Sakurai 3.21, Cartesian eigenbasis representation

[Silicon-Based]

Homework Statement

I struggle to understand the given argument for how to represent the three states in the cartesian eigenbasis. I tried writing it out but with no success. I would appreciate if someone could walk me through it step-by-step.

Relevant Equations

Given in picture.

 

[BvU]

step-by-step

All steps ? Do you understand the first step (from (2) to the 'we have' triplet) ?

 

[Silicon-Based]

All steps ? Do you understand the first step (from (2) to the 'we have' triplet) ?

What I'm confused about is how the representations were obtained from the closure relation. I understand everything before it. I don't see how to get rid of the inner products after making use of the closure relation.

 

[Abhishek11235]

What I'm confused about is how the representations were obtained from the closure relation. I understand everything before it. I don't see how to get rid of the inner products after making use of the closure relation.

Can you elaborate your problem in detail?

 

[Silicon-Based]

Can you elaborate your problem in detail?

For $N = 1 = n_x + n_y + n_z$ when you apply the completeness relation you get a sum states in coordinate basis for each $n_i=1$, for a total of three states, each with an inner product between the coordinate and spherical bases (the bra-kets on the very right in the completeness equation). I don't see how those turn into coefficients of $1/\sqrt{2}$ as presented in the final results.

 

[Abhishek11235]

I will give you sketch of the proof.I have forgotten orthogonality relations,so I derived them by myself using following procedure:

Now,I will translate the above problem into following:

$$|qlm>= c_1|100>+c_2|010>+c_3|001>$$

This is same as writing vector into its component form with coefficient and basis indicated.

Now,how do you determine those coefficients? Simply take the dot product with basis vectors and use orthonormality of basis vectors(Also,Use the relation that was derived just after equation (2).

Now,coming to your problem.
If you do above steps with each $|qlm>$ ,you won't get directly numbers. So,what we can do?

Well,Spherical harmonics are orthonormal! Using this we can find relation between coefficient. Again,after this you may not get directly numbers. So,then what?

We have beautiful relationship between spherical harmonics:

$$ Y_{l}^{-m}=(-1)^m[Y_{l}^{m}]*$$
(c.f Sakurai)

This can be used to find coefficients by elimination and solve your problem. A moment thought will show you that you have solved a variant of this problem when we want to expand $|S_x,+>$ in ##|S_z> form in 1st chapter of book.

 

[lucaschs]

I am trying to carry out the same relations and I have noticed that the expression (2) is only compatible if the state is acted \begin{align*}a_x^{\dagger}a_y\ket{n_xn_yn_z}\end{align*} with the conjugate complex, that is, by making
\begin{align*} \bra{n_xn_yn_z}a_xa_y^{\dagger}&=\bigg(a_x^{\dagger}a_y\ket{n_xn_yn_z}\bigg)^{\dagger}=\bigg(\sqrt{(n_x+1)n_y}\ket{n_x+1,n_y-1,n_z}\bigg)^{\dagger}\\
=&\sqrt{(n_x+1)n_y}\bra{n_x+1,n_y-1,n_z}
\end{align*}
So
\begin{equation*}
\bra{n_xn_yn_z}a_x^{\dagger}a_y=\bigg(a_xa_y^{\dagger}\ket{n_xn_yn_z}\bigg)^{\dagger}=
\sqrt{n_x(n_y+1)}\bra{n_x-1,n_y+1,n_z}
\end{equation*}
This is the form to find the expression (2).

Later,
\begin{align*}m\braket{100|01m}=-i\braket{010|01m}\hspace{2cm}N=1, n_x=1,n_y=n_z=0\end{align*} and so with the others relations


Starting with the eigenstate $$q=0,l=1,m=1$$ , it holds that the sum of terms is

\begin{equation*}
\ket{011}=\braket{100|011}\ket{100} +\braket{010|011}\ket{010}+\braket{001|011}\ket{001}
\end{equation*}
when $$m=1$$, so the relations are
\begin{align*}
\alpha&=\braket{100|011}=-i\braket{010|011}\\
\beta&=\braket{010|011}=i\braket{100|011}\\
\gamma&=\braket{001|011}=0
\end{align*}

For ortonormality of states , $$\braket{011|011}=1$$, such that
\begin{equation*}
1=|\braket{100|011}|^2+|\braket{010|011}|^2=|\alpha|^2+|\beta|^2
\end{equation*}
Observing,
\begin{align*}
\alpha=-i\beta\\
\beta=i\alpha
\end{align*}
Therefore,
\begin{equation*}
1=|\alpha|^2+|i\alpha|^2=2|\alpha|^2\implies \alpha=\frac{1}{\sqrt{2}}
\end{equation*}
Reemplacing this value of α and β, it is concluded that the state $$\ket{q=0,l=1,m=1}$$ is
\begin{equation*}
\ket{011}=\frac{1}{\sqrt{2}}\ket{100}+\frac
{i}{\sqrt{2}}\ket{010}.
\end{equation*}

I hope it will help. Greetings