3
$\begingroup$

I want to understand an experiment but I am struggling with the (basic) math/braket notation.

In the experiment two ions are entangled and separated into two wells $A$ and $B$. The spin state of the ions is thus $$\frac{1}{\sqrt{2}}\left[|\uparrow\rangle_A |\downarrow\rangle_B + |\downarrow\rangle_A |\uparrow\rangle_B \right]$$ As they want to operate a sideband transition next, they include the motional state of the ions as $$\frac{1}{\sqrt{2}}\left[|\uparrow\rangle_A |\downarrow\rangle_B +|\downarrow\rangle_A |\uparrow\rangle_B \right] |0\rangle_A |0\rangle_B$$ The sideband transition is only applied on the ion in well $A$ and they describe it as a rotation $$R(\theta,\phi)=\left(\begin{array}{r} \text {cos}(\theta/2) & -ie^{-i\phi}\text{sin}(\theta/2)\\ -ie^{i\phi}\text{sin}(\theta/2) & \text{cos}(\theta/2)\\ \end{array}\right)$$ in the basis $\left(\begin{array}{r} 1\\0 \end{array}\right)=|\uparrow\rangle|1\rangle$, $\left(\begin{array}{r} 0\\1 \end{array}\right)=|\downarrow\rangle|0\rangle$.

After applying the sideband transition $R_A(\pi,0)$ they get the state:

$$|f\rangle=\frac{1}{\sqrt{2}}|\uparrow\rangle_A \left[|\downarrow\rangle_B|0\rangle_A - i|\uparrow\rangle_B |1\rangle_A \right]|0\rangle_B$$

I am new to the braket notation and when I try to do the same calculation I end up with a different final state. Can someone please write down how applying $R_A(\pi,0)$ results in the final state $|f\rangle$? Thank you so much!

$\endgroup$
2
$\begingroup$

I think the question will be much more clear if you specify some of the remaining basis vectors, for instance the $\vert{\uparrow 0}\rangle$. I recommend to write the state as follows.

$$\vert{i}\rangle=\dfrac{1}{\sqrt{2}}(\vert{\uparrow 0}\rangle_A\vert{\downarrow 0}\rangle_B+\vert{\downarrow 0}\rangle_A\vert{\uparrow 0}\rangle_B)$$

Note that it lives in a Hilbert space which is the direct product of two (or more) Hilbert spaces i.e $$\mathcal{H}=\mathcal{H}_A\otimes\mathcal{H}_B$$

Then you should understand the rotation operator as $$R(\theta,\phi)\equiv R_A(\theta,\phi)\otimes \mathbb{1}_B$$ where $\mathbb{1}_B$ is the identity operator, so that $R(\theta,\phi)$ only acts on $\mathcal{H}_A$.

Hence :

$$R(\theta,\phi)\vert{i}\rangle=\dfrac{1}{\sqrt{2}}(R_A(\theta,\phi)\vert{\uparrow 0}\rangle_A\vert{\downarrow 0}\rangle_B+R_A(\theta,\phi)\vert{\downarrow 0}\rangle_A\vert{\uparrow 0}\rangle_B)=\vert{f}\rangle$$

Then by direct computation you should check that

$$R_A(\pi,0)\vert{\uparrow 0}\rangle_A=\vert{\uparrow 0}\rangle_A$$

$$R_A(\pi,0)\vert{\downarrow 0}\rangle_A=-i\vert{\uparrow 1}\rangle_A$$

For the second line I checked and it holds but you should check the first line.

EDIT: After reading the comment and taking a deeper look at the problem I realized that there is a little bit more here.

1) Note that $\mathcal{H}_{A}=\mathcal{H}_{s=1/2}\otimes \mathcal{H}_{\text{Fock Space}}$ and same for $\mathcal{H}_B$. The matrix representations of this oeprators are infinite dimensional matrices in the basis $\big\lbrace \vert \uparrow \rangle,\vert \downarrow \rangle \big \rbrace \otimes \big\lbrace \vert 0 \rangle,\vert 1 \rangle,\ldots \big \rbrace$.

2) The operator $R_A(\pi,0)$ rotates the basis vecotrs.

$$R_A(\pi,0)\vert{\downarrow 0}\rangle_A=-i\vert{\uparrow 1}\rangle_A$$ $$R_A(\pi,0)\vert{\uparrow 1}\rangle_A=-i\vert{\downarrow 0}\rangle_A$$

But note that it does not touch the basis vector $\vert \uparrow 0 \rangle$! In order to see it, consider the (finite dimensional) subspace of $\mathcal{H}_A$ spanned by the basis vectors:

$$\big\lbrace \vert \uparrow \rangle,\vert \downarrow \rangle \big \rbrace \otimes \big\lbrace \vert 0 \rangle,\vert 1 \rangle \big \rbrace=\big\lbrace \vert \uparrow 0 \rangle,\vert \uparrow 1 \rangle, \vert \downarrow 0 \rangle,\vert \downarrow 1 \rangle \big \rbrace.$$ The matrix representation of $R_A(\pi,0)$ in this subspace is:

\begin{equation} R_A(\pi,0)=\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 0 & -i & 0\\ 0 & -i & 0 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} . \end{equation} And the basis vectors can be taken as

$$ \vert \uparrow 0\rangle=\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix}, \vert \uparrow 1\rangle=\begin{pmatrix} 0 \\ 1 \\ 0 \\ 0\end{pmatrix}, \vert \downarrow 0\rangle=\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0\end{pmatrix}, \vert \downarrow 1\rangle=\begin{pmatrix} 0 \\ 0 \\ 0 \\ 1\end{pmatrix}. $$

So that all properties hold. I hope it also clarifies the question in the comment! C:

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you very much! This already makes the general procedure a lot clearer. I tried to find a vector expression for $|\uparrow 0 \rangle_A$ by solving $R_A(\pi,0)|\uparrow 0 \rangle_A=|\uparrow 0 \rangle_A$. However, the only solution is the trivial solution. In fact, I am confused why the authors only specified $(10)=|↑⟩|1⟩$, $(01)=|↓⟩|0⟩$ although we are dealing with tensor states and thus a complete basis would be $(10) \otimes(10)$, $(10) \otimes(01)$, $(01) \otimes(10)$, $(01) \otimes(01)$. But then how is the rotation $R_A(\pi,0)$ defined on those basis vectors? $\endgroup$ – doncarlos31415 May 29 at 10:50
  • $\begingroup$ I edited my answer because I couldn't answer your question in the comments! Cheers! Hope it helps!! $\endgroup$ – vin92 May 29 at 16:13
  • $\begingroup$ Thank you soo much for your detailed answer!! :) $\endgroup$ – doncarlos31415 May 29 at 18:00

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.