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!


1 Answer 1


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:

  • 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? $\endgroup$ Commented May 29, 2020 at 10:50
  • $\begingroup$ I edited my answer because I couldn't answer your question in the comments! Cheers! Hope it helps!! $\endgroup$
    – vin92
    Commented May 29, 2020 at 16:13
  • $\begingroup$ Thank you soo much for your detailed answer!! :) $\endgroup$ Commented May 29, 2020 at 18:00

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