Matrix Representation for Combined Ladder Operators

[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.

 

[DrClaude]

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.

 

[Raptor112]

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.

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?

 

[vela]

What basis are you working with in your project?

 

[Raptor112]

What basis are you working with in your project?

The Fock states

 

[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_{+})

 

[Raptor112]

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_{+})

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

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

 

[Fightfish]

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

 

[Raptor112]

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

$(M^{\dagger})^{\dagger}= M$
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?

 

[Fightfish]

If <br /> M =<br /> \begin{bmatrix}<br /> A &amp; B \\<br /> C &amp; D \\<br /> \end{bmatrix},<br />
what is M^{\dagger}?

 

[Raptor112]

If <br /> M =<br /> \begin{bmatrix}<br /> A &amp; B \\<br /> C &amp; D \\<br /> \end{bmatrix},<br />
what is M^{\dagger}?

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.

 

[DrClaude]

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.

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?

 

[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.

 

[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_{+}.

 

[Raptor112]

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_{+}.

That makes sense. Thanks

 

[Raptor112]

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_{+}.

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)?

 

[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.

 

[Raptor112]

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.

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}

 

[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.