I've got this problem to solve:

Consider the next electromagnetic field: $|A\rangle=\frac{1}{\sqrt{N!}}\left ( \int e^{-k^2} \boldsymbol{a}^\dagger_{k,+} \right )^N |0\rangle$. Find the mean value energy density in function of position.

My attempt is this: the Hamiltonian for the field is $H=\hbar\omega(\boldsymbol{a^\dagger a}+\frac{1}{2})$, then the mean value of energy would be $\langle H \rangle = \langle A|H|A\rangle$. With this value divided by a certain volume (I don't know which one) obtain the density.

My problem is how to calculate $\langle H \rangle = \langle A|H|A\rangle$. I don't know how to operate with the creator operator under the integral, under an nth power! Can anybody give me a clue? thanks.

  • $\begingroup$ I've seen that the energy density in classic EM theory is $\eta=\epsilon_0A^2 $ so I need to calculate $<A|A>$ but i dont know how. $\endgroup$ – Bruno Jul 20 '17 at 17:01

This isn't quite what you're being asked to do. The energy density is a very specific quantity and it is tightly linked with the squares of the force fields, as $$ u(\mathbf r) = \frac{\epsilon_0}{2}\mathbf E(\mathbf r)^2 + \frac{1}{2\mu_0}\mathbf B(\mathbf r)^2, $$ where the electric fields are promoted to operators in terms of the creation and annihilation operators via $$ \hat{\mathbf E}(\mathbf r) = \sum_{s=1}^2\int\mathrm d\mathbf k\left[ \boldsymbol{\mathcal E}_{0}(\mathbf k,s) e^{i\mathbf k\cdot\mathbf r} \hat{a}_{\mathbf{k},s} + \boldsymbol{\mathcal E}_{0}^*(\mathbf k,s) e^{-i\mathbf k\cdot\mathbf r} \hat{a}_{\mathbf{k},s}^\dagger \right] $$ for some normalization constant $\boldsymbol{\mathcal E}_{0}(\mathbf k,s)$ that will be pinned down in your textbook.

That means that you will have a fair amount of heavy lifting to do with the Fourier transform and the $k$ integral in your state, but as far as the bosonic part of Hilbert space goes it will boil down to a vacuum expectation value of the form $$ \left \langle 0\middle | \hat{a}^N (\alpha \hat a+\alpha^*\hat a^\dagger)^2 \left(\hat a^\dagger\right)^N \middle | 0\right\rangle , $$ and this is where the true quantum optics comes in. Here you've essentially got three terms:

  • one in $\alpha^2 \hat a^2$, which you can prove by induction will produce a zero expectation value, and similarly
  • one in $\alpha^*{{\hat a}^\dagger}^2$, which also vanishes, and finally
  • one in $|\alpha|^2(\hat a\hat a^\dagger + \hat a^\dagger \hat a)=|\alpha|^2(2\hat a\hat a^\dagger -1)$, through the canonical commutation relations, which will reduce to the norm of the states ${\hat a^\dagger}^N|0\rangle$ and ${\hat a^\dagger}^{N+1}|0\rangle$, and which is therefore easy to evaluate.

Everything else is just algebraic trudging which is uninstructive unless you do it yourself.

Oh, and as a footnote: I would strongly encourage you to fix the quantum state you were given so that the creation operator reads $\int e^{-\frac12 k^2\sigma^2} \boldsymbol{a}^\dagger_{k,+} \mathrm dk$, where $\sigma$ is a length, so that you don't unwittingly de-dimensionalize the problem. Doing this will enable you to retain dimensional analysis as a sanity check on your calculations, and you'll thank yourself later for that.

  • $\begingroup$ Thanks for the quick answer!. But i can't figure out where use the em field expression given in the problem. $\endgroup$ – Bruno Jul 20 '17 at 18:13
  • $\begingroup$ You're not given an EM field, you're given a quantum state. I've described a recipe to build an operator. Now build the expectation value of that operator in the quantum state you were given. $\endgroup$ – Emilio Pisanty Jul 20 '17 at 18:14
  • $\begingroup$ I can't follow these sentence "but as far as the bosonic part of Hilbert space goes it will boil down to a vacuum expectation value". Can you clarify to me? $\endgroup$ – Bruno Jul 20 '17 at 18:58
  • $\begingroup$ It means that if you actually sit down and do the calculation, then you will decouple all the $\hat a_k$ (i.e. you will only have $\hat a$s for a single $k$ inside the expectation value) and the energy will reduce to an integral over $k$ of expectation values of the form I've given. However, there is no substitute for actually putting in the algebra and working through it ─ this is your homework, and it's not going to do itself. $\endgroup$ – Emilio Pisanty Jul 20 '17 at 19:17

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