Unitary Operator on a Tensor Product

In summary, Fredrik is having trouble deriving a result in a book. He tried using the identity e^{\alpha A \otimes I_n} = e^{\alpha A \otimes I_n} to obtain: e^{\alpha A \otimes I_n} {} | c_{n,1} |1 \rangle \otimes |n \rangle \rangle = e^{\alpha A} \otimes I_n {} |c_{n,1} {}|1\rangle \otimes |n \rangle \rangle = c_{n,1} e^{\alpha A}{}|1\rangle \otimes |n \
  • #1
barnflakes
156
4
Having a little trouble deriving a result in a book.

If I have an operator of the form [tex]e^{\alpha A \otimes I_n}[/tex]

Where alpha is a complex constant, A a square hermitian matrix and I the identity matrix.

Now if I want to operator that on a tensor product, say for instance [tex]c_{n,1} |1 \rangle \otimes |n \rangle[/tex] then how would I do that?

I firstly used the identity [tex]e^{\alpha A \otimes I_n} = e^{\alpha A \otimes I_n} [/tex] to obtain:
[tex]e^{\alpha A \otimes I_n} {} | c_{n,1} |1 \rangle \otimes |n \rangle \rangle = e^{\alpha A} \otimes I_n {} |c_{n,1} {}|1\rangle \otimes |n \rangle \rangle = c_{n,1} e^{\alpha A}{}|1\rangle \otimes |n \rangle [/tex]

but my book gets [tex]c_{n,1} e^{\alpha}|1 \rangle \otimes |n \rangle [/tex] with no further explanation. By the way the form of the matrix [tex] A = \begin{pmatrix} 1 & 0 & 0 \\0 & 0 & 0 \\0 & 0 & -1 \end{pmatrix} [/tex] if it helps to know it.

Also I am aware you can represent [tex]e^{\alpha A}[/tex] in the form of an infinite series but I don't see how that helps here. In fact I tried it and I didn't know where I should cut the series off at, and it gave me coefficients of alpha rather than e^alpha. Oh and the n's are being summed over, not sure if that makes any difference.
 
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  • #2
[tex]e^{A\otimes B}|\alpha\rangle\otimes|\beta\rangle=\sum_{n=0}^\infty \frac{1}{n!}(A\otimes B)^n|\alpha\rangle\otimes|\beta\rangle=\sum_{n=0}^\infty \frac{1}{n!}\underbrace{(A\otimes B)\cdots(A\otimes B)}_{n\text{ factors}}|\alpha\rangle\otimes|\beta\rangle[/tex]

[tex]=\sum_{n=0}^\infty \frac{1}{n!}\underbrace{(A\otimes B)\cdots(A\otimes B)}_{n-1\text{ factors}}A|\alpha\rangle\otimes B|\beta\rangle=\dots=\sum_{n=0}^\infty \frac{1}{n!}A^n|\alpha\rangle\otimes B^n|\beta\rangle[/tex]

Now what do you get when you set B=I?
 
  • #3
Thank you for your reply Fredrik, however I'm still stuck.

Using your definition [tex]\sum_{n=0}^\infty \frac{1}{n!}A^n|\alpha\rangle\otimes B^n|\beta\rangle[/tex] applied to the above expression, I obtain:

[tex]\sum_{k=0}^\infty \frac{1}{k!}(\alpha A)^k|1\rangle\otimes|n\rangle[/tex]

I cannot quite see how to turn that into [tex] e^{\alpha}|1 \rangle \otimes |n \rangle [/tex]

What happens to the matrix A? The vector [tex]|1\rangle[/tex] is an eigenvector of A with eigenvalue 1.

So can I express the above as:

[tex]\sum_{k=0}^\infty \frac{1}{k!}(\alpha A)^k|1\rangle \otimes|n\rangle = \sum_{k=0}^\infty \frac{1}{k!}(\alpha)^k A^k|1\rangle \otimes|n \rangle [/tex] and then work out [tex]A^k |1\rangle[/tex]

However, I don't know how to work out [tex]A^k |1\rangle[/tex] ? I know that [tex]A |1\rangle = 1 |1\rangle = |1\rangle [/tex] but how can I use this to calculate the action of A^k?
 
  • #4
Hold on, I have solved it!

[tex]A^k = A.A...A[/tex] k times therefore [tex]A^k |1 \rangle = |1\rangle[/tex]

Thank you Fredrik :)
 

Related to Unitary Operator on a Tensor Product

1. What is a unitary operator on a tensor product?

A unitary operator on a tensor product is a linear transformation that preserves the inner product between two vectors in the tensor product space. It is a special type of operator that is used in quantum mechanics to model the evolution of quantum systems.

2. How is a unitary operator on a tensor product different from a regular unitary operator?

A unitary operator on a tensor product is a composition of unitary operators on individual spaces, while a regular unitary operator acts on a single space. This means that a unitary operator on a tensor product operates on multiple quantum systems simultaneously, while a regular unitary operator only operates on one quantum system at a time.

3. What is the significance of unitary operators on tensor products in quantum computing?

Unitary operators on tensor products are fundamental to quantum computing as they are used to perform quantum gates, which are the building blocks of quantum algorithms. These operators allow for the manipulation and entanglement of multiple qubits, which enable quantum computers to solve certain problems much faster than classical computers.

4. Can any unitary operator be expressed as a tensor product of smaller unitary operators?

Yes, any unitary operator can be decomposed into a tensor product of smaller unitary operators. This is known as the Schmidt decomposition and it is a unique representation of a unitary operator on a tensor product space.

5. How do unitary operators on tensor products relate to entanglement in quantum systems?

Unitary operators on tensor products are used to create and manipulate entanglement between quantum systems. Entanglement is a phenomenon where two or more quantum systems become correlated in such a way that their quantum states cannot be described independently. Unitary operators on tensor products allow for the creation of entangled states, which are crucial for many quantum computing and communication applications.

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