Limit to SHO / EM field analogy

1. Oct 20, 2012

snoopies622

For a simple harmonic oscillator, the creation and annihilation operators can be expressed as linear combinations of the position and momentum operators,

$$\hat {a} = \sqrt { \frac {m \omega} {2 \hbar} } ( \hat {x} + \frac {i \hat {p} } { m \omega } )$$

$$\hat {a} ^{\dagger} = \sqrt { \frac {m \omega} {2 \hbar} } ( \hat {x} - \frac {i \hat {p} } { m \omega } )$$

where the position and momentum refer, of course, to the position and momentum of the oscillating particle.

Since creation and annihilation operators also appear when quantizing the electromagnetic field, I was wondering if they can be similarly broken down into $\hat{x}$ and $\hat{p}$ operators, and if these operators correspond to anything physical in this case as well.

2. Oct 20, 2012

dextercioby

In the quantum theory of the electromagnetic field, the x's become merely scalar parameters and p's you get in the parameter spaces if you do a Fourier transformation of the x's space.
The a and a^dagger are related to the fields themselves, which are both operators on the Fock space and distributions over the Schwartz space (typically).

3. Oct 20, 2012

snoopies622

Thanks, Dextercioby. How then does one arrive at the $\hat {a}$ and $\hat {a} ^{\dagger}$ operators for the EM field? What is their mathematical definition?

4. Oct 21, 2012

dextercioby

Their definition follows from the commutation relations they must obey (and which come from the quantization of the classical Dirac brackets, if you use a reduced phase space which results from fixing the gauges) and they are then implemented as linear operators on the Fock space.

5. Oct 21, 2012

snoopies622

I must admit I've never been clear about this. They have no strict mathematical definitions beyond their commutation relations (as the position and momentum operators do in ordinary quantum mechanics)?

6. Oct 21, 2012

dextercioby

The strict mathematical definition would be: they are a set of operator valued distributions acting on the Fock space and subject to the commutation relations typical for the so-called "canonical quantization".

7. Oct 21, 2012

snoopies622

Hmmm...

8. Oct 21, 2012

dextercioby

Would other definition would you expect ? Let's see: what's the definition of $\hat{x}$ operator for a point particle in one dimension and described by non-specially relativistic quantum mechanics ?

9. Oct 21, 2012

snoopies622

In the position basis, multiply by x. No?

10. Oct 21, 2012

dextercioby

Yes, but to get to position basis, you need a statement about the general operator, irrespective of basis (or of abstract space's realizations) and a lot of math. That's what I gave you above about the a and a^dagger.

11. Oct 21, 2012

snoopies622

Interesting. So then instead of $x$ and $$-i \hbar \frac {\partial} {\partial x}$$, it is equivalent to define the position and momentum operators simply by saying that $[ \hat {x} , \hat {p}_x ] = i \hbar$? Does this really contain the same amount of information?

Last edited: Oct 21, 2012
12. Oct 22, 2012

dextercioby

Yes, by the Stone-von Neumann theorem, up to unitary equivalence the Schrödinger realization of x and p is the unique representation of Weyl unitaries (1927) of the [xˆ,pˆx]=iℏ (originally due to Born and Jordan, 1925).

13. Oct 22, 2012

snoopies622

And so one can therefore derive the
$\hat {x}= x$ and $$\hat {p} = -i \hbar \frac {\partial} {\partial x}$$
representations based solely on the $[ \hat {x} , \hat {p}_x ] = i \hbar$ relation?

(I'm wondering if something like this can be done with the creation and annihilation operators — starting with their commutation relations and arriving at a form similar to the individual ones for the position and momentum operators above.)

14. Oct 22, 2012

dextercioby

Yes. The original derivation by Schrödinger (1926) was heuristic (guess), the rigorous results by Stone and von Neumann proved it was the right one.

15. Oct 23, 2012

snoopies622

All good Dextercioby, thank you. One last question if I may: The commutation relations between $\hat {a}$ and $\hat {a}^{\dagger}$ are exactly the same in both cases, are they not? That is, whether one is talking about the ordinary quantum harmonic oscillator or the quantization of the electromagnetic field.

16. Oct 23, 2012

dextercioby

Not exactly, they differ through the delta's. For LHO there's a delta Kronecker, while for the e-m fields, the delta is a delta Dirac.

17. Oct 23, 2012

snoopies622

Got it. Thanks again.