Compatible Observables
Consider a system whose state vector $|\psi>$ and two observables $A$ and $B$ are given by
$|\psi>=(1/\sqrt{5})\begin{pmatrix}-i\\2\\0\end{pmatrix}$; $A=\begin{pmatrix}1&i&1\\-i&0&0\\1&0&0\end{pmatrix}$; $B=\begin{pmatrix}3&0&0\\0&1&i\\0&-i&0\end{pmatrix}$
a) Are $A$ and $B$ compatible? Which set of operators ${A},{B},{A,B}$ form a complete set of commuting operators?
b) If $A$ is measured first and $B$ immediately afterward, find the probability of obtaining -1 for $A$ and 3 for $B$.
c) Now find the probabilities of getting the same values if the order of measurements is reversed.
a)
A and B are compatible if
$AB=BA$
$AB=\begin{pmatrix}1&i&1\\-i&0&0\\1&0&0\end{pmatrix}\begin{pmatrix}3&0&0\\0&1&i\\0&-i&0\end{pmatrix}=\begin{pmatrix}3&0&-1\\-3&0&0\\3&0&0\end{pmatrix}$
$BA=\begin{pmatrix}3&0&0\\0&1&i\\0&-i&0\end{pmatrix}\begin{pmatrix}1&i&1\\-i&0&0\\1&0&0\end{pmatrix}=\begin{pmatrix}3&3i&3\\0&0&0\\-1&0&0\end{pmatrix}$
Observables A and B are not compatible.
b)
${A,B}$ is NOT a CSCO (complete set of commuting operators) since $A$ and $B$ do not commute
The eigenvalues $x$ of $A$ are
$det (A-x*I)=0$ or $\begin{vmatrix}1-x&i&1\\-i&-x&0\\1&0&-x)\end{vmatrix}=x^2 (1-x)+x+x=-x^3+x^2+2x=0$
$x_1=2$, $x_2=-1$, $x_3=0$
So that $A$ is nondegenerate and also its eigenvectors. Therefore ${A}$ is a CSCO.
The nondegenerate eigenvectors $X_{1,2,3}$ of $A$ comes from the equation
$(A-x*I)*X=0$
$X_1=(1/2) \begin{pmatrix}2\\-i\\1\end{pmatrix}$ ; $X_2=(1/√3)\begin{pmatrix}-1\\-i\\1\end{pmatrix}$ ; $X_3=(1/√2)\begin{pmatrix}0\\i\\1\end{pmatrix}$
The vectors $X_{1,2,3}$ are also orthogonal since $A$ is hermitic $A=(A^T )^*$
The eigenvalues of $B$ are
$det (B-y*I)=0$, which means $y_1=3$, $y_{2,3}=(1/2)±(√5/2)$
So that $B$ is non degenerate and also its eigen vectors. Therefore ${B}$ is a CSCO
The non degenerate eigen vectors $Y_{1,2,3}$ of $B$ comes from the equation
$(B-y*I)*Y=0$
$Y_1=\begin{pmatrix}1\\0\\0\end{pmatrix}$; $Y_2=(2/\sqrt{9+2\sqrt{5}}) *\begin{pmatrix}0\\1/2 i(1+√5)\\1\end{pmatrix}$ ; $Y_3=(2/\sqrt{9-2\sqrt{5}}) *\begin{pmatrix}0\\(1/2) i(1-√5)\\1\end{pmatrix}$
b)
The possible results of measuring $A$ first are its eigenvalues (0,-1 and 2). The non degenerate eigen vectors of $A$ are $X_{1,2,3}$ (see above) . The vectors $X_{1,2,3}$ are also orthogonal since $A$ is hermitic $A=(A^T )^*$
Recall that the state of the system is
$|ψ>=(1/√5)*\begin{pmatrix}-i\\2\\0\end{pmatrix}$ and $<ψ|ψ>=(1/5)*(i,2,0)\begin{pmatrix}-i\\2\\0\end{pmatrix}=(1/5)*(1+4+0)=1$
The probability of obtaining the value $x_2=-1$ after a $A$ is measured first is
$P(x_2)=\frac{|<X_2 |ψ>|^2}{(<ψ|ψ>)}=\left |\frac{1}{√15}*(-1,+i,1)\begin{pmatrix}-i\\2\\0\end{pmatrix} \right |^2=\frac{1}{15}*|i+2i+0|^2=\frac{9}{15}$
After measuring A the state of the system changes to
$|ϕ>=|X_2><X_2 |ψ>=\frac{1}{√3}\begin{pmatrix}-1\\-i\\1\end{pmatrix}\frac{1}{√15}(-1,+i,1)\begin{pmatrix}-i\\2\\0\end{pmatrix}=\frac{3}{3√5}\begin{pmatrix}-i\\1\\i\end{pmatrix}=\frac{1}{√5}\begin{pmatrix}-i\\1\\i\end{pmatrix}$
$<ϕ|ϕ> =1/5*(i,1,-i)\begin{pmatrix}-i\\1\\i\end{pmatrix}=3/5$
The probability of finding the value $y_1=$3 after $B$ is measured after $A$ is thus
$P’ (y_1 )=\frac{|<Y_1 |ϕ>|^2}{<ϕ|ϕ>}=\frac{5}{3}\left|\frac{1}{√5}(1,0,0)\begin{pmatrix}-i\\1\\i\end{pmatrix} \right |^2=\frac{5}{3} \left |-\frac{i}{√5} \right |^2=\frac{1}{3}$
c)
The probability of obtaining the value $y_1=3$ after $B$ is measured first is
$P(y_1)=\frac{|<Y_1 |ψ>|^2}{<ψ|ψ>}=\left |\frac{1}{√5}(1,0,0)\begin{pmatrix}-i\\2\\0\end{pmatrix} \right |^2=\frac{1}{5}*|-i|^2=\frac{1}{5}$
After measuring $B$ the state of the system is changed to
$|ϕ>=|Y_1><Y_1 |ψ>=\begin{pmatrix}1\\0\\0\end{pmatrix}*\frac{1}{√5} (1,0,0)\begin{pmatrix}-i\\2\\0\end{pmatrix}=\frac{1}{√5} \begin{pmatrix}-i\\0\\0\end{pmatrix}$
$<ϕ|ϕ> =1/5$
The probability to obtain the value $x_2=-1$ when $A$ is measured after $B$ is
$P’ (x_2)=\frac{(|<X_2 |ϕ>|^2)}{(<ϕ|ϕ>)}=5 \left |\frac{1}{√3}\frac{1}{√5}(-1,i,1)\begin{pmatrix}-i\\0\\0\end{pmatrix} \right |^2=\frac{1}{3}|i|^2=\frac{1}{3}$
