# How to operate on product state?

• KFC
In summary, the conversation discusses the application of an operator, represented by a 2x2 matrix, on a Fock state |n\rangle and any other state |x\rangle. If |x\rangle is a spinor, the operator can be applied by matrix multiplication. If |x\rangle is not a spinor, the operator simply acts on |n\rangle. The product state of |n\rangle and |x\rangle is represented by a 2-component vector, where |n\rangle is the first component and |x\rangle is the second component.

#### KFC

Suppose a proudct is form by a Fock state $$|n\rangle$$ and any other state $$|x\rangle$$, i.e.

$$|\phi\rangle = |n\rangle|x\rangle$$

If an operator defined as

$$A = \left( \begin{matrix} \alpha\hat{a}\hat{a}^\dagger & \beta\hat{a}^\dagger\hat{a} \\ \gamma\hat{a}^\dagger\hat{a} & \kappa\hat{a}\hat{a}^\dagger \end{matrix} \right)$$
where $$\hat{a}$$ and $$\hat{a}^\dagger$$ is creation and annilation operator will only opeate on Fock state. So how A operate on $$|n\rangle|x\rangle$$?

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It sounds like you're asking about the states of a spin-1/2 particle in a Fock space (e.g. a harmonic oscillator). In that case, the state $$|\chi\rangle$$ (what you've written $$|x\rangle$$) is actually a spinor, basically a two-component vector:

$$|\chi\rangle = \left(\begin{matrix} a \\ b\end{matrix}\right)$$

where $$\langle\chi |\chi\rangle = 1$$. Then the operator A, a 2x2 matrix, can be applied to that 2-component vector by the usual rules of matrix multiplication.

$$A|\phi\rangle = \left( \begin{matrix} \alpha\hat{a}\hat{a}^\dagger & \beta\hat{a}^\dagger\hat{a} \\ \gamma\hat{a}^\dagger\hat{a} & \kappa\hat{a}\hat{a}^\dagger \end{matrix} \right)|n\rangle\left(\begin{matrix} a \\ b\end{matrix}\right) = \left( \begin{matrix} (\alpha a\hat{a}\hat{a}^\dagger + \beta b\hat{a}^\dagger\hat{a})|n\rangle \\ (\gamma a\hat{a}^\dagger\hat{a} + \kappa b\hat{a}\hat{a}^\dagger)|n\rangle \end{matrix}\right) = \left( \begin{matrix} \alpha a\sqrt{n + 1} + \beta b\sqrt{n} \\ \gamma a\sqrt{n} + \kappa b\sqrt{n + 1} \end{matrix} \right)|n\rangle$$

If $$|x\rangle$$ is not a 2-component vector, then I don't know that there's much you can do with it. You'd probably just "distribute" $$|n\rangle$$ over all the elements of the matrix A and wind up with something like

$$A|\phi\rangle = \left( \begin{matrix} \alpha\hat{a}\hat{a}^\dagger|n\rangle & \beta\hat{a}^\dagger\hat{a}|n\rangle \\ \gamma\hat{a}^\dagger\hat{a}|n\rangle & \kappa\hat{a}\hat{a}^\dagger|n\rangle \end{matrix} \right)|x\rangle = \left( \begin{matrix} \alpha\sqrt{n + 1} & \beta\sqrt{n} \\ \gamma\sqrt{n} & \kappa\sqrt{n + 1} \end{matrix} \right)|x\rangle$$

I guess the point is, if the creation and annihilation operators only act on $$|n\rangle$$, you just leave $$|x\rangle$$ alone. It's fairly simple.

diazona said:
It sounds like you're asking about the states of a spin-1/2 particle in a Fock space (e.g. a harmonic oscillator). In that case, the state $$|\chi\rangle$$ (what you've written $$|x\rangle$$) is actually a spinor, basically a two-component vector:

$$|\chi\rangle = \left(\begin{matrix} a \\ b\end{matrix}\right)$$

where $$\langle\chi |\chi\rangle = 1$$. Then the operator A, a 2x2 matrix, can be applied to that 2-component vector by the usual rules of matrix multiplication.

$$A|\phi\rangle = \left( \begin{matrix} \alpha\hat{a}\hat{a}^\dagger & \beta\hat{a}^\dagger\hat{a} \\ \gamma\hat{a}^\dagger\hat{a} & \kappa\hat{a}\hat{a}^\dagger \end{matrix} \right)|n\rangle\left(\begin{matrix} a \\ b\end{matrix}\right) = \left( \begin{matrix} (\alpha a\hat{a}\hat{a}^\dagger + \beta b\hat{a}^\dagger\hat{a})|n\rangle \\ (\gamma a\hat{a}^\dagger\hat{a} + \kappa b\hat{a}\hat{a}^\dagger)|n\rangle \end{matrix}\right) = \left( \begin{matrix} \alpha a\sqrt{n + 1} + \beta b\sqrt{n} \\ \gamma a\sqrt{n} + \kappa b\sqrt{n + 1} \end{matrix} \right)|n\rangle$$

Thank you so much. However, I found it quite confusing to understand the following operation, in above calculation, it seems that you write

$$|x\rangle = \left(\begin{matrix}a \\ b \end{matrix}\right)$$
and
$$|n\rangle|x\rangle = \left(\begin{matrix}a|n\rangle \\ b|n\rangle \end{matrix}\right)$$

so

$$\left( \begin{matrix} \alpha\hat{a}\hat{a}^\dagger & \beta\hat{a}^\dagger\hat{a} \\ \gamma\hat{a}^\dagger\hat{a} & \kappa\hat{a}\hat{a}^\dagger \end{matrix} \right)|n\rangle\left(\begin{matrix} a \\ b\end{matrix}\right) = \left( \begin{matrix} (\alpha a\hat{a}\hat{a}^\dagger + \beta b\hat{a}^\dagger\hat{a})|n\rangle \\ (\gamma a\hat{a}^\dagger\hat{a} + \kappa b\hat{a}\hat{a}^\dagger)|n\rangle \end{matrix}\right)$$

But I readed a text, in which the product state of two states is given by
$$|n\rangle|x\rangle = |n\rangle \otimes |x\rangle = \left(\begin{matrix}\alpha|x\rangle \\ \beta|x\rangle \end{matrix}\right)$$
where $$|n\rangle = \left(\begin{matrix}\alpha \\ \beta\end{matrix}\right)$$

I wonder if
$$\left(\begin{matrix}\alpha|x\rangle \\ \beta|x\rangle \end{matrix}\right) \equiv \left(\begin{matrix}a|n\rangle \\ b|n\rangle \end{matrix}\right)$$

?

## 1. What is a product state?

A product state is a quantum state that can be represented as a tensor product of individual states. This means that the overall state is a combination of the states of multiple systems, rather than a single system.

## 2. How do you operate on a product state?

To operate on a product state, you must first apply the desired operation on each individual state. Then, you can combine the results using the tensor product operation. This will give you the resulting product state after the operation has been applied.

## 3. What are the benefits of using product states?

Product states allow for the manipulation and analysis of multiple systems simultaneously. They also make it easier to understand and compute complex quantum systems, as they can be broken down into simpler components.

## 4. Can product states be entangled?

Yes, product states can be entangled. Entanglement occurs when the individual systems in a product state are correlated in a way that their combined state cannot be expressed as a product state. This phenomenon is a key feature of quantum mechanics and has potential applications in quantum information processing.

## 5. How are product states used in quantum computing?

Product states are used in quantum computing as the basis for quantum algorithms and calculations. They allow for the manipulation of multiple qubits simultaneously, which is necessary for many quantum algorithms to work effectively. Product states also play a key role in the development of quantum error correction codes.