## Time evolution of spin half probability and dispersion

October 15, 2015 phy1520 , , ,

## Question: Time evolution of spin half probability and dispersion ([1] pr. 2.3)

A spin $$1/2$$ system $$\BS \cdot \ncap$$, with $$\ncap = \sin \beta \xcap + \cos\beta \zcap$$, is in state with eigenvalue $$\Hbar/2$$, acted on by a magnetic field of strength $$B$$ in the $$+z$$ direction.

## (a)

If $$S_x$$ is measured at time $$t$$, what is the probability of getting $$+ \Hbar/2$$?

## (b)

Evaluate the dispersion in $$S_x$$ as a function of t, that is,

\label{eqn:spinTimeEvolution:20}
\expectation{\lr{ S_x – \expectation{S_x}}^2}.

## (c)

Check your answers for $$\beta \rightarrow 0, \pi/2$$ to see if they make sense.

## (a)

The spin operator in matrix form is
\label{eqn:spinTimeEvolution:40}
\begin{aligned}
S \cdot \ncap
&=
\frac{\Hbar}{2} \lr{ \sigma_z \cos\beta + \sigma_x \sin\beta } \\
&=
\frac{\Hbar}{2} \lr{ \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \cos\beta + \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \sin\beta } \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos\beta & \sin\beta \\
\sin\beta & -\cos\beta
\end{bmatrix}.
\end{aligned}

The $$\ket{S \cdot \ncap ; + }$$ eigenstate is found from

\label{eqn:spinTimeEvolution:60}
\lr{ S \cdot \ncap – \Hbar/2}
\begin{bmatrix}
a \\
b
\end{bmatrix}
= 0,

or

\label{eqn:spinTimeEvolution:80}
\begin{aligned}
0
&=
\lr{ \cos\beta – 1 } a + \sin\beta b \\
&=
\lr{ -2 \sin^2(\beta/2) } a + 2 \sin(\beta/2) \cos(\beta/2) b \\
&=
\lr{ – \sin(\beta/2) } a + \cos(\beta/2) b,
\end{aligned}

or

\label{eqn:spinTimeEvolution:100}
\ket{ S \cdot \ncap ; + }
=
\begin{bmatrix}
\cos(\beta/2) \\
\sin(\beta/2) \\
\end{bmatrix}.

The Hamiltonian is

\label{eqn:spinTimeEvolution:120}
H
= – \frac{e B}{m c} S_z
= – \frac{e B \Hbar}{2 m c} \sigma_z,

so the time evolution operator is

\label{eqn:spinTimeEvolution:140}
U
= e^{-i H t/\Hbar}
= e^{ \frac{i e B t }{2 m c} \sigma_z }.

Let $$\omega = e B/(2 m c)$$, so

\label{eqn:spinTimeEvolution:160}
\begin{aligned}
U
&=
e^{i \sigma_z \omega t} \\
&=
\cos(\omega t) + i \sigma_z \sin(\omega t) \\
&=
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}
\cos(\omega t)
+
i \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix} \sin(\omega t) \\
&=
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}.
\end{aligned}

The time evolution of the initial state is

\label{eqn:spinTimeEvolution:180}
\begin{aligned}
\ket{S \cdot \ncap ; + }(t)
&=
U \ket{S \cdot \ncap ; + }(0) \\
&=
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) \\
\sin(\beta/2) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix}.
\end{aligned}

The probability of finding the state in $$\ket{S \cdot \xcap ; + }$$ at time $$t$$ (i.e. measuring $$S_x$$ and finding $$\Hbar/2$$) is

\label{eqn:spinTimeEvolution:200}
\begin{aligned}
\Abs{\braket{S \cdot \xcap ; + }{S \cdot \ncap ; + }}^2
&=
\Abs{\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
\end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix}
}^2 \\
&=
\inv{2}
\Abs{
\cos(\beta/2) e^{i \omega t} +
\sin(\beta/2) e^{-i \omega t} }^2 \\
&=
\inv{2} \lr{ 1 + 2 \cos(\beta/2) \sin(\beta/2) \cos(2 \omega t) } \\
&=
\inv{2} \lr{ 1 + \sin(\beta) \cos( 2 \omega t) }.
\end{aligned}

## (b)

To calculate the dispersion first note that

\label{eqn:spinTimeEvolution:300}
S_x^2
= \lr{ \frac{\Hbar}{2} }^2 \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}^2
= \lr{ \frac{\Hbar}{2} }^2,

so only the first order expectation is non-trivial to calculate. That is

\label{eqn:spinTimeEvolution:320}
\begin{aligned}
\expectation{S_x}
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos(\beta/2) e^{-i \omega t} &
\sin(\beta/2) e^{i \omega t}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
\cos(\beta/2) e^{i \omega t} \\
\sin(\beta/2) e^{-i \omega t} \\
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
\cos(\beta/2) e^{-i \omega t} &
\sin(\beta/2) e^{i \omega t}
\end{bmatrix}
\begin{bmatrix}
\sin(\beta/2) e^{-i \omega t} \\
\cos(\beta/2) e^{i \omega t} \\
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\sin(\beta/2) \cos(\beta/2) \lr{ e^{-2 i \omega t} + e^{ 2 i \omega t} } \\
&=
\frac{\Hbar}{2} \sin\beta \cos( 2 \omega t ).
\end{aligned}

This gives

\label{eqn:spinTimeEvolution:340}
\boxed{
\expectation{(\Delta S_x)^2}
=
\lr{ \frac{\Hbar}{2} }^2 \lr{ 1 – \sin^2\beta \cos^2( 2 \omega t ) }.
}

## (c)

For $$\beta = 0$$, $$\ncap = \zcap$$, and $$\beta = \pi/2$$, $$\ncap = \xcap$$. For the first case, the state is in an eigenstate of $$S_z$$, so must evolve as

\label{eqn:spinTimeEvolution:220}
\ket{S \cdot \ncap ; + }(t) = \ket{S \cdot \ncap ; + }(0) e^{i \omega t}.

The probability of finding it in state $$\ket{S \cdot \xcap ; + }$$ is therefore

\label{eqn:spinTimeEvolution:240}
\begin{aligned}
\Abs{
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
e^{i \omega t} \\
0
\end{bmatrix}
}^2
&=
\inv{2} \Abs{ e^{i\omega t} }^2 \\
&=
\inv{2} \\
&=
\inv{2} \lr{ 1 + \sin(0) \cos(2 \omega t) }.
\end{aligned}

This matches \ref{eqn:spinTimeEvolution:200} as expected.

For $$\beta = \pi/2$$ we have

\label{eqn:spinTimeEvolution:260}
\begin{aligned}
\ket{S \cdot \xcap ; + }(t)
&=
\inv{\sqrt{2}}
\begin{bmatrix}
e^{i \omega t} & 0 \\
0 & e^{-i \omega t}
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix} \\
&=
\inv{\sqrt{2}}
\begin{bmatrix}
e^{i \omega t} \\
e^{-i \omega t}
\end{bmatrix}.
\end{aligned}

The probability for the $$\Hbar/2$$ $$S_x$$ measurement at time $$t$$ is
\label{eqn:spinTimeEvolution:280}
\begin{aligned}
\Abs{
\inv{2}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
e^{i \omega t} \\
e^{-i \omega t}
\end{bmatrix}
}^2
&=
\inv{4} \Abs{ e^{i \omega t} + e^{-i \omega t} }^2 \\
&=
\cos^2(\omega t) \\
&=
\inv{2}\lr{ 1 + \sin(\pi/2) \cos( 2 \omega t )}.
\end{aligned}

Again, this matches the expected value.

For the dispersions, at $$\beta = 0$$, the dispersion is

\label{eqn:spinTimeEvolution:360}
\lr{\frac{\Hbar}{2}}^2

This is the maximum dispersion, which makes sense since we are measuring $$S_x$$ when the initial state is $$\ket{S \cdot \zcap ; + }$$. For $$\beta = \pi/2$$ the dispersion is

\label{eqn:spinTimeEvolution:380}
\lr{\frac{\Hbar}{2}}^2 \sin^2 ( 2 \omega t ).

This starts off as zero dispersion (because the initial state is $$\ket{ S \cdot \xcap ; + }$$, but then oscillates.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## Question: Heisenberg picture spin precession ([1] pr. 2.1)

For the spin Hamiltonian

\label{eqn:heisenbergSpinPrecession:20}
H = -\frac{e B}{m c} S_z = \omega S_z,

express and solve the Heisenberg equations of motion for $$S_x(t), S_y(t)$$, and $$S_z(t)$$.

The equations of motion are of the form

\label{eqn:heisenbergSpinPrecession:40}
\begin{aligned}
\frac{dS_i^\textrm{H}}{dt}
&= \inv{i \Hbar} \antisymmetric{S_i^\textrm{H}}{H} \\
&= \inv{i \Hbar} \antisymmetric{U^\dagger S_i U}{H} \\
&= \inv{i \Hbar} \lr{U^\dagger S_i U H – H U^\dagger S_i U } \\
&= \inv{i \Hbar} U^\dagger \lr{ S_i H – H S_i } U \\
&= \frac{\omega}{i \Hbar} U^\dagger \antisymmetric{ S_i}{S_z } U.
\end{aligned}

These are

\label{eqn:heisenbergSpinPrecession:60}
\begin{aligned}
\frac{dS_x^\textrm{H}}{dt} &= -\omega U^\dagger S_y U \\
\frac{dS_y^\textrm{H}}{dt} &= \omega U^\dagger S_x U \\
\frac{dS_z^\textrm{H}}{dt} &= 0.
\end{aligned}

To completely specify these equations, we need to expand $$U(t)$$, which is

\label{eqn:heisenbergSpinPrecession:80}
\begin{aligned}
U(t)
&= e^{-i H t /\Hbar} \\
&= e^{-i \omega S_z t /\Hbar} \\
&= e^{-i \omega \sigma_z t /2} \\
&= \cos\lr{ \omega t/2 } -i \sigma_z \sin\lr{ \omega t/2 } \\
&=
\begin{bmatrix}
\cos\lr{ \omega t/2 } -i \sin\lr{ \omega t/2 } & 0 \\
0 & \cos\lr{ \omega t/2 } + i \sin\lr{ \omega t/2 }
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix}.
\end{aligned}

The equations of motion can now be written out in full. To do so seems a bit silly since we also know that $$S_x^\textrm{H} = U^\dagger S_x U, S_y^\textrm{H} U^\dagger S_x U$$. However, if that is temporarily forgotten, we can show that the Heisenberg equations of motion can be solved for these too.

\label{eqn:heisenbergSpinPrecession:100}
\begin{aligned}
U^\dagger S_x U
&=
\frac{\Hbar}{2}
\begin{bmatrix}
e^{i\omega t/2} & 0 \\
0 & e^{-i\omega t/2}
\end{bmatrix}
\begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t/2} \\
e^{-i\omega t/2} & 0
\end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix},
\end{aligned}

and
\label{eqn:heisenbergSpinPrecession:120}
\begin{aligned}
U^\dagger S_y U
&=
\frac{\Hbar}{2}
\begin{bmatrix}
e^{i\omega t/2} & 0 \\
0 & e^{-i\omega t/2}
\end{bmatrix}
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{i\Hbar}{2}
\begin{bmatrix}
0 & -e^{i\omega t/2} \\
e^{-i\omega t/2} & 0
\end{bmatrix}
\begin{bmatrix}
e^{-i\omega t/2} & 0 \\
0 & e^{i\omega t/2}
\end{bmatrix} \\
&=
\frac{i \Hbar}{2}
\begin{bmatrix}
0 & -e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix}.
\end{aligned}

The equations of motion are now fully specified

\label{eqn:heisenbergSpinPrecession:140}
\begin{aligned}
\frac{dS_x^\textrm{H}}{dt} &=
-\frac{i \Hbar \omega}{2}
\begin{bmatrix}
0 & -e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
\frac{dS_y^\textrm{H}}{dt} &=
\frac{\Hbar \omega}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
\frac{dS_z^\textrm{H}}{dt} &= 0.
\end{aligned}

Integration gives

\label{eqn:heisenbergSpinPrecession:160}
\begin{aligned}
S_x^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} + C \\
S_y^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & -i e^{i\omega t} \\
i e^{-i\omega t} & 0
\end{bmatrix} + C \\
S_z^\textrm{H} &= C.
\end{aligned}

The integration constants are fixed by the boundary condition $$S_i^\textrm{H}(0) = S_i$$, so

\label{eqn:heisenbergSpinPrecession:180}
\begin{aligned}
S_x^\textrm{H} &=
\frac{\Hbar}{2}
\begin{bmatrix}
0 & e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
S_y^\textrm{H} &=
\frac{i \Hbar}{2}
\begin{bmatrix}
0 & – e^{i\omega t} \\
e^{-i\omega t} & 0
\end{bmatrix} \\
S_z^\textrm{H} &= S_z.
\end{aligned}

Observe that these integrated values $$S_x^\textrm{H}, S_y^\textrm{H}$$ match \ref{eqn:heisenbergSpinPrecession:100}, and \ref{eqn:heisenbergSpinPrecession:120} as expected.

# References

[1] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics. Pearson Higher Ed, 2014.

## An observation about the geometry of Pauli x,y matrices

July 19, 2015 phy1520 , , , ,

## Motivation

The conventional form for the Pauli matrices is

\label{eqn:pauliMatrixXYgeometry:20}
\begin{aligned}
\sigma_x &=
\begin{bmatrix}
0 & 1 \\
1 & 0 \\
\end{bmatrix} \\
\sigma_y &=
\begin{bmatrix}
0 & -i \\
i & 0 \\
\end{bmatrix} \\
\sigma_z &=
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\end{aligned}.

In [1] these forms are derived based on the commutation relations

\label{eqn:pauliMatrixXYgeometry:40}
\antisymmetric{\sigma_r}{\sigma_s} = 2 i \epsilon_{r s t} \sigma_t,

by defining raising and lowering operators $$\sigma_{\pm} = \sigma_x \pm i \sigma_y$$ and figuring out what form the matrix must take. I noticed an interesting geometrical relation hiding in that derivation if $$\sigma_{+}$$ is not assumed to be real.

## Derivation

For completeness, I’ll repeat the argument of [1], which builds on the commutation relations of the raising and lowering operators. Those are

\label{eqn:pauliMatrixXYgeometry:60}
\begin{aligned}
\antisymmetric{\sigma_z}{\sigma_{\pm}}
&=
\sigma_z \lr{ \sigma_x \pm i \sigma_y }
-\lr{ \sigma_x \pm i \sigma_y } \sigma_z \\
&=
\antisymmetric{\sigma_z}{\sigma_x} \pm i \antisymmetric{\sigma_z}{\sigma_y} \\
&=
2 i \sigma_y \pm i (-2 i) \sigma_x \\
&= \pm 2 \lr{ \sigma_x \pm i \sigma_y } \\
&= \pm 2 \sigma_{\pm},
\end{aligned}

and

\label{eqn:pauliMatrixXYgeometry:80}
\begin{aligned}
\antisymmetric{\sigma_{+}}{\sigma_{-}}
&=
\lr{ \sigma_x + i \sigma_y } \lr{ \sigma_x – i \sigma_y }
-\lr{ \sigma_x – i \sigma_y } \lr{ \sigma_x + i \sigma_y } \\
&=
-i \sigma_x \sigma_y + i \sigma_y \sigma_x
– i \sigma_x \sigma_y + i \sigma_y \sigma_x \\
&= 2 i \antisymmetric{ \sigma_y }{\sigma_x} \\
&= 2 i (-2i) \sigma_z \\
&= 4 \sigma_z
\end{aligned}

From these a matrix representation containing unknown values can be assumed. Let

\label{eqn:pauliMatrixXYgeometry:100}
\sigma_{+} =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}.

The commutator with $$\sigma_z$$ can be computed

\label{eqn:pauliMatrixXYgeometry:120}
\begin{aligned}
\antisymmetric{\sigma_z}{\sigma_{+}}
&=
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}

\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\\
&=
\begin{bmatrix}
a & b \\
-c & -d
\end{bmatrix}

\begin{bmatrix}
a & -b \\
c & -d
\end{bmatrix} \\
&=
2
\begin{bmatrix}
0 & b \\
-c & 0
\end{bmatrix}
\end{aligned}

Now compare this with \ref{eqn:pauliMatrixXYgeometry:60}

\label{eqn:pauliMatrixXYgeometry:140}
2
\begin{bmatrix}
0 & b \\
-c & 0
\end{bmatrix}
=
2 \sigma_{+}
=
2
\begin{bmatrix}
a & b \\
d & d
\end{bmatrix}.

This shows that $$a = 0$$, and $$d = 0$$. Similarly the $$\sigma_z$$ commutator with the lowering operator is

\label{eqn:pauliMatrixXYgeometry:160}
\begin{aligned}
\antisymmetric{\sigma_z}{\sigma_{-}}
&=
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\begin{bmatrix}
0 & -c^\conj \\
b^\conj & 0
\end{bmatrix}

\begin{bmatrix}
0 & -c^\conj \\
b^\conj & 0
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\\
&=
\begin{bmatrix}
0 & -c^\conj \\
-b^\conj & 0
\end{bmatrix}

\begin{bmatrix}
0 & c^\conj \\
b^\conj & 0
\end{bmatrix} \\
&=
-2
\begin{bmatrix}
0 & c^\conj \\
b^\conj & 0
\end{bmatrix}
\end{aligned}

Again comparing to \ref{eqn:pauliMatrixXYgeometry:60}, we have
\label{eqn:pauliMatrixXYgeometry:180}
-2
\begin{bmatrix}
0 & c^\conj \\
b^\conj & 0
\end{bmatrix}
= – 2 \sigma_{-}
= – 2
\begin{bmatrix}
0 & -c^\conj \\
b^\conj & 0
\end{bmatrix},

so $$c = 0$$. Computing the commutator of the raising and lowering operators fixes $$b$$

\label{eqn:pauliMatrixXYgeometry:200}
\begin{aligned}
\antisymmetric{\sigma_{+}}{\sigma_{-}}
&=
\begin{bmatrix}
0 & b \\
0 & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 & 0 \\
b^\conj & 0 \\
\end{bmatrix}

\begin{bmatrix}
0 & 0 \\
b^\conj & 0 \\
\end{bmatrix}
\begin{bmatrix}
0 & b \\
0 & 0 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
\Abs{b}^2 & 0 \\
0 & 0
\end{bmatrix}

\begin{bmatrix}
0 & 0
0 & -\Abs{b}^2 \\
\end{bmatrix} \\
&=
\Abs{b}^2
\begin{bmatrix}
1 & 0 \\
0 & -1 \\
\end{bmatrix}
\\
&=
\Abs{b}^2 \sigma_z.
\end{aligned}

From \ref{eqn:pauliMatrixXYgeometry:80} it must be that $$\Abs{b}^2 = 4$$, so the most general form of the raising operator is

\label{eqn:pauliMatrixXYgeometry:220}
\sigma_{+}
=
2
\begin{bmatrix}
0 & e^{i \phi} \\
0 & 0
\end{bmatrix}.

## Observation

The conventional choice is to set $$\phi = 0$$, but I found it interesting to see the form of $$\sigma_x, \sigma_y$$ without that choice. That is

\label{eqn:pauliMatrixXYgeometry:240}
\begin{aligned}
\sigma_x
&= \inv{2} \lr{ \sigma_{+} + \sigma_{-} } \\
&=
\begin{bmatrix}
0 & e^{i \phi} \\
e^{-i \phi} & 0 \\
\end{bmatrix}
\end{aligned}

\label{eqn:pauliMatrixXYgeometry:260}
\begin{aligned}
\sigma_y
&= \inv{2 i} \lr{ \sigma_{+} – \sigma_{-} } \\
&=
\begin{bmatrix}
0 & -i e^{i \phi} \\
-i e^{-i \phi} & 0 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & e^{i (\phi – \pi/2) } \\
e^{-i (\phi – \pi/2)} & 0 \\
\end{bmatrix}.
\end{aligned}

Notice that the Pauli matrices $$\sigma_x$$ and $$\sigma_y$$ actually both have the same form as $$\sigma_x$$, but the phase of the complex argument of each differs by $$90^\circ$$. That $$90^\circ$$ separation isn’t obvious in the standard form \ref{eqn:pauliMatrixXYgeometry:20}.

It’s a small detail, but I thought it was kind of cool that the orthogonality of these matrix unit vector representations is built directly into the structure of their matrix representations.

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

## Update to old phy356 (Quantum Mechanics I) notes.

It’s been a long time since I took QM I. My notes from that class were pretty rough, but I’ve cleaned them up a bit.

The main value to these notes is that I worked a number of introductory Quantum Mechanics problems.

These were my personal lecture notes for the Fall 2010, University of Toronto Quantum mechanics I course (PHY356H1F), taught by Prof. Vatche Deyirmenjian.

The official description of this course was:

The general structure of wave mechanics; eigenfunctions and eigenvalues; operators; orbital angular momentum; spherical harmonics; central potential; separation of variables, hydrogen atom; Dirac notation; operator methods; harmonic oscillator and spin.

This document contains a few things

• My lecture notes.
Typos, if any, are probably mine(Peeter), and no claim nor attempt of spelling or grammar correctness will be made. The first four lectures had chosen not to take notes for since they followed the text very closely.
• Notes from reading of the text. This includes observations, notes on what seem like errors, and some solved problems. None of these problems have been graded. Note that my informal errata sheet for the text has been separated out from this document.
• Some assigned problems. I have corrected some the errors after receiving grading feedback, and where I have not done so I at least recorded some of the grading comments as a reference.
• Some worked problems associated with exam preparation.