# Using tensor products in the bra-ket notation

I'm trying to find the expectation value of the operator $$\hat W(x_1,x_2)=\hat x_1 \hat x_2$$ with respect to the eigenstates of a system composed of two one dimensional quantum harmonic oscillators. The eigenstate of the total system will be $$|n_1n_2\rangle=|n_1\rangle \otimes|n_2\rangle$$, with $$|n_1\rangle$$, $$|n_2\rangle$$ the eigenstates of each individual oscillator, so the expectation value will be

$$\left((|n_1\rangle \otimes|n_2\rangle )^\dagger,\hat W(|n_1\rangle \otimes|n_2\rangle) \right)$$ Two questions have arisen to me with this:

• Is the bra corresponding to a ket formed by a tensor product just the tensor product of the bras, $$\left(|n_1 n_2\rangle \right)^\dagger=\left(|n_1\rangle \otimes|n_2\rangle \right)^\dagger=\langle n_1| \otimes \langle n_2| =\langle n_1 n_2|$$?
• Are operators corresponding to different Hilbert spaces associative with respect to the tensor product of different states? That is, $$\hat x_1 \hat x_2 (|n_1\rangle \otimes|n_2\rangle)=\hat x_1 |n_1\rangle \otimes\hat x_2 |n_2\rangle?$$
• How do the inner products behave with respect to the tensor products? Would it be just $$\big(\langle n_1| \otimes \langle n_2|\big) \big(\hat x_1 |n_1\rangle \otimes\hat x_2 |n_2\rangle \big)= \langle n_1|\hat x_1 |n_1\rangle\otimes\ \langle n_2|\hat x_2 |n_2\rangle$$?

The answer to your first question is yes, see for example equations $$(1.32)-(1.36)$$ in these lecture notes.

To answer the second question, consider a bipartite Hilbert space $$\mathscr{H} \equiv \mathscr{H}_1 \otimes \mathscr{H}_2$$ and let $$o_1$$ and $$o_2$$ denote operators on $$\mathscr{H}_1$$ and $$\mathscr{H}_2$$, respectively. We then can define the action of $$o_1$$ and $$o_2$$ on $$\mathscr{H}$$ by \begin{align} O_1 &\equiv o_1 \otimes \mathbb{I}_2 \\ O_2 &\equiv \mathbb{I}_1 \otimes o_2 \quad , \end{align} where $$\mathbb{I}_i$$ for $$i=1,2$$ denotes the identity operator on $$\mathscr{H}_i$$.

Now let $$|\varphi_i\rangle \in \mathscr{H}_i$$ and $$\mathscr{H} \ni|\varphi\rangle \equiv |\varphi_1\rangle \otimes |\varphi_2\rangle$$. We compute \begin{align} O_1 |\varphi\rangle &= o_1 |\varphi_1\rangle \otimes \mathbb{I}_2 |\varphi_2\rangle\\ O_2 |\varphi\rangle &= \mathbb{I}_1 |\varphi_1\rangle \otimes o_2 |\varphi_2\rangle \quad . \end{align} Consequently, by applying both operators successively, we obtain: $$O_1 \, O_2 |\varphi\rangle= O_2\, O_1 |\varphi\rangle = o_1 |\varphi_1\rangle \otimes o_2 |\varphi_2\rangle \quad .$$

Additionally, for $$O\equiv o_1 \otimes o_2$$ we have $$O^\dagger = o_1^\dagger \otimes o_2^\dagger$$.

Regarding the third question, note that for an inner product on $$\mathscr{H}$$ it holds that $$(\varphi_1 \otimes \varphi_2 , \phi_1 \otimes \phi_2)_{\mathscr{H}} = (\varphi_1,\phi_1)_{\mathscr{H}_1}\,(\varphi_2,\phi_2)_{\mathscr{H}_2} \quad .$$ Defining $$\phi_i \equiv o_i \varphi_i$$ yields an expression for the expectation value of $$O_1\,O_2$$.

A more detailed explanation is given in the above linked lecture notes, equations $$(1.26)-(1.31)$$ or also in the Wikipedia link provided in the other answer.

• How could we justify that $\mathbb{I}_2$ doesn't act on $|\varphi_1\rangle$? May 22, 2021 at 15:31
• @Invenietis Well, its action on elements of $\mathscr{H}_1$ is, in general, not even well-defined. Just as an example, if $\mathrm{dim}\, \mathscr{H}_1 = 2$ and $\mathrm{dim} \,\mathscr{H}_2 = 3$, then, roughly speaking, $|\varphi_1\rangle$ is a vector with $2$ entries, but $\mathbb{I}_2$ is a $3\times 3$ matrix. And something like $\mathbb{I}_2 |\varphi_1\rangle$ is not well-defined. In general, as the other answer also points out, we have $o_1 \otimes o_2 \left( |\varphi_1\rangle \otimes |\varphi_2\rangle\right) = o_1|\varphi_1\rangle \otimes o_2|\varphi_2\rangle$. May 22, 2021 at 16:13
• @Invenietis One designates, one does not "justify"; this is what the subscripts mean. Appreciating this, one simply skips tensor product symbols, as they are implicit! May 22, 2021 at 16:15
• @Invenietis Does this answer the question? May 22, 2021 at 16:37
• @Invenietis Yes! May 22, 2021 at 16:43

Regarding your second question: Yes, and that's actually the defining property of $$\hat x_1 \hat x_2$$:

Let $$V,V',W,W'$$ be vector spaces over a field $$F$$ (for example, $$V'$$ and $$W'$$ can be Hilbert spaces with vector subspaces $$V$$ and $$W$$). If $$A\colon V\to V'$$ and $$B\colon W\to W'$$ are linear functions, the function \begin{align} V\times W&\to V'\otimes W'\\ (v,w)&\mapsto(Av)\otimes(Bw) \end{align} is bilinear and by the universal property of the tensor product, it extends to a unique linear function $$A\otimes B\colon V\otimes W\to V'\otimes W'$$ satisfying $$(A\otimes B)(v\otimes w)=(Av)\otimes(Bw)$$ for all $$(v,w)\in V\times W$$.

Warning: If $$H_1$$ and $$H_2$$ are Hilbert spaces, the vector space $$H_1\otimes H_2$$ together with the unique inner product satisfying $$\langle v_1\otimes v_2|w_1\otimes w_2\rangle=\langle v_1|w_1\rangle\langle v_2|w_2\rangle$$ is not necessarely a new Hilbert space, which is why we usually consider the Hilbert tensor product $$H_1\hat{\otimes} H_2$$.