# Wigner function of cat state

1. Aug 5, 2016

### Idoubt

1. The problem statement, all variables and given/known data
I am trying to calculate the wigner function for the even coherent state or cat state gives by,
$$| \psi \rangle = N_+ \left( |\alpha \rangle + | -\alpha \rangle \right),$$
where $|\alpha \rangle$ is a coherent state and $|N_+|^2 = \dfrac{1}{ 2 + 2e^{-2|\alpha|^2}}$. I am doing this is a slightly round about manner, I am first calculating the Glauber-Sudarshan P distribution for the state and then calculating the Wigner function from there. I have obtained an expression but it is not agreeing with expressions I have found in reference books for the Wigner function of a even coherent state.
2. Relevant equations
To find the P distribution given a state $\rho$,

$$P_{GS}(\gamma) = \frac{e^{|\gamma |^2}}{\pi^2} \int \langle -\beta | \rho | \beta \rangle e^{| \beta |^2}e^{\beta^* \gamma - \beta \gamma^*} d^2\beta$$

The Wigner function can be obtained from the P distrubtion as,
$$W(\beta) = \frac{2}{\pi} \int P_{GS}(\gamma) e^{-2|\gamma - \beta|^2} d^2\gamma$$

Expected answer in real variables (such that $\beta = q + ip$) with $\alpha$ real, from Introduction to quantum optics by Gerry & Knight,
$$W(q, p) = \frac{1}{\pi (1 + \exp (-2\alpha^2))} \left\lbrace \exp (-2(q-\alpha)^2 -2p^2) + \exp (-2(q + \alpha)^2 -2p^2) \right. \\ \left. + 2\exp (-2q^2 - 2p^2)\cos (4p\alpha)\right\rbrace$$

Inner product of coherent states is given by,

$$\langle \alpha | \beta \rangle = \exp \left( -|\alpha|^2 / 2 - |\beta|^2 / 2 + \alpha^* \beta \right)$$
3. The attempt at a solution

For us, $\rho = |N_+|^2 ( | \alpha \rangle \langle \alpha | + | -\alpha \rangle \langle -\alpha | + | -\alpha \rangle \langle \alpha | + | \alpha \rangle \langle -\alpha |).$
$$P_{GS}(\gamma) = \sum\limits_{i = 1}^4 P_i ,$$
with,
\begin{align*} P_1 &= \frac{|N_+|^2 e^{|\gamma |^2}}{\pi^2} \int \langle -\beta | \alpha \rangle \langle \alpha | \beta \rangle e^{| \beta |^2}e^{\beta^* \gamma - \beta \gamma^*} d^2\beta \\ &= \frac{|N_+|^2 e^{|\gamma|^2 - |\alpha|^2 }}{\pi^2} \int \exp(\beta^*(\gamma - \alpha) - \beta (\gamma^* - \alpha^*)) d^2\beta \\ &= |N_+|^2 e^{|\gamma|^2 - |\alpha|^2 } \delta^{(2)} ( \gamma - \alpha) \\ &= |N_+|^2 \delta^{(2)} ( \gamma - \alpha) \end{align*}

$P_2$ the term corresponding to $|-\alpha \rangle \langle -\alpha |$ can be obtained from $P_1$ by replacing $\alpha$ with $-\alpha$, giving
$$P_2 = |N_+|^2 \delta^{(2)} ( \gamma + \alpha)$$

For the cross term $|-\alpha \rangle \langle \alpha |$,
\begin{align*} P_3 &= \frac{|N_+|^2 e^{|\gamma |^2}}{\pi^2} \int \langle -\beta | -\alpha \rangle \langle \alpha | \beta \rangle e^{| \beta |^2}e^{\beta^* \gamma - \beta \gamma^*} d^2\beta \\ &= \frac{|N_+|^2 e^{|\gamma|^2 - |\alpha|^2 }}{\pi^2} \int \exp(\beta^*(\gamma + \alpha) - \beta (\gamma^* - \alpha^*)) d^2\beta \\ &= \frac{|N_+|^2 e^{|\gamma|^2 - |\alpha|^2 }}{\pi^2} \int \exp (2\beta^*\alpha) \exp(\beta^*(\gamma - \alpha) - \beta (\gamma^* - \alpha^*)) d^2\beta \\ &= \frac{|N_+|^2 e^{|\gamma|^2 - |\alpha|^2 }}{\pi^2} \int \exp (-2\alpha \frac{\partial}{\partial \alpha}) \exp(\beta^*(\gamma - \alpha) - \beta (\gamma^* - \alpha^*)) d^2\beta \\ &= \frac{|N_+|^2 e^{|\gamma|^2 - |\alpha|^2 }}{\pi^2} \exp (-2\alpha \frac{\partial}{\partial \alpha}) \int \exp(\beta^*(\gamma - \alpha) - \beta (\gamma^* - \alpha^*)) d^2\beta \\ &= |N_+|^2 e^{|\gamma|^2 - |\alpha|^2 } \exp (-2\alpha \frac{\partial}{\partial \alpha})\delta^{(2)} ( \gamma - \alpha) \\ &= |N_+|^2 \exp (-2\alpha \frac{\partial}{\partial \alpha})\delta^{(2)} ( \gamma - \alpha) \end{align*}
Again, $P_4$ can be obtained from $P_3$ by replacing $\alpha$ by $-\alpha$.
$$P_4 = |N_+|^2 \exp (-2\alpha \frac{\partial}{\partial \alpha})\delta^{(2)} ( \gamma + \alpha)$$
Finally putting the 4 terms together the P distribution is given by,

$$P_{GS}(\gamma) = |N_+|^2 \left\lbrace 1 + \exp ( -2\alpha\frac{\partial}{\partial \alpha}) \right\rbrace \left[ \delta^{(2)} ( \gamma - \alpha) + \delta^{(2)} ( \gamma + \alpha) \right]$$

The wigner function can be obtained using the relation given in the begining as,
$$W(\beta) = \frac{2 |N_+|^2}{\pi} \left\lbrace1 + \exp(-2\alpha\frac{\partial}{\partial \alpha}) \right\rbrace \left[ e^{-2|\alpha - \beta|^2} + e^{-2|\alpha + \beta|^2} \right]$$

Now,
$$\exp(-2\alpha\frac{\partial}{\partial \alpha}) e^{-2|\alpha - \beta|^2} = \exp (4(\alpha^* - \beta^*)\alpha)e^{-2(\alpha - \beta)(\alpha^* - \beta^*)} = e^{2(\alpha + \beta )(\alpha^* - \beta^*)}$$
replacing $\alpha$ by $- \alpha$ again gives,

$$\exp(-2\alpha\frac{\partial}{\partial \alpha}) e^{-2|\alpha + \beta|^2} = e^{2(\alpha - \beta )(\alpha^* + \beta^*)}$$

Now the wigner function can be written as,

$$W(\beta) = \frac{2 |N_+|^2}{\pi} \left[ e^{-2|\alpha - \beta|^2} + e^{-2|\alpha + \beta|^2} + e^{2(\alpha + \beta )(\alpha^* - \beta^*)} + e^{2(\alpha - \beta )(\alpha^* + \beta^*)} \right]$$

I will now write this function in terms of real variables to compare with the answer I already have. Taking,
$\alpha = q_0 + ip_0$ and $\beta = q + ip$, and substituting for #|N_+|^2#. Note also that the last two terms are complex conjugates of each other and so can be written as twice the real part of each.
$$W(q, p) = \frac{1}{\pi (1 + \exp( -2(q_0^2 + p_0^2 ) ) } \left[ e^{-2(q - q_0)^2}e^{-2(p - p_0)^2} + e^{-2(q + q_0)^2}e^{-2(p + p_0)^2} \\ + 2\Re ( e^{2 \lbrace (q_0 - q) + i(p_0 - p) \rbrace \lbrace (q_0 + q ) +i(p_0 + p) \rbrace }) \right]$$

Simplifying the last term gives,

$$W(q, p) = \frac{1}{\pi (1 + \exp( -2(q_0^2 + p_0^2 ) ) } \left[ e^{-2(q - q_0)^2}e^{-2(p - p_0)^2} + e^{-2(q + q_0)^2}e^{-2(p + p_0)^2} \\ + 2 e^{2 [(q_0^2 + p_0^2) - (q^2 + p^2) ] }\cos 4(p_0q - q_0p) \right]$$

Now to reproduce the result I have, I should substitute, $q_0 = \alpha$ and $p_0 = 0$
This gives,

$$W(q, p) = \frac{1}{\pi (1 + \exp( -2 \alpha^2 ) } \left[ e^{-2(q - \alpha)^2}e^{-2p^2} + e^{-2(q + \alpha)^2}e^{-2p^2} \\ + 2 e^{2 [\alpha^2 - (q^2 + p^2) ] }\cos(4\alpha p) \right]$$

$$W(q, p) = \frac{1}{\pi (1 + \exp (-2\alpha^2))} \left\lbrace \exp (-2(q-\alpha)^2 -2p^2) + \exp (-2(q + \alpha)^2 -2p^2) \right. \\ \left. + 2\exp (-2q^2 - 2p^2)\cos (4p\alpha)\right\rbrace$$

The 3rd term doesn't agree to a factor of $e^{2\alpha^2}$. I am unable to find my error, any help would be appreciated :-).

2. Aug 10, 2016