# Schrodinger Cat states in terms of Displacement Operator

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• deepalakshmi
In summary, 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?The unitary transformation from the vacuum to a coherent state with complex parameter ##\alpha## is$$\hat{U}(\alpha)=\exp(\alpha \hat{b}^{\dagger}-\alpha^* \hat{b})=\exp \left (-\frac{|\alpha|^2}{2} \right) \exp(\alpha \hat{b}^{\dagger}) \exp(-\alpha^* \hat deepalakshmi TL;DR Summary How can I write even coherent state in terms of Displacement Operator? The coherent state can be written in terms of e^(αb†+α∗b)|0>. But how the even coherent state i.e. |α>+|-α> can be written in terms of displacement operator? The unitary transformation from the vacuum to a coherent state with complex parameter ##\alpha## is$$\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. sysprog and deepalakshmi Are you saying that even coherent state cannot be written as Displacement operator? What do you mean by "displacement operator"? deepalakshmi said: 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. vanhees71 said: 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 and vanhees71 |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. 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/ Thanks for latex guide.Then, you can't write it as cosh term because 2 is missing 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.

deepalakshmi

## 1. What is a Schrodinger Cat state?

A Schrodinger Cat state is a quantum state that exists in a superposition of two or more classical states. It is named after the famous thought experiment by physicist Erwin Schrodinger, where a cat in a sealed box can be both alive and dead at the same time.

## 2. What is the significance of the Displacement Operator in Schrodinger Cat states?

The Displacement Operator is a mathematical operator in quantum mechanics that describes the displacement of a quantum state in phase space. In the context of Schrodinger Cat states, it is used to describe the superposition of the classical states of the cat being both alive and dead at the same time.

## 3. How is the Displacement Operator used in Schrodinger Cat states?

The Displacement Operator is used to manipulate the quantum state of the cat in the thought experiment. It can be used to shift the cat's state from being alive to dead, or vice versa, and to create the superposition of both states simultaneously.

## 4. Can Schrodinger Cat states be observed in real life?

No, Schrodinger Cat states are purely theoretical and cannot be observed in real life. They are used as a thought experiment to illustrate the principles of quantum mechanics and the concept of superposition.

## 5. How are Schrodinger Cat states relevant to quantum computing?

Schrodinger Cat states are relevant to quantum computing because they demonstrate the principles of superposition and entanglement, which are essential for quantum computing operations. They also serve as a useful tool for understanding and developing quantum algorithms.

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