Matrix Representation for Combined Ladder Operators

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1. Jan 12, 2016

Raptor112

Due to the definition of spin-up (in my project ),
\begin{eqnarray}
\sigma_+ =
\begin{bmatrix}
0 & 2 \\
0 & 0 \\
\end{bmatrix}
\end{eqnarray}
as opposed to
\begin{eqnarray}
\sigma_+ =
\begin{bmatrix}
0 & 1 \\
0 & 0 \\
\end{bmatrix}
\end{eqnarray}
and the annihilation operator is
\begin{eqnarray}
\hat{a} =
\begin{bmatrix}
0 & \sqrt{1} & 0 & 0 & \dots\\
0 & 0 & \sqrt{2} & 0 &\dots\\
0 & 0 & 0 & \sqrt{3} & \dots\\
0 & 0 & 0 & 0 &\dots\\
\vdots & \vdots & \vdots & \vdots&\ddots\\
\end{bmatrix}
\end{eqnarray}

The matrix elememts of \begin{eqnarray} \hat{a}\hat{\sigma_+} \end{eqnarray} were given to me and are:

\begin{eqnarray}
\hat{a}\hat{\sigma_+} =
\begin{bmatrix}
0 & 0 & 0 & 0 & 0\\
0 & 0 & 2\sqrt{1} & 0 &0\\
0 & 0 & 0 & 0 &0\\
0 & 0& 0 & 0& 2\sqrt{2} \\
0 & 0& 0 & 0 &0\\
\end{bmatrix} \end{eqnarray}

From this I need to find out what the matrix elements of
\begin{eqnarray}
\hat{a^{\dagger}}\hat{\sigma_-}
\end{eqnarray}
are?

I suppose the issue is I don't know how to represent the atomic raising/lowering operator for dimenstions greater than 2.

2. Jan 12, 2016

Staff: Mentor

For any operator $\hat{A}$, you find its matrix representation in the basis $|n\rangle$ by calculating the matrix elements using
$$A_{mn} = \langle m | \hat{A} | n \rangle$$
You simply need to apply this to the ladder operators.

3. Jan 12, 2016

Raptor112

As $\hat{a}$ in my case is larger than a 2*2 matrix how does one find the elements for $\hat{\sigma}_+$ for a matrix of dimensions greater than 2*2?

4. Jan 12, 2016

vela

Staff Emeritus
What basis are you working with in your project?

5. Jan 12, 2016

Raptor112

The Fock states

6. Jan 13, 2016

Fightfish

The ordering of the basis is very important, especially when you are working with tensor product states as in here. I would hazard a guess that you are working on atom-light interactions in some sort of cavity? (like Jaynes-Cummings?)
In that case, your basis is a composite of the Fock states of the photon field and the spin of the atom: $\left|n\right\rangle \otimes \left|\sigma\right\rangle$. Once you determine how the basis in Eq. (5) are ordered, you should be able to get the matrix elements of $\hat{a}^{\dagger} \sigma_{-}$
(There is actually a very fast method of doing so if you can see the relationship between $\hat{a}^{\dagger} \sigma_{-}$ and $\hat{a} \sigma_{+}$)

7. Jan 13, 2016

Raptor112

$(\sigma_- a^{\dagger})^{\dagger} = a \sigma_+$
and
$(\sigma_+ a)^{\dagger} = a^{\dagger} \sigma_-$

but how does that give $a^{\dagger} \sigma_-$ from $a \sigma_+$

8. Jan 13, 2016

Fightfish

Given a matrix $M$, how is $M^{\dagger}$ related to it?

9. Jan 13, 2016

Raptor112

$(M^{\dagger})^{\dagger}= M$
so from:
$(\sigma_+ a)^{\dagger} = a^{\dagger} \sigma_-$
$((\sigma_-)^{\dagger} (a^{\dagger})^{\dagger})^{\dagger} = a^{\dagger} \sigma_-$
but then you just go in circles, I am I missing something?

10. Jan 13, 2016

Fightfish

If $$M = \begin{bmatrix} A & B \\ C & D \\ \end{bmatrix},$$
what is $M^{\dagger}$?

11. Jan 13, 2016

Raptor112

I think the question assumes that I know the matrix elements of $\sigma_+$, for a matrix of dimensions greater than 2*2 , which I don't.

12. Jan 13, 2016

Staff: Mentor

The $\sigma_+$ matrix you gave in the OP is 2×2, so there is something I don't understand. Could you give more details about what you are working on?

13. Jan 13, 2016

Raptor112

I am looking at the dynamics between a qubit and an electromagentic field in a cavity. I need to find the expectation value of $a^{\dagger} \sigma_-$ by contracting the density matrix and taking the trace. The dimensions of density matrix, in the project, can vary depending how many levels of the qubit and and how many levels of the cavity is chosen. All I need, is to set up $a^{\dagger} \sigma_-$ and the rest is done.

Last edited: Jan 13, 2016
14. Jan 13, 2016

Fightfish

You already have the matrix elements for $a \sigma_{+}$, which you have recognised to be the Hermitian conjugate of $a^{\dagger} \sigma_{-}$.
The "big" matrix you see for $a \sigma_{+}$ is not the matrix element of "$\sigma_{+}$ for more than 2 x 2 dimensions", which you seem to be confused about. The matrix given in the problem is the tensor product of $a$ and $\sigma_{+}$.

15. Jan 13, 2016

Raptor112

That makes sense. Thanks

16. Jan 13, 2016

Raptor112

So:
\begin{bmatrix}
0 & \sqrt{1} & 0 & 0 & \\
0 & 0 & \sqrt{2} & 0 \\
0 & 0 & 0 & \sqrt{3} \\
0 & 0 & 0 & 0 \\
\end{bmatrix} $\otimes$ \begin{bmatrix} 0 & 2 & \\ 0 & 0 & \\ \end{bmatrix} =

\begin{bmatrix}
0 & 0 & 0 & 2 \sqrt{1} & 0& 0 \dots \\
0 & 0 & 0 & 0 & 0 & 0 \dots \\
0 & 0 & 0 & 0 & 0 & 2\sqrt{2} \dots \\
0 & 0 & 0 & 0 & 0 & 0 \dots \\
\vdots & \vdots & \vdots & \vdots & \vdots&\ddots\\
\end{bmatrix}
which is not the same as eqn(5)?

Last edited: Jan 13, 2016
17. Jan 13, 2016

Fightfish

That is because the basis in Eqn (5) was reordered (for some unknown reason)
The direct tensor product you got in your previous post has the basis order $|0,+\rangle, |0,-\rangle,|1,+\rangle,|1,-\rangle \cdots$, while the basis for Eqn (5) has the order $|0,-\rangle, |0,+\rangle,|1,-\rangle,|1,+\rangle \cdots$.
Both are equally valid - as long as you keep proper track of the basis order.

18. Jan 14, 2016

Raptor112

So
$a^{\dagger} \sigma_-$=

\begin{bmatrix}
0 & 0 & 0 & 0 & 0& 0 &\dots \\
0 & 0 & 0 & 0 & 0 & 0 &\dots \\
0 & 2\sqrt{1}& 0 & 0 & 0 & 0& \dots \\
0 & 0 & 0 & 0 & 0 & 0 & \dots \\
0 & 0 & 0 & 2\sqrt{2} & 0& 0 &\dots \\
\vdots & \vdots & \vdots & \vdots & \vdots & \vdots&\ddots\\
\end{bmatrix}

19. Jan 15, 2016

Fightfish

Yup, that seems correct to me.
It'll probably be worthwhile to spend some time reviewing / learning about tensor product spaces, since you seemed to be confused at some points.

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