5
$\begingroup$

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 $\big(\langle n_1| \otimes \langle n_2|\big) \big(\hat x_1 |n_1\rangle \otimes\hat x_2 |n_2\rangle \big)= \langle n_1|\hat x_1 |n_1\rangle\otimes\ \langle n_2|\hat x_2 |n_2\rangle$?
$\endgroup$
1
9
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ How could we justify that $\mathbb{I}_2$ doesn't act on $|\varphi_1\rangle$? $\endgroup$ – Invenietis May 22 at 15:31
  • 1
    $\begingroup$ @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 $. $\endgroup$ – Jakob May 22 at 16:13
  • 1
    $\begingroup$ @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! $\endgroup$ – Cosmas Zachos May 22 at 16:15
  • 1
    $\begingroup$ @Invenietis Does this answer the question? $\endgroup$ – Jakob May 22 at 16:37
  • 1
    $\begingroup$ @Invenietis Yes! $\endgroup$ – Jakob May 22 at 16:43
5
$\begingroup$

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 $H_1\hat{\otimes} H_2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.