6
$\begingroup$

Suppose we have a system of bosons represented by their occupation numbers $$\tag{1} | n_1, n_2, ..., n_\alpha, ... \rangle$$ Then we can define creation and annihilation operators $$\tag{2} a_\alpha^\dagger| n_1, n_2, ..., n_\alpha, ... \rangle = \sqrt{n_\alpha+1} | n_1, n_2, ..., n_\alpha+1, ... \rangle$$ $$\tag{3} a_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle = \sqrt{n_\alpha} | n_1, n_2, ..., n_\alpha-1, ... \rangle$$ This is nice because the number operator is just $a_\alpha^\dagger a_\alpha$. However, would it be sensible to define an alternate set of operators to work with? $$\tag{4} b_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle = | n_1, n_2, ..., n_\alpha+1, ... \rangle$$ $$\tag{5} c_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle = \begin{cases} | n_1, n_2, ..., n_\alpha-1, ... \rangle & n_\alpha>0 \\ 0 & n_\alpha=0 \end{cases}$$ $$\tag{6} N_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle = n_\alpha| n_1, n_2, ..., n_\alpha, ... \rangle $$ Why don't we work with these operators? The bosonic creation and annihilation operators $a_\alpha^\dagger$ and $a_\alpha$ were defined to mimic the harmonic oscillator's raising and lowering operators ($x \pm i p$), but is there any compelling reason to keep the $\sqrt{n_\alpha+1}$ and $\sqrt{n_\alpha}$ factors?

I suppose $a_\alpha^\dagger$ and $a_\alpha$ obey nice properties such as $[a_\alpha,a_\alpha^\dagger]=1$ and the fact that they are Hermitian adjoints of each other. What are the analogous relationships that $b_\alpha$ and $c_\alpha$ would obey?

$\endgroup$
  • 1
    $\begingroup$ For yet another take on these ideas see physics.stackexchange.com/a/90078/26076 where I show that your (2) and (3) arise naturally from the idea of finding the most general quantum system whose measurement statistics vary sinusoidally with time. $\endgroup$ – WetSavannaAnimal Dec 13 '13 at 8:39
7
$\begingroup$

Here are three properties that would make your definitions awkward.

You can think of $a^\dagger\,a$ as the LU (lower triangular, upper triangular or Cholesky) decomposition of the number observable. Actually, it's not the unique Cholesky factorisation but it is the one found by the outer product version of the algorithm. Your definition would not have this property.

As a result, the Lie bracket between your two $b,\,c$ operators would be a bit nasty. In energy eigen co-ordinates (i.e. so that a quantum harmonic oscillators state is given by an infinite column vector of probability amplitudes to be in each of the number states) it would be:

$$[b,\,c]={\rm diag}[1,\,0,\,0,\,\cdots]$$

and the Hamiltonian would be complicateder:

$$\hat{H} = \hbar\,\omega\left(c\,b\,N + \frac{1}{2}\right) = \hbar\,\omega\left(N\,c\,b + \frac{1}{2}\right)$$

and there's no simple way to write the Hamiltonian in terms of $b\,c ={\rm diag}[0,\,1,\,1,\,1,\,\cdots]$.

All this would make rewriting operator and observable expressions in normal ordering very awkward indeed.

Lastly, there is a rather elegant way of writing down a general coherent state of the quantum harmonic oscillator, through the so-called displacement operator:

$$\left|\left.\alpha\right>\right. = \exp\left(\alpha\,a^\dagger - \alpha^* \, a\right)\left|\left.0\right>\right.$$

which shifts the quantum ground state to a coherent one with amplitude (i.e. displacement from the origin in $x-p$ phase space) $\alpha$. This highly useful formula would be much awkwarder in your notation.

$\endgroup$
  • $\begingroup$ An excellent answer, but why do we care to have an LU decomposition of $N$? It's diagonal in the occupation number representation anyway. And $[b,c]=-|0 \rangle \langle 0|$ is just a projection operator onto the ground state, so it's not so nasty. Also, I think the Hamiltonian would just be $H=\hbar \omega (N+1/2)$ (we don't need the $cb$ thing in it), which is not complicated at all. $\endgroup$ – ChickenGod Nov 30 '13 at 6:19
  • $\begingroup$ However, you are absolutely right in that $a$ and $a^\dagger$ are central to the construction and theory of a coherent state, which would be awkward with $b$ and $c$. $\endgroup$ – ChickenGod Nov 30 '13 at 6:24
  • 1
    $\begingroup$ @ChickenGod The LU decomposition's consquences are very useful for normal ordering. Sure, the Lie bracket is the projection operator as you say, but this makes normal ordering formulas a bit awkward. Try re-ordering something like $a^n {a^\dagger}^m$: you can do it but it sure makes life a great deal easier if you can swap things around by subtracting them from the identity (as when $[a, a^\dagger] = 1$). The displacement operator may or may not be useful to you: in quantum optics it is cannot be done without. None of this is, of course, set in stone: notations never are: .... $\endgroup$ – WetSavannaAnimal Nov 30 '13 at 6:29
  • $\begingroup$ @ChickenGod .... but these are a few reasons I can think of why the present notation is handy. The Hamiltonian formula likewise: often you can simply use the rnumber operator as you do, but there are also times where the decomposition is useful. $\endgroup$ – WetSavannaAnimal Nov 30 '13 at 6:31
  • $\begingroup$ Ah yes, I forgot about normal ordering. In that case, I can see why $a a^\dagger = 1 + a^\dagger a$ is superior to $b c = 1 - | 0 \rangle \langle 0 |$. $\endgroup$ – ChickenGod Dec 1 '13 at 4:35
3
$\begingroup$

The problem is that the physical states have positive occupation numbers $n_1, n_2,...$.

With your operators, you have, for instance :

$ c_\alpha| n_1, n_2, ..., 0, ... \rangle = | n_1, n_2, ..., -1, ... \rangle$.

This gives you a totally unphysical state, so you would have to add by-hand constraints like $n_1 \geq 0, n_2 \geq 0,...$, .

With the operators $a_{\alpha}$, you do not have this problem, the no-existence of states with negative occupation numbers is completely natural, because of the term $\sqrt{n_\alpha}$

$\endgroup$
  • $\begingroup$ +1 Most excellent point. I was actually "adding the constraints in by hand" in my own answer without even realizing I was doing it. Indeed I've just realised that I've always subconsciously been doing it with problems of this kind: I'm ashamed to say that I'd never even really noticed what you say before: the $\sqrt{n}$ shuts the "downstepping" of the annihilation operator down automatically at $n=0$. $\endgroup$ – WetSavannaAnimal Nov 30 '13 at 1:03
  • $\begingroup$ Right. I actually wanted to add the extra constraint that $c_\alpha$ acting on an $n_\alpha = 0$ state gives $0$, but I forgot to type that in. Thanks though! $\endgroup$ – ChickenGod Nov 30 '13 at 5:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.