# Wigner function of cat state

## Homework Statement

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.

## Homework 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,

$$\langle \alpha | \beta \rangle = \exp \left( -|\alpha|^2 / 2 - |\beta|^2 / 2 + \alpha^* \beta \right)$$

## 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,

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

The expected answer is,

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

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 :-).

## Answers and Replies

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Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.

I guess there are two problems with your strategy. First, is that the P function is a function of \gamma whereas \alpha has a fixed value given by your state. Then i guess it would be correct to use the partial derivative with respect to \gamma instead to with respect to \alpha in order to get rid of the integral for finding P by virtue of the integral definition of the 2-dimensional dirac delta function. The second problem with this strategy will follow once you proceed with the derivative with respect to \gamma. Since the Wigner function is an integral over \gamma domain, you will have the integral of the nth derivative of the two dimensional distribution. However, given that the integral is two dimensional as well, whereas the derivative is only with respect to \gamma (and not a series of second order derivatives with respect to \gamma and \gamma^*), you cannot express it as the value of the derivatives of the integrand as Gerry, Knight eq. (3.104) suggests.

I would suggest to get the Wigner function using a different strategy. Find the Weyl characteristic function and proceed directly to the Wigner function as Gerry, Knight eq. (3.136) first line implies.

PeroK
Homework Helper
Gold Member
I guess there are two problems with your strategy. First, is that the P function is a function of \gamma whereas \alpha has a fixed value given by your state. Then i guess it would be correct to use the partial derivative with respect to \gamma instead to with respect to \alpha in order to get rid of the integral for finding P by virtue of the integral definition of the 2-dimensional dirac delta function. The second problem with this strategy will follow once you proceed with the derivative with respect to \gamma. Since the Wigner function is an integral over \gamma domain, you will have the integral of the nth derivative of the two dimensional distribution. However, given that the integral is two dimensional as well, whereas the derivative is only with respect to \gamma (and not a series of second order derivatives with respect to \gamma and \gamma^*), you cannot express it as the value of the derivatives of the integrand as Gerry, Knight eq. (3.104) suggests.

I would suggest to get the Wigner function using a different strategy. Find the Weyl characteristic function and proceed directly to the Wigner function as Gerry, Knight eq. (3.136) first line implies.
Maybe you're three years too late to help the OP.

Maybe you're three years too late to help the OP.
I guess you helped him right on time.

PeroK
Homework Helper
Gold Member
I guess you helped him right on time.
As you can see, no one helped him. Nevertheless, I hope he graduated successfully.

In general, it would be useful if PF had functionality to notify the user when responding to an older thread. I suspect you didn't know how old the question was when you replied.

Note, also, if you click on the OP's name you can see when he/she was last active. In this case, the OP was last seen in Aug 2016, so they are unlikely to see your response.

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