# What are raising & lowering operators

1. May 17, 2013

### Roodles01

A little behind in this subject, but I understand raising & lowering operators to almost be factors of the Hamiltonian operator.

raising - Ahat dagger = 1/√2 (x/a - a*(δ/δx))
lowering - Ahat = 1/√2 (x/a + a*(δ/δx))

I also have the Hamiltonian as;
Hhat = (Ahat dagger * Ahat + 1/2) hbar ω0

Is this an identity defining the Hamiltonian?

Help!
A little pointer in the right direction or an explanation to provide a bit of clarity would be great.
Thank you.

2. May 17, 2013

### CompuChip

I prefer to look at the definition of the ladder operators in terms of commutator relations, and define the Hamiltonian as usual - e.g. by quantizing the classical harmonic oscillator Hamiltonian. You can then derive an expression ladder operators in terms of position and momentum operators, and either rewrite the Hamiltonian to show that it is all consistent.

Let me try to elaborate.

The idea of defining raising and lowering operators is quite general, see e.g. this Wikipedia page. Basically it says that for certain operators it is possible to find operators which raise or lower the quantum number. For example, if you have energy eigenstates, then H|n> = En|n> and you will be able to find operators that take |n> to either |n - 1> or |n + 1>. Another example is in quantum field theory, where N is the number of particles with eigenstates N|n> = n|n>, and you can find operators to go to state |n - 1> or |n + 1> (basically creating or destroying particles, this is why they are often called creation and destruction operators). They are even nice enough to give a|0> = 0 - i.e. destroying the ground state gives zero.

So if you want to define these operators for the energy eigenstates $E_n = \hbar \omega (n + \tfrac12)$ of the harmonic oscillator, you are imposing - as per the Wiki page I linked to - that
$$(*) \qquad [H, a] = -\hbar \omega a$$
and therefore, since H is Hermitian,
$$(**) \qquad [H, a^\dagger] = (-\hbar \omega a)^\dagger = \hbar \omega a^\dagger$$
(Now I'm not entirely sure, but I think these relations actually fix a and therefore a-dagger already. But that's not really relevant now.)

Of course, since the operators really raise and lower the state, you should get $a^\dagger a \left|n\right\rangle \propto \left|n\right\rangle$ and so in particular
$$H a^\dagger a \left|n\right\rangle \propto H \left|n\right\rangle = E_n \left|n\right\rangle.$$
Indeed, playing around with the commutation relations ("pulling" H "through" the a and a-dagger so you can apply it to |n>) you will find exactly that.

Now let's look at the Hamiltonian. We have said we are looking at the harmonic oscillator, but we haven't written it in any particular form yet. Of course we cannot change the operator itself, but we can change the way it looks when we write it down. The most common way is as
$$H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac12 m \omega^2 x^2$$
Now there's a few things you can do. One is you can "factor" this. This is slightly tricky, because you're working with derivatives, and d/dx and x don't commute (e.g. look at [d/dx, x] f(x) = d(x f(x))/dx - x df/dx = ... = f(x) \neq 0). However if you are careful you will see that you can write the equation as
$$H = \tfrac12 \hbar \omega \left[ \left( -\frac{d}{dx} + x \right) \left( \frac{d}{dx} + x \right) \right]$$
so you can define operators
$$b^\dagger = -\frac{d}{dx} + x, b = \frac{d}{dx} + x$$
and check that $b^\dagger$ is actually the conjugate of b. Now if you compute [H, b] you will find that you get almost exactly the right commutation relation - you're off by a constant which you can sweep into the definition and if you do this correctly you will get two operators with exactly the commutation relations (*)-(**).

Another way to go about this is do the canonical replacement $p = -i\hbar \frac{d}{dx}$ (first quantization) so that H becomes
$$H = \frac{1}{2m} p^2 + \frac{1}{2} m \omega^2 x^2$$
(note I should write some hats here, but I'm sloppy so I'll leave them off. In fact I've already done that earlier).
Now the factorization is easier though you still have to keep track of the commutation relations $[x, p] = -i\hbar$ (I think, it's getting late). Alternatively, you can now say: "OK, so I have this Hamiltonian in terms of x and p, and I want to find ladder operators. Since the only non-trivial commutations are between x and p, I can just write $a = b x + c p$ and compute $[H, a] = b [H, x] + c [H, p] \propto - b p + c x$. This should be $-\hbar \omega a \propto b x + c p$ by equation (*) so you can determine constants b and c from here. Once you have done this, you rewrite the Hamiltonian in terms of the operators, as you know what $a^\dagger a \left|n\right\rangle$ yields as well as $H \left|n\right\rangle$.

It's really tired and I'm sure I got some of the signs and i's wrong, but I hope it is sufficiently clear.

Last edited: May 17, 2013