Quantized Radiation Field: Where Am I Wrong?

[kof9595995]

For a quantized radiation field(radiation gauge),the vector potential takes the form:
$$A(x,t) = \sum\limits_{k,\alpha } {\sqrt {1/\omega } } [{\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over \varepsilon } ^\alpha }{a_{k,\alpha }}{e^{ - i(\omega t - kx)}} + {\mathord{\buildrel{\lower3pt\hbox{\scriptscriptstyle\frown}} \over \varepsilon } ^\alpha }a_{k,\alpha }^\dag {e^{i(\omega t - kx)}}]$$
up to some multiplicative constant which is not relevant to my question.
then it seems the expectation value of the field is
$$< {n_{k,\alpha }}|A|{n_{k,\alpha }} > = 0$$
However I thought this expectation value should give us a classical plane wave instead of 0, since the number state $$|{n_{k,\alpha }} >$$ represents a monochromatic wave with a definite momentum.
So where am I wrong?

 

[The_Duck]

I believe the quantum mechanical analogs of classical plane waves are http://en.wikipedia.org/wiki/Coherent_state" .

 

[kof9595995]

I see, then is it still reasonable, for example in the treatment of stimulated and spontaneous radiation, to assume the initial and final state of radiation are occupation number state, while in reality we usually excite an atom by a classical source(laser)?
By the way I'm now also confused by the physical difference between the sates |n_k1>+|n_k2> and|n_k1,n_k2> , can you explain it to me?

 

[A. Neumaier]

I see, then is it still reasonable, for example in the treatment of stimulated and spontaneous radiation, to assume the initial and final state of radiation are occupation number state, while in reality we usually excite an atom by a classical source(laser)?

In quantum optics one treats emission via coherent states, not via occupation number states.
The latter have a very nonclassical statistics.

I
By the way I'm now also confused by the physical difference between the states |n_k1>+|n_k2> and|n_k1,n_k2> , can you explain it to me?

the first is a state of a single particle in a superposition of different momenta (as produced by a beam splitter, say), the second a (tensor product) state of two particles with different momenta (as produced by parametric downconversion, say).

 

[kof9595995]

I
the first is a state of a single particle in a superposition of different momenta (as produced by a beam splitter, say), the second a (tensor product) state of two particles with different momenta (as produced by parametric downconversion, say).

What do you mean? Both of them are many-particle states(say n>1), aren't they?

 

[A. Neumaier]

What do you mean? Both of them are many-particle states(say n>1), aren't they?

No. Only the second (n=2); the first is a single-particle state.

 

[kof9595995]

I guess I have a huge misunderstanding on this issue: what exactly is a "single-paticle state" ? I thought the occupation number gives the number of particles, if not, what's the meaning of occupation number?

 

[A. Neumaier]

I guess I have a huge misunderstanding on this issue: what exactly is a "single-paticle state" ? I thought the occupation number gives the number of particles, if not, what's the meaning of occupation number?

Oh, according to what you had written, I thought that n is a unit vector in the direction of the momentum.

Ignoring polarization, a single photon is described by a superposition of states |p> of different momenta p, an n-photon state is a superposition of states |p_1,...,p_n> with one momentum vector p_k for each photon. if the setting is monochromatic with frequency omega then p_k=omega*n_k with unit vectors n_k. Thus there are no occupation numbers in this notation. (Occupation numbers make sense only for single-state systems.)

If you meant something different then please explain your notation in sufficient detail.

 

[kof9595995]

Ok I followed the notations from Sakurai, |n_k> (ignoring polariztion) denotes the eigenstate of a number operator N=(a*_k )a_k, where a* and a are ladder operators.
Anyway why can't we interpret |n_k1,n_k2> as there are n_k1 photons with wavevector k1 and n_k2 photons with wavevector k2？
And if you have a copy of Sakurai's advanced QM you can check he clearly interprets the state like this(page 26,"photon states")

 

[A. Neumaier]

Ok I followed the notations from Sakurai, |n_k> (ignoring polariztion) denotes the eigenstate of a number operator N=(a*_k )a_k, where a* and a are ladder operators.
Anyway why can't we interpret |n_k1,n_k2> as there are n_k1 photons with wavevector k1 and n_k2 photons with wavevector k2？
And if you have a copy of Sakurai's advanced QM you can check he clearly interprets the state like this(page 26,"photon states")

I don't have Sakurai's book. But your explanation is enough to see what you mean.

In this case, |n_k1>+|n_k2> is a superposition of a state with n_k1 photons with wave vector k1 and a state with n_k2 photons with wave vector k2, thus with an indefinite number of photons, while |n_k1,n_k2> is a tensor product state consisting of n_k1 photons with wave vector k1 and n_k2 photons with wave vector k2, thus containing the definite number of n_k1 + n_k2 photons. (But the first is a very unnatural situation, and the second is natural only if the n's are 0,1, or 2.)

 

[kof9595995]

Ok, just to check if I understand this issue correctly:
A classical monochromatic wave with wave vector k is given by a coherent state |c_k> ("c" for "coherent"), so to construct a wave packet it's better to use |c_k> as basis and do the superposition $$\sum\limits_k {{f_k}|{c_k} > }$$, actually in this case "|n_k1>+|n_k2>" can appear as part of the superposition since each |c_k> is a linear combination of |n_k> w.r.t different n's.
And by the way why did you say "second is natural only if the n's are 0,1, or 2"

 

[A. Neumaier]

Ok, just to check if I understand this issue correctly:
A classical monochromatic wave with wave vector k is given by a coherent state |c_k> ("c" for "coherent"), so to construct a wave packet it's better to use |c_k> as basis and do the superposition $$\sum\limits_k {{f_k}|{c_k} > }$$, actually in this case "|n_k1>+|n_k2>" can appear as part of the superposition since each |c_k> is a linear combination of |n_k> w.r.t different n's.

Coherent states are superpositions of |0_k>, |1_k>,...|n_k>,... (all n) with the _same_ k.
Only _these_ superpositions are natural.
Typical experiments involving linear optical devices produce superpositions of coherent states of different k (one for each beam direction and color). Nonlinear optics produces also squeezed states
and other more complex states. None is of the simple forms you discussed.

And by the way why did you say "second is natural only if the n's are 0,1, or 2"

Because already these are very difficult to produce, and the difficulty increases rapidly with occupation number.

 

[kof9595995]

Thanks