Limit to SHO / EM field analogy

[snoopies622]

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

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

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

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.

 

[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).

 

[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?

 

[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.

 

[snoopies622]

Their definition follows from the commutation relations they must obey...

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)?

 

[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".

 

[snoopies622]

Hmmm...

 

[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 ?

 

[snoopies622]

In the position basis, multiply by x. No?

 

[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.

 

[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?

 

[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).

 

[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.)

 

[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.

 

[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.

 

[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.

 

[snoopies622]

Got it. Thanks again.