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 $\big(\langle n_1| \otimes \langle n_2|\big) \big(\hat x_1 |n_1\rangle \otimes\hat x_2 |n_2\rangle \big)= \langle n_1|\hat x_1 |n_1\rangle\otimes\ \langle n_2|\hat x_2 |n_2\rangle$?

2 Answers 2


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.

  • $\begingroup$ How could we justify that $\mathbb{I}_2$ doesn't act on $|\varphi_1\rangle$? $\endgroup$
    – Invenietis
    Commented May 22, 2021 at 15:31
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    $\begingroup$ @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 $. $\endgroup$ Commented May 22, 2021 at 16:13
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    $\begingroup$ @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! $\endgroup$ Commented May 22, 2021 at 16:15
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    $\begingroup$ @Invenietis Does this answer the question? $\endgroup$ Commented May 22, 2021 at 16:37
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    $\begingroup$ @Invenietis Yes! $\endgroup$ Commented 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 $H_1\hat{\otimes} H_2$.


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