Limit to SHO / EM field analogy

In summary, the commutation relations between the position and momentum operators in the ordinary quantum harmonic oscillator and the quantization of the electromagnetic field are the same, but the delta's between them are different.
  • #1
snoopies622
846
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For a simple harmonic oscillator, the creation and annihilation operators can be expressed as linear combinations of the position and momentum operators,

[tex]
\hat {a} = \sqrt { \frac {m \omega} {2 \hbar} } ( \hat {x} + \frac {i \hat {p} } { m \omega } )
[/tex]

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



[/tex]

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 [itex] \hat{x} [/itex] and [itex] \hat{p} [/itex] operators, and if these operators correspond to anything physical in this case as well.
 
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  • #2
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
Thanks, Dextercioby. How then does one arrive at the [itex] \hat {a} [/itex] and [itex] \hat {a} ^{\dagger} [/itex] operators for the EM field? What is their mathematical definition?
 
  • #4
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
dextercioby said:
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)?
 
  • #6
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
Hmmm...
 
  • #8
Would other definition would you expect ? Let's see: what's the definition of [itex] \hat{x} [/itex] operator for a point particle in one dimension and described by non-specially relativistic quantum mechanics ?
 
  • #9
In the position basis, multiply by x. No?
 
  • #10
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
Interesting. So then instead of [itex]x[/itex] and [tex] -i \hbar \frac {\partial} {\partial x} [/tex], it is equivalent to define the position and momentum operators simply by saying that [itex] [ \hat {x} , \hat {p}_x ] = i \hbar [/itex]? Does this really contain the same amount of information?
 
Last edited:
  • #12
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
And so one can therefore derive the
[itex]\hat {x}= x[/itex] and [tex] \hat {p} = -i \hbar \frac {\partial} {\partial x} [/tex]
representations based solely on the [itex] [ \hat {x} , \hat {p}_x ] = i \hbar [/itex] 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
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
All good Dextercioby, thank you. One last question if I may: The commutation relations between [itex] \hat {a}[/itex] and [itex] \hat {a}^{\dagger} [/itex] 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
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
Got it. Thanks again.
 

Related to Limit to SHO / EM field analogy

1. What is the SHO / EM field analogy?

The SHO / EM field analogy is a comparison between the behavior of a simple harmonic oscillator (SHO) and an electromagnetic (EM) field. It suggests that the equations governing the motion of an SHO and the propagation of EM waves are similar.

2. What are the main similarities between SHO and EM fields?

The main similarities between SHO and EM fields include the concept of oscillation, the presence of a restoring force, and the ability to store and transfer energy. Both also exhibit periodic behavior and can be described by differential equations.

3. How is the SHO / EM field analogy useful in physics?

The SHO / EM field analogy is useful in physics because it allows us to apply our understanding of SHO to better understand EM fields. This can help in solving problems and making predictions about the behavior of EM waves.

4. Are there any limitations to the SHO / EM field analogy?

Yes, there are limitations to the SHO / EM field analogy. For example, it does not fully capture the complexity of EM fields, which can have multiple components and interact with each other in ways that an SHO does not. It also does not account for the effects of magnetic fields on EM waves.

5. Can the SHO / EM field analogy be extended to other systems?

Yes, the SHO / EM field analogy can be extended to other systems, such as mechanical systems with multiple degrees of freedom or quantum systems. However, the level of similarity between these systems and EM fields may vary and may not always be applicable.

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