rule to the scattering matrix element, e.g. the delta function represented by a line, the interaction represented as a vertex...etc. You will see that the action that moving all the annihilation operators to the right decomposes the scattering matrix element into sum of connected pieces of diagrams.

For each connected scattering matrix element, we can argue from the basic topological theorem that the number of vertices and lines would satisfy certain relation. From this we could prove that the connected scattering matrix element can only contain exactly one spatial momentum delta function! which is required by cluster decomposition principle, i.e. distant experiments yield uncorrelated results.

(I'm reading Weinberg's QFT book, my status is at chap 4. The Cluster Decomposition Principle is addressed in chapter four, it's interesting.
Actually, I have a little study group here, anybody who interested in reading Weinberg's book is welcome to discuss with me.)
(BTW, Everybody is welcome to correct my concepts~)

5. Jun 21, 2008

### ballzac

Is there an 'over my head' emoticon?

6. Jun 21, 2008

### Fredrik

Staff Emeritus
Don't worry about it. He's talking about quantum field theories where operators very similar to yours show up as creation and annihilation operators. They take n-particle states to (n+1)- or (n-1)-particle states, unless an annihilation operator acts on the vaccum (the 0-particle state) in which case the result is zero. Anyway, you were clearly talking about solving the Schrödinger equation with a harmonic oscillator potential, so you can forget about this for a couple of years.