Time evolution on a tensor product space

Question

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?

Answer 1

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).