Minimizing angular momentum uncertainties

[mrbetadine]

Homework Statement

Consider a physical system of fixed angular momentum $l$. The state of the system is in the subspace spanned by $2l+1$ eigenvectors $|l,m\rangle$ of $L_z$ ($-l\leq m\leq +l$). Find the state $|\psi_0\rangle$ of the system for which $(\Delta L_x)^2+(\Delta L_y)^2+(\Delta L_z)^2$ is minimal.

Relevant Equations

n/a

\begin{align*}
&(\Delta L_x)^2+(\Delta L_y)^2+(\Delta L_z)^2\\
={}&\langle L_x^2\rangle-\langle L_x \rangle^2+\langle L_y^2\rangle-\langle L_y \rangle^2+\langle L_z^2\rangle-\langle L_z \rangle^2\\
={}&\langle L_x^2+L_y^2+L_z^2 \rangle-(\langle L_x \rangle^2+\langle L_y \rangle^2+\langle L_z \rangle^2)\\
={}&l(l+1)\hbar^2-(\langle L_x \rangle^2+\langle L_y \rangle^2+\langle L_z \rangle^2).
\end{align*}
To minimize $(\Delta L_x)^2+(\Delta L_y)^2+(\Delta L_z)^2$ is equivalent to maximizing
$$ \langle L_x \rangle^2+\langle L_y \rangle^2+\langle L_z \rangle^2$$

How should I proceed?
I have also obtained an alternative expression
$$\langle L_x \rangle^2+\langle L_y \rangle^2+\langle L_z \rangle^2=\langle L_+\rangle\langle L_-\rangle+\langle L_z\rangle^2 $$
where
$$ L_\pm=L_x\pm i L_y$$
but it does not help much. It would be easy if we restrict ourselves to finding states of the form $|l,m\rangle$.

*The hint I got indicates that $|\psi_0\rangle$ is the solution to the equations
\begin{align*}
(L_x+iL_y)|\psi_0\rangle&=0\\
L_z|\psi_0\rangle&=l\hbar|\psi_0\rangle.
\end{align*}

 

[anuttarasammyak]

By the hint
$$<L_x>=<L_y>=0$$
$$<L_z>=l\hbar$$

 

[JimWhoKnew]

There may be a simple and elegant solution to the problem, but I can't see it currently. So I'll suggest a brute force approach.

Note that if $~|\psi_0\rangle~$ minimizes the uncertainty, then any state derived from it by a rotation is just as good.

Any normalized state in the $l$ subspace can be spanned as$$|\psi\rangle=\sum^l_{m=-l}\left(\alpha_m+i\beta_m\right)|l,m\rangle\quad,$$where $~\{\alpha_m,\beta_m\}~$ are $~4l+2~$ real coefficients.
As you concluded in post #1, minimizing the uncertainty is the same as maximizing$$f:=\langle L_+\rangle\langle L_-\rangle+\langle L_z\rangle^2 $$(which can now be expressed explicitly as a real function of the real $~\{\alpha_m,\beta_m\}~$), subjected to the normalization constraint$$\chi:=\langle\psi|\psi\rangle-1=\sum\left(\alpha_m^2+\beta_m^2\right)-1=0\quad.$$The method of Lagrange multipliers is usually applicable in cases like this.

 

[julian]

We wish to maximise $\langle L_x \rangle^2 + \langle L_y \rangle^2 + \langle L_z \rangle^2$. Write

\begin{align*}
| \psi_0 \rangle = \sum_{m=-l}^l a_m |l,m\rangle
\end{align*}

I get

\begin{align*}
& | \langle L_+\rangle |^2 + \langle L_z\rangle^2 =
\nonumber \\
& = \hbar^2\, \left| \sum_{m=-l}^{l-1} a_{m+1}^* a_m \sqrt{l(l+1)-m(m+1)} \right|^2 + \hbar^2\, (\sum_{m=-l}^l |a_m|^2m)^2
\end{align*}

We want to maximise this subject to $\sum_{m=-l}^l |a_m|^2 = 1$.

The value of the first sum is maximised when all terms have the same phase. Choose $a_m = |a_m| e^{-i m \phi - i \phi_0}$, then

\begin{align*}
& | \langle L_+\rangle |^2 + \langle L_z\rangle^2 =
\nonumber \\
& = \hbar^2\, \left( \sum_{m=-l}^{l-1} x_{m+1} x_m \sqrt{l(l+1)-m(m+1)} \right)^2 + \hbar^2\, (\sum_{m=-l}^l x_m^2m)^2
\end{align*}

where $x_m = |a_m|$. We want to maximise this subject to $\sum_{m=-l}^l x_m^2 = 1$.

Let us maximise $f(x_m) = \hbar \sqrt{S^2+M^2}$, where

\begin{align*}
S = \sum_{m=-l}^{l-1} x_{m+1} x_m \sqrt{l(l+1)-m(m+1)} , \quad M = \sum_{m=-l}^l x_m^2m
\end{align*}

Method of Lagrange multipliers gives the eigenvector equation

\begin{align*}
\frac{\partial}{\partial x_m} f (x_m) - \lambda \frac{\partial}{\partial x_m} (\sum_{m=-l}^l x_m^2 - 1) = 0
\end{align*}

Thus

\begin{align*}
\hbar\, \frac{S}{\sqrt{S^2+M^2}} \frac{\partial S}{\partial x_m} + \hbar\, \frac{M}{\sqrt{S^2+M^2}} \frac{\partial M}{\partial x_m} - \lambda 2 x_m = 0
\end{align*}

Define

\begin{align*}
\sin \alpha = \frac{S}{\sqrt{S^2+M^2}} , \quad \cos \alpha = \frac{M}{\sqrt{S^2+M^2}} ,
\end{align*}

then the eigenvector equation becomes

\begin{align*}
\hbar\, \sin \alpha \frac{\partial S}{\partial x_m} + \hbar\, \cos \alpha \frac{\partial M}{\partial x_m} - \lambda 2 x_m = 0
\end{align*}

Perform $\frac{\partial S}{\partial x_m}$ and $\frac{\partial M}{\partial x_m}$.

Calculate $\langle l,m | \, e^{-i \phi L_z/\hbar} L_{x} e^{i \phi L_z/\hbar} | \psi_0 \rangle$.

Use $e^{-i \phi L_z/\hbar} L_{x} e^{i \phi L_z/\hbar} = \cos \phi L_{x} + \sin \phi L_{y}$. Substituting into the eigenvector equation, we obtain

\begin{align*}
\sin \alpha (\cos \phi L_{x} + \sin \phi L_{y}) \,|\psi_0 \rangle + \cos \alpha L_{z} \,|\psi_0 \rangle = \lambda |\psi_0 \rangle
\end{align*}

or

\begin{align*}
\vec{L} \cdot \vec{n} |\psi_0 \rangle = \lambda |\psi_0 \rangle
\end{align*}

where $\vec{n} = (\sin \alpha \cos \phi , \sin \alpha \sin \phi, \cos \alpha)$. Without loss of generality, you can consider

\begin{align*}
L_z |\psi_0 \rangle = \lambda |\psi_0 \rangle
\end{align*}

and find the eigenvector and eigenvalue that maximises $\langle L_x \rangle^2 + \langle L_y \rangle^2 + \langle L_z \rangle^2$. For eigenstates of $L_{z}$, $\langle l,m |L_x | l,m \rangle = \langle l,m | L_y | l,m \rangle = 0$ and $\langle l,m | L_z | l,m \rangle^2 = m^2$. Obviously the states $| l,l \rangle$ and $| l,-l \rangle$ maximises $\langle L_x \rangle^2 + \langle L_y \rangle^2 + \langle L_z \rangle^2$.

States that also maximises $\langle L_x \rangle^2 + \langle L_y \rangle^2 + \langle L_z \rangle^2$, are obtained from the state $| l,l \rangle$ by rotation.

 

[JimWhoKnew]

Very nice. Thanks, @julian.

Actually, it is easier to show by variation that $~|l,l\rangle~$ (or its rotations) yield a maximum for$$f:=\langle L_+\rangle\langle L_-\rangle+\langle L_z\rangle^2\quad.$$The shortcoming of this variation approach is that we still have to prove that this maximum is global.

Consider the variation$$|\psi\rangle=\left(1-\varepsilon^2\sum^{l-1}_{m=-l}\left|a_m\right|^2\right)^{1/2}|l,l\rangle+\varepsilon\sum^{l-1}_{m=-l}a_m|l,m\rangle\quad,$$where $~\varepsilon\ll 1~$. Then
1. The lowest order correction to $f$ is proportional to $~\varepsilon^2~$.
2. The contribution of the left term of $f$ to the $~a_{l-1}$-dependence of the $~\varepsilon^2~$ correction, is canceled by the contribution of the right term.
3. For $~m\le l-2~$, only the (diagonal) right term of $f$ contributes to the $~\varepsilon^2~$ correction, and these contributions are either zero (rotation) or negative.