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:

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

  • 1
    $\begingroup$ 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? $\endgroup$
    – user178876
    May 2, 2019 at 4:13
  • $\begingroup$ @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. $\endgroup$
    – leob
    May 2, 2019 at 4:29
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    $\begingroup$ math3ma.com/blog/the-tensor-product-demystified $\endgroup$
    – mmesser314
    May 2, 2019 at 4:44
  • 1
    $\begingroup$ This post (v2) seems to belong on Mathematics. $\endgroup$
    – Qmechanic
    May 2, 2019 at 10:17

1 Answer 1


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.

  • $\begingroup$ Hello, Valter Moretti, what are the technicalities for the infinitely-dimentional case? $\endgroup$
    – Sl0wp0k3
    Jun 21, 2022 at 14:00
  • $\begingroup$ Perhaps this should be obvious to me, but the standard product rule proof would require I be able to pass the limit “through” the tensor product. Should this be obvious to me? $\endgroup$
    – EE18
    Aug 9 at 2:31

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