What Is the Coherent State for This EM Field?

I'm not sure if I understand your argument. It is true that a coherent state is an eigenstate of the annihilation operator a, but the expectation value of the electric field will still depend on the state, since it is given by\langle \psi | \hat{\vec{E}}|\psi \rangle=\sum _{k,\lambda } \sqrt{\frac{2 \omega _k}{V}} \textbf{Im}\left(\alpha e^{-i k r} \epsilon _{x_{k,\lambda }}\right).So, the task is to find a coherent state such that the expectation value matches the given function \vec{f}(t,\vec{x}). I hope this clarifies it
  • #1
Raz91
21
0
[itex]\Huge[/itex]

Homework Statement



Consider a state of the EM field which satisfies
[itex]\left\langle \textbf{E}_x(\vec{r})\right\rangle =f(\vec{r})[/itex]

Find a coherent state which satises these expectation values.

Homework Equations



[itex] \textbf{E}(\textbf{r})=\frac{i}{\sqrt{2 V}}\sum _{\textbf{k},\lambda } \sqrt{\omega _k}\left(e^{-i \textbf{k}\textbf{ r}} a^{\dagger }{}_{\textbf{k},\lambda } \hat{\epsilon }^*{}_{\textbf{k},\lambda }+e^{-i \textbf{k}\textbf{r}} a_{\textbf{k},\lambda } \hat{\epsilon }_{\textbf{k},\lambda }\right)[/itex]

Coherent State :

[itex]a|\alpha \rangle =\alpha |\alpha \rangle[/itex]

The Attempt at a Solution



I tried to calculate this , but i just don't understand what am I suppose to prove here?
isn't it trivial that the expectation value will be a function of r (vector) ?

I've got this :
[itex]
\left\langle \textbf{E}_x(r)\right\rangle =\sum _{k,\lambda } \sqrt{\frac{2 \omega _k}{V}} \textbf{Im}\left(\alpha e^{-i k r} \epsilon _{x_{k,\lambda }}\right)[/itex]Thank you !
 
Physics news on Phys.org
  • #2
The point is to find a coherent state such that the expectation value of the electric-field components (operators) take the given (classical) field [itex]f[/itex].
 
  • #3
I still don't understand where is it given ?
it's just a "new name" for <Ex> , isn't it?

of course the expectation value won't be an operator... so I don't see what's so special here or what should I do ...

or f(r) is a known function in Electrodynamics that i should know ?

Thank u ...
 
  • #4
No, it's not a known function. You just assume a function [itex]\vec{f}(t,\vec{x})[/itex] and look for a coherent state [itex]|\psi[/itex] of the electromagnetic field such that
[tex]\langle \psi | \hat{\vec{E}}|\psi \rangle=\vec{f}(t,\vec{x}).[/tex]
 
  • #5
but according to the defination of the electric field , any coherent state will lead to such an expectation value because it's an eigen-state of the annihilation operator.
 
Last edited:
  • #6
As u can see , my result is depended on r (vector) for an arbitrary coherent state |alpha>...
 

FAQ: What Is the Coherent State for This EM Field?

What are coherent states in quantum fields?

Coherent states are quantum states that are used to describe the behavior of a quantum field. They are mathematical representations of the classical concept of a harmonic oscillator, and are characterized by a well-defined average value for the field and a minimum uncertainty in the field's measurement.

How are coherent states different from other quantum states?

Unlike other quantum states, coherent states have a well-defined average value for the field and a minimum uncertainty in the field's measurement. This means that they behave in a more classical and predictable manner, making them useful for studying the behavior of quantum fields in various physical systems.

How are coherent states created?

Coherent states can be created through a process called "squeezing," which involves manipulating the quantum field in a way that reduces the uncertainty in its measurement. This can be achieved through various techniques, such as using lasers or applying electromagnetic fields.

What are the applications of coherent states in quantum fields?

Coherent states have a wide range of applications in quantum field theory, such as in quantum optics, quantum computing, and quantum information processing. They are also used in the study of Bose-Einstein condensates and in understanding the behavior of quantum systems in strong and weak fields.

How do coherent states relate to the Heisenberg uncertainty principle?

Coherent states are unique in that they minimize the uncertainty in the field's measurement, while still obeying the Heisenberg uncertainty principle. This means that, unlike other quantum states, the uncertainties in the field's position and momentum are not equal, and one can be reduced at the expense of the other. Coherent states provide a way to study this trade-off in more detail.

Back
Top