# Time evolution on a tensor product space

I'm reading "Introduction to Quantum Field Theory for Mathematicians, Lecture Notes, sourav Chatterjee (https://souravchatterjee.su.domains//qft-lectures-combined.pdf), p.21 and stuck at understanding some equality.

In his lecture note, p.7, he states Stone's Theorem as

There is a one-to-one correspondence between one parameter strongly continuous unitary groups of operators on $$\mathcal{H}$$ (Hilbert space) and self-adjoint operators on $$\mathcal{H}$$. Given $$U$$, the corresponding self-adjoint operator $$A$$ is defined as $$Ax = \lim_{t\to 0}\frac{U(t)x-x}{it}$$ with $$\mathcal{D}(A) = \{x : \text{the above limit exists} \}$$.

And he propose Postulate 5 of quantum mechanics :

P5. If the system is not affected by external influences then its state evolves in time as $$\psi_t = U(t)\psi$$ for some strongly continuous unitary group $$U$$ that only depends on the system(and not on the state)

By Stone's theorem, there exsists a unique self-adjoint operator $$H$$ corresponding to $$U(t)$$. It is called the 'Hamiltonian'.

And in his note, p.21, he saids that

"Suppose that the state of a single particle evolves according to the unitary group $$(U(t))_{t\in \mathbb{R}}$$. Then the time evolution on $$\mathcal{H}^{\otimes n}$$ of $$n$$ non-interacting particles, also denoted by $$U(t)$$, is defined as

$$U(t)(\psi_1 \otimes \cdots \otimes \psi_n) := (U(t)\psi_1)\otimes(U(t)\psi_2)\otimes \cdots \otimes (U(t)\psi_n)$$

and extended by linearity."

Then, he calculated an action of the hamiltonian on $$\psi_1 \otimes \cdots \otimes \psi_n$$ by

My question is, why the underlined equality is true?

Can anyone help?

• Is the case $n=1$ clear? Commented Aug 3, 2022 at 12:37

Note that with these definitions the operator associated by Stone's theorem to $$U$$ is not exactly $$H$$ but apparently $$-H.$$
Remember $$c(a\otimes b)=(ca)\otimes b=a\otimes (cb).$$ You can distribute $$-1/it$$ into the sum and onto the correct factor. $$\lim_{t\to0}\sum_{j=1}^nU(t)\psi_1\otimes\cdots\otimes\frac{U_j(t)\psi_j-\psi_j}{-it}\otimes\cdots\otimes\psi_n$$
As $$t\to0,$$ $$U(t)\to1,$$ so in each term of the sum we have $$U(t)\psi_1\otimes\cdots\otimes U(t)\psi_{j-1}\to\psi_1\otimes\cdots\otimes\psi_{j-1},$$ and we also get $$\frac{U_j(t)\psi_j-\psi_j}{-it}\to H\psi_j$$ by definition of $$H.$$
A nicer way to view this is to note that the definition of $$H$$ is just $$-iH=U'(0)$$ where $$'$$ denotes the derivative (treating $$U$$ as a function from $$\mathbb R$$ to operators). Then the $$n$$-particle unitary $$U_n(t)=U(t)\otimes\cdots\otimes U(t)$$ (where the extension of $$\otimes$$ to operators is understood) produces the $$n$$-particle Hamiltonian $$H_n(t)=(H\otimes1\otimes\cdots\otimes1)+(1\otimes H\otimes1\otimes\cdots\otimes1)+\cdots+(1\otimes\cdots\otimes1\otimes H)$$ by the product rule (because $$\otimes$$ has the algebraic properties of a product).
• Yes. for the below remark, you used that $H_n = i U_n'(0)$. Thank you~ Commented Aug 4, 2022 at 1:16