Schrodinger Cat states in terms of Displacement Operator

[deepalakshmi]

TL;DR Summary

How can I write even coherent state in terms of Displacement Operator?

The coherent state can be written in terms of e^(αb†+α∗b)|0>. But how the even coherent state i.e. |α>+|-α> can be written in terms of displacement operator?

 

[vanhees71]

The unitary transformation from the vacuum to a coherent state with complex parameter $\alpha$ is
$$\hat{U}(\alpha)=\exp(\alpha \hat{b}^{\dagger}-\alpha^* \hat{b})=\exp \left (-\frac{|\alpha|^2}{2} \right) \exp(\alpha \hat{b}^{\dagger}) \exp(-\alpha^* \hat{b}).$$
This gives
$$|\Phi(\alpha) \rangle=\hat{U}(\alpha)|0 \rangle=\exp \left (-\frac{|\alpha|^2}{2} \right) \sum_{n=0}^{\infty} \frac{(\alpha \hat{b}^{\dagger})^n}{n!} |0 \rangle = \exp \left (-\frac{|\alpha|^2}{2} \right) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle,$$
where $|n \rangle$ are the number eigenstates. Of course $|\Phi(\alpha) \rangle + |\Phi(-\alpha) \rangle$ is not a coherent state, because it's not an eigenstate of the annihilation operator, $\hat{b}$. It's just a superposition of two coherent states.

 

[deepalakshmi]

Are you saying that even coherent state cannot be written as Displacement operator?

 

[vanhees71]

What do you mean by "displacement operator"?

 

[Cthugha]

The coherent state can be written in terms of e^(αb†+α∗b)|0>. But how the even coherent state i.e. |α>+|-α> can be written in terms of displacement operator?

You know how to write a coherent state in terms of the displacement operator and you know that the even cat state is a superposition of two coherent states with different displacements. These two points should enable you to describe the even cat state in terms of displacement operators in a very basic way.

If you then work out the math, you will find that you can define a non-exponential displacement operator $\hat{D}_+=cosh(\alpha \hat{a}^\dagger -\alpha^\ast \hat{a})$ that creates an even cat state out of the vacuum.

Of course $|\Phi(\alpha) \rangle + |\Phi(-\alpha) \rangle$ is not a coherent state, because it's not an eigenstate of the annihilation operator, $\hat{b}$. It's just a superposition of two coherent states.

That is of course fully correct. One might still note that such states are commonly referred to as two-photon coherent states in the sense that they are eigenstates of $\hat{a}^2$

 

[deepalakshmi]

|α>+|-α>= e^(αb†-α∗b)|0>+e^-(αb†-α∗b)|0>.{This is what I am getting}
If there is { e^(αb†-α∗b)|0>+e^-(αb†-α∗b)|0>}/2 then I can write it in terms of cosh. But here there is no term like this. plus I am not good at mathematics.

 

[vanhees71]

Except for the factor of 2 it's a operator-cosh function as you proved yourself. BTW it would be great if you could use LaTeX, because it's very hard to read formulas in the way you typed them. You find a short description pressing the button "LaTeX Guide" directly below the dashboard:

https://www.physicsforums.com/help/latexhelp/

 

[deepalakshmi]

Thanks for latex guide.Then, you can't write it as cosh term because 2 is missing

 

[vanhees71]

Well, you can write
$$|\alpha \rangle+|-\alpha \rangle=2\cosh(\alpha \hat{b}^{\dagger}-\alpha^* \hat{b})|0 \rangle.$$
Note that the overall factor is irrelevant, because a pure state in QT is given by a ray and not a vector in Hilbert space. To evaluate probabilities you have to normalize it anyway, by defining the overall factor such that $\langle \Psi|\Psi \rangle=1$, and even this description fixes the overall factor only up to a phase factor $\exp(\mathrm{i} \varphi)$ with $\varphi \in \mathbb{R}$, but this doesn't matter since all observable quantities you calculate within QT don't depend on such an overall phase factor.