# Quick Question About Tensor Products

1. Jul 19, 2008

### e(ho0n3

In the expression

$$U(|a \rangle \otimes |b \rangle)$$

is U distributive over the tensor product or does it behave like the dot product with scalars, i.e.

$$U(|a \rangle \otimes |b \rangle) = U|a \rangle \otimes |b \rangle = |a \rangle \otimes U|b \rangle$$

2. Jul 19, 2008

### mrandersdk

It depends on what U is?

3. Jul 19, 2008

### e(ho0n3

Right. U is a unitary matrix.

4. Jul 19, 2008

### n_bourbaki

I hope I can see through the unimportant, and confusing, QM notation to say that you can define the action of a linear map on V on V\otimes V in at least 3 ways. It is up to the context to determine which you mean.

5. Jul 19, 2008

### e(ho0n3

This is in the context of quantum computing.

6. Jul 19, 2008

### n_bourbaki

That is not the meaning of the word "context" I was aiming for. What is the definition given in the source this occurs? That is an elementary but necessary question. The action could be on the left, the right or diagonal. (But none of these is precisely what you wrote: a left and right action will not in general coincide as you implied.) None of these is necessarily canonical.

7. Jul 19, 2008

### e(ho0n3

From what I'm reading, the action is on the right. I thought there would be some standard properties that linear operators obey over tensor products. Weird.

8. Jul 20, 2008

### mrandersdk

Actually the only right thing then is

$$U(|a \rangle \otimes |b \rangle)$$

but as the others say, it depends on the context, because often the operator is defined in such a way that it is clear that it is only working on the first ket, or the second ket, and then one would write it just like above and then it would be implicit that

$$U(|a \rangle \otimes |b \rangle) = U|a \rangle \otimes |b \rangle$$

or in the second case

$$U(|a \rangle \otimes |b \rangle) = |a \rangle \otimes U|b \rangle$$

but this is because one actually mean

$$U = 1 \otimes U$$

or

$$U = U \otimes 1$$

which is a ***** to write all the time so the shorthand notation is used. But in general a operator isn't distributive over the tensor product.

9. Jul 20, 2008

### n_bourbaki

In order for it to be distributive, one would need something like a coalgebra structure, or at least a comultiplication map from X to X\otimes X.

10. Jul 20, 2008

### e(ho0n3

I think I understand now. Thanks.