# 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