\def\bra#1{\hbox{$\left<#1\right|$}}
\def\ket#1{\hbox{$\left|#1\right>$}}
\def\braket#1#2{\hbox{$\left<#1|#2\right>$}}
\def\bmit{\bf}
\font\cmbig=cmr10 scaled\magstep1
\def\dg{^{\dag}}
\parindent=0pt
\hfill{\cmbig 8.05 Problem Set 2}
\smallskip\hrule\smallskip
\hfill{\cmbig Andrew Marc Greene}
\vskip.25in
\centerline{\vbox{\hsize=3.5in\hbox{\it Is this $A\dg$ I see before me,}
\hbox{\it The handle toward my hand?}
\hfill\hbox{--- Macbeth, Act II, Scene 1}}}
\vskip.5in

{\bf Problem 1.}  Let {\bf A} be an arbitrary linear operator
(not necessarily Hermitian), prove
$$\bra{\psi}A\dg A\ket{\psi} \ge 0$$

\medskip
$$\eqalign{
&\bra{\psi}A\dg A\ket{\psi} \cr
=&\sum_n \bra{\psi}A\dg\ket{n} \bra{n}A\ket{\psi} \cr
=&\sum_n \left(\bra{n}A\ket{\psi}\right)^* \ket{n}A\bra{\psi} \cr
=&\sum_n \left|\bra{n}A\ket{\psi}\right|^2 \cr
\ge&0}$$

\bigskip
{\bf Problem 2.} Consider a three-dimensional space of kets.  Choose a
certain orthonormal basis, and denote the basis kets as \ket1, \ket2,
\ket3.  In this basis, the operators {\bf A} and {\bf B} are
represented by the given matrices.

{\bf (a)} What are the eigenvalues of {\bf A}?

$$\eqalign{
{\bmit A}\ket{c} &= c\ket{c} \cr
\left|\matrix{a-c&0&0 \cr 0&-a-c&0 \cr 0&0&-a-c \cr}\right| &= 0 \cr
(a-c)(-a-c)^2 = (a-c)(a+c)^2 &= 0 \cr
a &=\pm c }$$

\medskip
{\bf (b)} What are the eigenvalues of {\bf B}?

$$\eqalign{
{\bmit B}\ket{d} &= d\ket{d} \cr
\left|\matrix{b-d&0&0 \cr 0&-d&-ib \cr 0&ib&-d \cr}\right| &= 0 \cr
(b-d)(d^2-b^2) &= 0 \cr
(b-d)^2(b+d) &= 0 \cr
d &= \pm b }$$

\medskip
{\bf (c)} Show that {\bf A} and {\bf B} commute.

$$\eqalign{
& {\bmit AB}-{\bmit BA} \cr
=& \left(\matrix{a&0&0\cr0&-a&0\cr0&0&-a\cr}\right)
\left(\matrix{b&0&0\cr0&0&-ib\cr0&ib&0\cr}\right) -
\left(\matrix{b&0&0\cr0&0&-ib\cr0&ib&0\cr}\right)
\left(\matrix{a&0&0\cr0&-a&0\cr0&0&-a\cr}\right) \cr
=& \left(\matrix{ab&0&0\cr0&0&iab\cr0&-iab&0\cr}\right) -
\left(\matrix{ab&0&0\cr0&0&iab\cr0&-iab&0\cr}\right) \cr
=& \left(\matrix{0&0&0\cr0&0&0\cr0&0&0\cr}\right)
}$$

\medskip
{\bf (d)} Find a new set of orthonormal kets which are simultaneously
eigenkets of both {\bf A} and {\bf B}.  What are the eigenvalues of
{\bf A} and {\bf B} for these new eigenkets?

$$\eqalign{
\left(\matrix{a&0&0\cr0&-a&0\cr0&0&-a}\right)
\left(\matrix{1\cr w_2\cr w_3}\right) & = a
\left(\matrix{1\cr w_2\cr w_3}\right) \cr
\left(\matrix{a\cr-aw_2\cr-aw_3}\right) & =
\left(\matrix{a\cr aw_2\cr aw_3}\right) \Longrightarrow
\left(\matrix{1\cr0\cr0}\right)
\cr
& \vbox to .25in{\qquad} \cr
\left(\matrix{b&0&0\cr0&0&-ib\cr0&ib&0}\right)
\left(\matrix{x_1\cr1\cr x_3}\right) &= b
\left(\matrix{x_1\cr1\cr x_3}\right) \cr
\left(\matrix{b\,x_1\cr-ib\,x_3\cr ib}\right) &=
\left(\matrix{b\,x_1\cr b\cr b\,x_3}\right) \Longrightarrow
\sqrt{1\over2}\left(\matrix{0\cr1\cr -i}\right)
\cr
& \vbox to .25in{\qquad} \cr
\left(\matrix{b&0&0\cr0&0&-ib\cr0&ib&0}\right)
\left(\matrix{y_1\cr y_2\cr1}\right) &= -b
\left(\matrix{y_1\cr y_2\cr1}\right) \cr
\left(\matrix{by_1\cr-ib\cr iby_2}\right) &=
\left(\matrix{-by_1\cr-by_2\cr-b}\right) \Longrightarrow
\sqrt{1\over2}\left(\matrix{0\cr i\cr1}\right)
}$$

\bigskip
{\bf Problem 3.}  Consider a hypothetical quantum mechanical system
with only two independent states.  Suppose you are given an
orthonormal basis, consisting of two states, \ket1 and \ket2, with
$\braket11 = \braket22 = 1$ and $\braket12 = 0$.

Suppose {\bf A} is an operator which acts on these states as follows:
$$\eqalign{
{\bmit A}\ket1 &= \alpha\ket1 + \beta\ket2 \cr
{\bmit A}\ket2 &= \gamma\ket1}$$
where $\alpha$, $\beta$, and $\gamma$ are complex constants.

$$\eqalign{
A\left(\matrix{1\cr0}\right) = \alpha\left(\matrix{1\cr0}\right) +
\beta\left(\matrix{0\cr1}\right) &=
\left(\matrix{\alpha\cr\beta}\right) \cr
\alpha\left(\matrix{0\cr1}\right) = \gamma\left(\matrix{1\cr0}\right)
+ 0 \left(\matrix{0\cr1}\right) &=
\left(\matrix{\gamma\cr0}\right)
} $$
$$ A = \left(\matrix{\alpha & \gamma \cr \beta & 0}\right)
\hbox to 2in{\quad}
A\dg = \left(\matrix{\alpha^* & \beta^* \cr \gamma^* & 0}\right) $$

\medskip
{\bf (a)} What is $\bra1A\dg$?  What is $\bra2A\dg$?

$$\eqalign{
\bra1A\dg &= (1\quad 0)\left(\matrix{\alpha^* & \beta^* \cr \gamma^* &
0}\right) = (\alpha^*\quad \beta^*) \cr
\bra2A\dg &= (0\quad 1)\left(\matrix{\alpha^* & \beta^* \cr \gamma^* &
0}\right) = (\gamma^*\quad 0)
} $$

\medskip
{\bf (b)} What is $\bra1A$?  What is $\bra2A$?

$$\eqalign{
\bra1A &= (1\quad 0)\left(\matrix{\alpha & \gamma \cr \beta & 0}\right)
= (\alpha\quad \gamma) \cr
\bra2A &= (0\quad 1)\left(\matrix{\alpha & \gamma \cr \beta & 0}\right)
= (\beta\quad 0)
} $$

\medskip
{\bf (c)} What is $A\dg\ket1$?  What is $A\dg\ket2$?

$$\eqalign{
A\dg\ket1 &= \left(\matrix{\alpha^* & \beta^* \cr \gamma^* & 0}\right)
\left(\matrix{1\cr0}\right) =
\left(\matrix{\alpha^*\cr\gamma^*}\right) \cr
A\dg\ket2 &= \left(\matrix{\alpha^* & \beta^* \cr \gamma^* & 0}\right)
\left(\matrix{0\cr1}\right) =
\left(\matrix{\beta^*\cr0}\right)
}$$

\medskip
{\bf (d)} Find the condition on $\alpha$, $\beta$, and $\gamma$ such
that the operator {\bf A} is Hermitian.

$$A=A\dg \Longrightarrow \hbox{$\alpha$ is real, $\beta=\gamma\dg$}$$

\medskip
{\bf (e)} Find the eigenvectors and eigenvalues of {\bf A}.  Express
the eigenvectors as linear superpositions of \ket1 and \ket2.  Show
that if $\alpha$, $\beta$, and $\gamma$ have been chosen to make {\bf
A} Hermitian, then the eigenvalues of {\bf A} are real.

$$\eqalign{
\left|\matrix{\alpha-a&\gamma\cr\beta&-a}\right| &= 0 \cr
a^2-\gamma a - \beta\gamma &= 0 \cr
a &= {{\alpha\pm\sqrt{\alpha^2+4\beta\gamma}}\over2} \cr
& \vbox to .25in{\quad} \cr
\left(\matrix{\alpha&\gamma\cr\beta&0}\right)
\left(\matrix{a_{11}\cr a_{12}}\right) &=
{{\alpha+\sqrt{\alpha^2+4\beta\gamma}}\over2}
\left(\matrix{a_{11}\cr a_{12}}\right) \cr
2\left(\matrix{\alpha a_{11} + \gamma a_{12} \cr \beta a_{11}}\right)
&= \left(\alpha+\sqrt{\alpha^2+4\beta\gamma}\right)
\left(\matrix{a_{11}\cr a_{12}}\right) \cr
& \vbox to .25in{\quad} \cr
2\left(\matrix{\alpha a_{21} + \gamma a_{22} \cr \beta a_{21}}\right)
&= \left(\alpha-\sqrt{\alpha^2+4\beta\gamma}\right)
\left(\matrix{a_{21}\cr a_{22}}\right) \cr
}$$

The eigenvectors are therefore

$$\eqalign{
\ket{a_1} &= \left(\alpha+\sqrt{\alpha^2+4\beta\gamma}\right)\ket1 +
2\beta\ket2 \cr
\ket{a_2} &= \left(\alpha-\sqrt{\alpha^2+4\beta\gamma}\right)\ket1 +
2\beta\ket2
}$$

If $A=A\dg$ then $\beta\gamma\ge0$, then the radicand is positive and
the radical is real.

\medskip

{\bf (f)} Find the condition on $\alpha$, $\beta$, and $\gamma$ such
that the operator {\bf A} is unitary.

$$\eqalign{
A\dg A &= {\bf 1} \cr
\left(\matrix{\alpha^*&\beta^*\cr\gamma^*&0}\right)
\left(\matrix{\alpha&\beta\cr\gamma&0}\right) &=
\left(\matrix{1&0\cr0&1}\right)
} $$
$$\left.\matrix{
\alpha^*\alpha+\beta^*\beta &= 1 \cr
\alpha^*\gamma &= 0 \cr
\gamma^*\alpha &= 0 \cr
\gamma^*\gamma &= 1}\right\} \Longrightarrow
\left\{\matrix{
\alpha = 0 \cr
\left|\gamma\right| = 1 \cr
\left|\beta\right| = 1}\right.$$

\medskip

{\bf (g)} Solve the eigenvalue problem for the operator $A\dg$.  Show
that the eigenvalues of $A\dg$ are only equal to the eigenvalues found
in part (e) if $A$ is Hermitian.  What is the relation between the
eigenvalues of $A\dg$ and the eigenvalues of $A$ acting on bras?

$$\eqalign{
A\dg-\lambda I &= 0 \cr
\left|\matrix{\alpha^*-\lambda & \beta^* \cr \gamma^* &
-\lambda}\right| &= 0 \cr
\lambda^2 - \alpha^*\lambda - \beta^*\gamma^* &= 0 \cr
\lambda &= {{\alpha^*\pm\sqrt{(\alpha^*)^2+4\beta^*\gamma^*}}\over2}
}$$

The eigenvectors are therefore

$$\eqalign{
\ket{a_3} &= \left(\alpha^*+\sqrt{(\alpha^*)^2+4\beta^*\gamma^*}\right)\ket1 +
2\beta^*\ket2 \cr
\ket{a_4} &= \left(\alpha^*-\sqrt{(\alpha^2)^*+4\beta^*\gamma^*}\right)\ket1 +
2\beta^*\ket2
}$$

The hermitivity of $A$ is a sufficient but {\it not necessary}
condition for the equality of these eigenvalues to those found in
part (e).  Working backwards, we get
$$\alpha^*\pm\sqrt{(\alpha^*)^2+4\beta^*\gamma^*} =
\alpha\pm\sqrt{\alpha^2+4\beta\gamma},$$
which requires that $\alpha=\alpha^*, \beta^*\gamma^*=\beta\gamma$.

The eigenvalues' action on bras is thus the complex conjugate of those
in (e).

\medskip

{\bf (h)} Find the action of the operator $A\dg A$ on the states \ket1
and \ket2.  Is $A\dg A$ hermitian?

$$\eqalign{
A\dg A &= \left(\matrix{\alpha^*\alpha-\beta^*\beta & \alpha^*\gamma
\cr \gamma^*\alpha & \gamma^*\gamma}\right) \cr
A\dg A\ket1 &= \left(\matrix{\cdot&\cdot\cr\cdot&\cdot}\right)
\left(\matrix{1\cr0}\right) = \left(\matrix{
\alpha^*\alpha+\beta^*\beta \cr \gamma^*\alpha}\right) \cr
A\dg A\ket2 &= \left(\matrix{\cdot&\cdot\cr\cdot&\cdot}\right)
\left(\matrix{0\cr1}\right) = \left(\matrix{
\alpha^*\gamma\cr\gamma^*\gamma}\right)
}$$

Yes, it is hermitian, since $(A\dg A)_{12}=[(A\dg A)_{21}]^*$.

\medskip

{\bf (i)} Calculate $\left[A,A\dg\right]$.

$$\eqalign{
\left[A,A\dg\right] &= AA\dg - A\dg A \cr
&= \left(\matrix{\alpha&\gamma\cr\beta&0}\right)
\left(\matrix{\alpha^*&\beta^*\cr\gamma^*&0}\right) -
\left(\matrix{\alpha^*&\beta^*\cr\gamma^*&0}\right)
\left(\matrix{\alpha&\gamma\cr\beta&0}\right) \cr
&= \left(\matrix{\alpha\alpha^*+\gamma\gamma^*&\alpha\beta^* \cr
\beta\alpha^* & \beta\beta^*}\right) -
\left(\matrix{\alpha^*\alpha+\beta^*\beta&\alpha^*\gamma \cr
\gamma^*\alpha & \gamma^*\gamma}\right) \cr
&= \left(\matrix{\gamma\gamma^*-\beta\beta^* &
\alpha\beta^*-\alpha^*\gamma \cr \alpha^*\beta-\alpha\gamma^* &
\beta\beta^*-\gamma^*\gamma}\right)
}$$

\medskip

{\bf (j)} Define the operators $B=A+A\dg$ and $C=-i(A-A\dg)$.  Show
both $B$ and $C$ are Hermitian.  Calculate their commutator.

$$B\dg = (A+A\dg)\dg = A\dg+A = A+A\dg = B$$
$$C\dg = ((-i)(A-A\dg))\dg = (-i)^*(A-A\dg)\dg = i(A\dg-A) =
-i(A-A\dg) = C$$

$$\eqalign{
[B,C] &= BC-CB = (A+A\dg)(-i)(A-A\dg) - (-i)(A-A\dg)(A+A\dg) \cr
&= (-i)\left(AA-AA\dg+A\dg A-A\dg A\dg - AA+A\dg A-AA\dg-A\dg A\dg\right) \cr
&= (2i)\left(AA\dg-A\dg A\right) \cr
&= (2i)[A,A\dg]
}$$

\medskip

{\bf (k)} What are the limits on the accuracy with which the
observable associated with $B$ and $C$ can be measured in

{\sl i) the state \ket1?}

$$\Delta B \Delta C \ \hbox{in state \ket1} \ge
{i\over2}\bra1[B,C]\ket1$$
$$=\left({i\over2}\right)\left(2i\right)
\left(\gamma\gamma^*-\beta\beta^*\right) = \beta\beta^*-\gamma\gamma^*
$$

{\sl ii) the state \ket2?}

Similarly, 
$$\Delta B \Delta C \ge \left({1\over2}\right)[B,C]_{22} =
\gamma\gamma^*-\beta\beta^*$$

{\sl iii) the eigenstates of $A$?}

By treating each eigenstate as a superposition of \ket1 and \ket2, we
can weight the answers to (k.i) and (k.ii), above, by
$\big|\braket{A_1}n\big|^2$ for $n=1,2$.  But, to be quite honest,
I've had quite enough symbol-pushing for one problem set.

\medskip

{\bf Problem 4.  Gasiorowicz \S7 Problem 1.}  (restated)  Given $[A,A\dg]=1$, show that $$A\ket{n}=\sqrt{n}\ket{n-1}$$

$$\eqalign{
A\ket{n} &= A\left({1\over\sqrt{n!}}\left(A\dg\right)^n\ket0\right) \cr
&= {1\over\sqrt{n!}}AA\dg\left(A\dg\right)^{n-1}\ket0 \cr
&= {1\over\sqrt{n}}{1\over\sqrt{(n-1)!}}AA\dg\left(A\dg\right)^{n-1}\ket0 \cr
&= {1\over\sqrt{n}}AA\dg{1\over\sqrt{(n-1)!}}\left(A\dg\right)^{n-1}\ket0 \cr
&= {1\over\sqrt{n}}\left(AA\dg-AA\dg+A\dg A+1\right)\ket{n-1} \cr
&= {1\over\sqrt{n}}\left(A\dg A+1\right)\ket{n-1} \cr
&= {1\over\sqrt{n}}\big((n-1)+1\big)\ket{n-1} \cr
&= {n\over\sqrt{n}}\ket{n-1} \cr
&= \sqrt{n}\ket{n-1}
}$$

\medskip

{\bf Problem 5.  Gasiorowicz \S8 Problem 2.}  (restated) Let
$\widehat{P_{12}}$ be defined by $\bra{x}\widehat{P_{12}}\ket{1,2} =
\braket{x}{2,1}$.  Show that $\widehat{P_{12}}=\widehat{P\dg_{12}}$.

$$\eqalign{
& \bra{1,2}\widehat{P_{12}}\ket{x} \cr
=& \int\!\int d\alpha\,d\beta\ \bra{1,2}\widehat{P_{12}}\ket{\alpha,\beta}
\braket{\alpha,\beta}{x} \cr
=& \int\!\int d\alpha\,d\beta\ \braket{1,2}{\beta,\alpha}
\left(\braket{x}{\alpha,\beta}\right)^* \cr
=& \int\!\int d\alpha\,d\beta\ \delta_{1\beta}\delta_{2\alpha} 
\left(\braket{x}{\alpha,\beta}\right)^* \cr
=& \left(\braket{x}{2,1}\right)^* \cr
=& \left(\bra{x}\widehat{P_{12}}\ket{1,2}\right)^* \cr
=& \bra{1,2}\widehat{P\dg_{12}}\ket{x}
}$$







\bye
