Elementary QFT question, part 3

  • #1
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Main Question or Discussion Point

Suppose I have some sinusoidal electromagnetic waves, plane-shaped, moving in a vacuum with frequency f and wavelength [itex] \lambda [/itex]. I want to consider the energy content of a certain volume of space through which they are passing. Let's say it is a cube with side lengths [itex] \Delta x [/itex] where [itex] \Delta x << \lambda [/itex] .

According to classical physics, I can find the energy inside the cube by integrating the sum of the square of the electric field strength and the square of the magnetic field strength over the volume. Let's call the result of that computation [itex] E_1 [/itex] .

According to elementary quantum mechanics, the energy of these waves is being carried by photons, each with an energy content of hf. Let's define [itex] E_2 [/itex] to be that quantity, hf.

I'm wondering what will happen if I measure the energy quantity in my cube and [itex] E_1/E_2 [/itex] is not an integer.

When I look into the subject of the quantization of the electromagnetic field, one thing I keep finding mentioned is the quantum harmonic oscillator (QHO), and the idea that an EM field is like a big collection of QHO's.

I know that for any particular frequency f, a QHO can have only the energies

hf(n+1/2), where n=0,1,2...

So an oscillator corresponding to the frequency of my light waves can have several different energy values, but none of which is necessarily [itex] E_1 [/itex] .

I also know that in quantum mechanics a system at any given moment is represented by a vector, and this vector does not have to be an eigenvector of an energy operator, but can instead be a linear combination of them, and so for a given system and state one could have an energy expectation value of any real quantity in between two energy eigenvalues.

Therefore, there may be a QHO corresponding to frequency f - the frequency of my light waves - and a vector for it that has an energy expectation value of [itex] E_1 [/itex], as long as the coefficients are chosen correctly. (By coefficients I mean the coefficients of the energy eigenvectors such that their linear combination makes the desired state vector here.)

So - if my thinking isn't way off - that leaves only the question: how does one find the coefficients? I assume that since the EM waves are already sinusoidal, one can't use Fourier analysis to break them down any further, since in that sense each sine wave is itself already a basis vector.
 
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Answers and Replies

  • #2
Suppose I have some sinusoidal electromagnetic waves, plane-shaped, moving in a vacuum with frequency f and wavelength [itex] \lambda [/itex].
Here, you are describing classical electromagnetic waves with definite frequency (and polarization). In quantum mechanics, this is usually modeled using coherent states. Provided your wave is perfectly monochromatic, I propose that your state vector of the system is

[tex]|\text{EM}\rangle = \mathcal{N} e^{\alpha\, \hat{a}^\dag_{k,\epsilon}}|0\rangle\,.[/tex]
Here, [itex]N[/itex] is the normalization of your state, [itex]\alpha[/itex] controls the amplitude and phase of your electromagnetic wave, and [itex]\hat{a}^\dag_{k,\lambda}[/itex] creates a photon with wavenumber k and polarization [itex]\epsilon[/itex]. You may check that this gives the desired classical plane-wave by evaluating the expectation value [itex]\langle\text{EM}|\hat{A}_\mu(x)|\text{EM}\rangle[/itex].
I'm wondering what will happen if I measure the energy quantity in my cube and [itex] E_1/E_2 [/itex] is not an integer.
The way to compute [itex]E_1[/itex] in the quantum realm is to compute the expectation value of the Hamiltonian in the state [itex]|\text{EM}\rangle[/itex]. On the other hand, the possible values of [itex]E_2[/itex] come from the decomposition of the state-vector into energy eigenstates: these will be the Fock states, which are state of definite photon-number.
[tex]|\text{EM}\rangle=\mathcal{N}\sum_{n=0}^\infty\frac{\alpha^n}{\sqrt{n!}}|n\rangle_{k,\epsilon}=\mathcal{N}\left(|0\rangle + \alpha |1\rangle + \frac{\alpha^2}{\sqrt{2}}|2\rangle+\ldots \right)\,,[/tex]​
where I have simply written the coherent state as a Taylor series, and used [itex]|n\rangle=\frac{1}{\sqrt{n!}}(\hat{a}^\dag)^n|0\rangle[/itex]. Each state of definite photon number is also a state of definite energy, given by [itex]E_2=\hbar\omega_k n[/itex], where [itex]\omega\sim k[/itex] is the angular frequency of your monochromatic electromagnetic wave. In general, you should not expect the ratio [itex]E_1/E_2[/itex] to be an integer.
So - if my thinking isn't way off - that leaves only the question: how does one find the coefficients? I assume that since the EM waves are already sinusoidal, one can't use Fourier analysis to break them down any further, since in that sense each sine wave is itself already a basis vector.
Your question is answered above: the coefficients are governed by the Taylor expansion of the exponential as part of the definition of the coherent state (with the normalization included). The QM decomposition is not in terms of Fourier components, as you have noticed; rather, it is in terms of Fock states (states of definite photon number).

Important lesson: Classical electromagnetic waves are states of indefinite photon number.
 
  • #3
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Wow, thanks.
 
  • #4
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A follow-up, related to coherent states, if I may:

Is there a kind of measurement that corresponds to the creation and annihilation operators, in the way that the momentum operator corresponds to the act of measuring momentum, the position operator corresponds to the act of measuring position, etc. ? I know that they have to do with discrete energy changes in physical systems, but they are not the same thing as the energy operators themselves.

To put it another way, is there an act that one performs to a system to make it become an eigenstate of either of these operators? or does it turn into one for other reasons?
 
  • #5
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A follow-up, related to coherent states, if I may:

Is there a kind of measurement that corresponds to the creation and annihilation operators, in the way that the momentum operator corresponds to the act of measuring momentum, the position operator corresponds to the act of measuring position, etc. ? I know that they have to do with discrete energy changes in physical systems, but they are not the same thing as the energy operators themselves.

To put it another way, is there an act that one performs to a system to make it become an eigenstate of either of these operators? or does it turn into one for other reasons?
Coherent states of light (eigenstates of a corresponding annihilation operator) are usually produced in lasers using stimulated emission by electrons. But one cannot call this a measurement.

An experiment that produces a particular kind of state is called a _preparation_, not a measurement.
 
  • #6
As a reminder, observables are associated with Hermitian operators (this is debatable). Since the ladder operators are not Hermitian, one cannot associate a physical observable to it.

The momentum operator, on the other hand, is Hermitian.
 

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