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I am trying to show that squeezed states are non classical by showing that the Glauber-Sudarshan $P$ function takes negative values. A squeezed state is one for which one of the quadratures $\Delta X_1 < \frac{1}{2}$. In Introductory quantum optics by Gerry and Knight on page 152 they show that for the quadrature $X_1$ we have

$$ \langle (\Delta X_1)^2 \rangle = \frac{1}{4} \bigg\{1+ \int P(\alpha)[(\alpha +\alpha^*)-(\langle a \rangle- \langle a^\dagger \rangle)]^2d^2\alpha\bigg\}.$$

I am trying to derive this from the fact that for an observable $A$ and a density operator $\rho$,$\langle A \rangle = \mathrm{tr}(A \rho)$. We can express the density operator in terms of the Glauber-Sudarshan $P$ function as

$$ \rho = \int d^2 \alpha P(\alpha)|\alpha \rangle \langle \alpha |$$

where $ \{|\alpha \rangle \}$ are coherent states. So working through the algebra I find

\begin{equation} \begin{split} \langle (\Delta X_1)^2\rangle & = \mathrm{tr}\bigg[ \rho (\Delta X_1)^2 \bigg] \\ & = \mathrm{tr} \bigg[ \int \mathrm{d}^2 \alpha P(\alpha) |\alpha \rangle \langle \alpha | (\Delta X_1)^2 \bigg] \\ & = \int \mathrm{d}^2 \alpha P(\alpha) \langle \alpha | (\Delta X_1)^2 |\alpha \rangle \\ &= \frac{1}{4} \int \mathrm{d}^2 \alpha P(\alpha) \end{split} \end{equation} where I used the fact that $\langle \alpha | (\Delta X_1)^2 |\alpha \rangle = \frac{1}{4}$ for a coherent state $|\alpha \rangle$. The density matrix has unit trace so $\int \mathrm{d}^2 \alpha P(\alpha)=1$ and I get $(\Delta X_1)^2 =\frac{1}{4}$ which does not agree with Gerry and Knight. Where have I gone wrong?

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  • $\begingroup$ are you asking someone to check your calculation? Aside: the coherent state is NOT squeezed and in fact saturates the bound so for the coherent state $(\Delta X_1)^2$ should be $1/4$ (or am I misunderstanding?) $\endgroup$ – ZeroTheHero Apr 15 '18 at 20:17
  • $\begingroup$ Pending a look at the book, the signs on your first expression look extremely dubious. $\endgroup$ – Emilio Pisanty Apr 15 '18 at 20:33
  • $\begingroup$ @ZeroTheHero Yes I have used that fact in my calculation, I stated it at the end. $\rho$ in my calculation is a general state but i seem to have shown that $\Delta X_1 = \frac{1}{4} $ no matter what the state was which is obviously not the right answer but I cannot see what I have done wrong in my calculation. I am just looking for some help deriving the expression from the book. $\endgroup$ – Matt0410 Apr 15 '18 at 20:45
  • $\begingroup$ You clearly don't have the density matrix for a squeezed state as it is diagonal in the coherent state basis. $\endgroup$ – ZeroTheHero Apr 15 '18 at 20:49
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    $\begingroup$ $ΔX^2$ is not an operator average, so the whole behind your equation is wrong. You should compute $\left<X^2\right>$ and $\left<X\right>$ separately and then $ΔX^2=\left<X^2\right>-\left<X\right>$ $\endgroup$ – Frédéric Grosshans Apr 16 '18 at 9:39
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The notation used by Gerry and Knight is different from what you are thinking. Their definition is given as follows:

\begin{equation} \label{definition} \langle (\Delta X_1)^2\rangle = \langle X^2_1 \rangle - \langle X_1 \rangle^2 \end{equation}

Evaluating each term of the last equation, one have

$$\langle X^2_1 \rangle = \frac{1}{4}\left(1+\int d^2 \alpha \, P(\alpha) [(\alpha+\alpha^*)^2]\right) \, ,$$

and,

$$\langle X_1 \rangle^2 = \frac{1}{4}\int d^2 \alpha \, P(\alpha)[(\alpha+\alpha^*)(\langle a\rangle + \langle a^{\dagger} \rangle)] \, ,$$

where $\langle a\rangle \equiv \int d^2 \alpha \, P(\alpha) \, \alpha$ and $\langle a^\dagger \rangle \equiv \int d^2 \alpha \, P(\alpha) \, \alpha^*$. Now, summing and subtracting the term $(\langle a\rangle + \langle a^{\dagger} \rangle)^2$ in the last equation and computing $\langle (\Delta X_1)^2\rangle$ reads

\begin{equation} \langle (\Delta X_1)^2 \rangle = \frac{1}{4} \bigg\{1+ \int P(\alpha)[(\alpha +\alpha^*)-(\langle a \rangle- \langle a^\dagger \rangle)]^2d^2\alpha\bigg\}. \end{equation}

Because of the term inside the square bracket is always positive and from the condition that \langle $(\Delta X_1)^2 < 1/4$, follows that $P(\alpha)$ must be negative at least in some regions of the phase space.

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  • $\begingroup$ in the second expression, does the integration extend to outside the big brackets? It does not look right. $\endgroup$ – flippiefanus Sep 27 at 4:07
  • $\begingroup$ No, it doesn't. I made a mistake when I was writing the expression. Now it is correct. Thank you, flippiefanus. $\endgroup$ – Alexssandre de Oliveira J Sep 28 at 14:11

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