# Derivative of tensor product of quantum states

Recently I asked a question over at the math stack exchange:

However I figured I'd ask here too, seeing as the question originated in a physics course I'm doing, and the context may change the answer.

I know similar questions have been asked on the stack exchange, but all the ones I've found haven't had answers.

I have just started working with tensor products, but have had to learn them a little on the run. I was wondering how the derivative acts on a tensor product.

That is, what is: $$\frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big)$$ ?

Is it: $$\frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big) = \Big( \frac{\mathrm{d}}{\mathrm{d}x} f \left( x \right) \Big) \otimes g \left( x \right) + f \left( x \right) \otimes \Big( \frac{\mathrm{d}}{\mathrm{d}x} g \left( x \right) \Big)$$ or $$\frac{\mathrm{d}}{\mathrm{d}x} \big( f \left( x \right) \otimes g \left( x \right) \big) = \Big( \frac{\mathrm{d}}{\mathrm{d}x} f \left( x \right) \Big) \otimes \Big( \frac{\mathrm{d}}{\mathrm{d}x} g \left( x \right) \Big)$$ or something else?

Specifically I came across this question while working through a quantum measurement exercise (non-homework). In this I came across a time derivative of the tensor product of two quantum states:

$$\partial_t \left( \vert \varphi \rangle \otimes \vert \phi \rangle \right)$$

• You can answer the question yourself if you consider the case that, say, $g$ doesn't depend on $x$. The second option would tell you that you get 0. Do you think that this makes sense? – user178876 May 2 '19 at 4:13
• @marmot Ah great, that seems to make sense. Does this mean the first option is correct? Unfortunately in the course I'm currently taking we weren't even given a definition of a tensor product, so I have no intuition yet about how they behave. We've barely had its properties defined! I've worked with what I think are some pretty odd, unintuitive structures before, so I'm always a bit cautious about new stuff like this. – leob May 2 '19 at 4:29
• math3ma.com/blog/the-tensor-product-demystified – mmesser314 May 2 '19 at 4:44
• This post (v2) seems to belong on Mathematics. – Qmechanic May 2 '19 at 10:17

## 1 Answer

Barring technicalities, relevant only in the infinite dimensional case, the tensor product satisfies all properties of the standard product of numbers sufficient to prove Leibnitz' rule of derivative operator. Using the same proof as for usual functions and usual product, you see that classical rule is still valid wit the tensor product. Just try.