exponential

Hyperbolic sine representation of mth Fibonacci number

July 26, 2024 math and physics play , , , , ,

[Click here for a PDF version of this post]

I saw a funky looking formula for the mth Fibonacci number on twitter
\begin{equation}\label{eqn:fibonacci_sinh:20}
F_m = \frac{2}{\sqrt{5} i^m} \sinh\lr{ m \ln\lr{i\phi} },
\end{equation}
where
\begin{equation}\label{eqn:fibonacci_sinh:60}
\phi = \frac{ 1 + \sqrt{5} }{2},
\end{equation}
is the golden ratio.

This certainly doesn’t look like it’s a representation of the sequence
\begin{equation}\label{eqn:fibonacci_sinh:40}
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \cdots
\end{equation}
We can verify that it works in Mathematica, as seen in fig. 1.

fig. 1. Verification of hyperbolic sine representation of mth Fibonacci numbe

Recall that we previously found this formula for the mth Fibonacci number
\begin{equation}\label{eqn:fibonacci_sinh:80}
F_m = \inv{\sqrt{5}} \lr{ \phi^m – { \bar{\phi}}^m },
\end{equation}
where \( \bar{\phi} \) is the conjugate of the golden ratio
\begin{equation}\label{eqn:fibonacci_sinh:100}
\bar{\phi} = \frac{ 1 – \sqrt{5} }{2}.
\end{equation}

Let’s see how these are equivalent. First observe that the golden conjugate is easily related to the inverse of the golden ratio
\begin{equation}\label{eqn:fibonacci_sinh:120}
\begin{aligned}
\inv{\phi}
&=
\frac{2}{1 + \sqrt{5}} \\
&=
\frac{2\lr{ 1 – \sqrt{5}} }{1^2 – \lr{\sqrt{5}}^2 } \\
&=
-\frac{1 – \sqrt{5} }{2} \\
&=
-\bar{\phi}.
\end{aligned}
\end{equation}
Substitution gives
\begin{equation}\label{eqn:fibonacci_sinh:140}
F_m = \inv{\sqrt{5}} \lr{ \phi^m – \lr{\frac{-1}{\phi}}^m }.
\end{equation}
Multiplying by \( i^m \), we have
\begin{equation}\label{eqn:fibonacci_sinh:160}
\begin{aligned}
i^m F_m
&= \inv{\sqrt{5}} \lr{ i^m \phi^m – \inv{(-i)^m} \lr{\frac{-1}{\phi}}^m } \\
&= \inv{\sqrt{5}} \lr{ \lr{ i \phi} ^m – \lr{i \phi}^{-m} } \\
\end{aligned}
\end{equation}

We can write any exponent in terms of \( e \)
\begin{equation}\label{eqn:fibonacci_sinh:180}
a^m = e^{\ln a^m} = e^{m \ln a},
\end{equation}
so
\begin{equation}\label{eqn:fibonacci_sinh:200}
\begin{aligned}
i^m F_m
&= \inv{\sqrt{5}} \lr{ e^{m \ln \lr{ i \phi}} – e^{-m \ln\lr{i \phi} } } \\
&= \inv{\sqrt{5}} 2 \sinh\lr{ m \ln \lr{ i \phi } },
\end{aligned}
\end{equation}
as we wanted to show. It’s a bit strange looking, but we see why it works.

Lorentz transformations in Space Time Algebra (STA)

December 12, 2020 math and physics play , , , , , , , , , , , , , , , , , ,

[If mathjax doesn’t display properly for you, click here for a PDF of this post]

Motivation.

One of the remarkable features of geometric algebra are the complex exponential sandwiches that can be used to encode rotations in any dimension, or rotation like operations like Lorentz transformations in Minkowski spaces. In this post, we show some examples that unpack the geometric algebra expressions for Lorentz transformations operations of this sort. In particular, we will look at the exponential sandwich operations for spatial rotations and Lorentz boosts in the Dirac algebra, known as Space Time Algebra (STA) in geometric algebra circles, and demonstrate that these sandwiches do have the desired effects.

Lorentz transformations.

Theorem 1.1: Lorentz transformation.

The transformation
\begin{equation}\label{eqn:lorentzTransform:580}
x \rightarrow e^{B} x e^{-B} = x’,
\end{equation}
where \( B = a \wedge b \), is an STA 2-blade for any two linearly independent four-vectors \( a, b \), is a norm preserving, that is
\begin{equation}\label{eqn:lorentzTransform:600}
x^2 = {x’}^2.
\end{equation}

Start proof:

The proof is disturbingly trivial in this geometric algebra form
\begin{equation}\label{eqn:lorentzTransform:40}
\begin{aligned}
{x’}^2
&=
e^{B} x e^{-B} e^{B} x e^{-B} \\
&=
e^{B} x x e^{-B} \\
&=
x^2 e^{B} e^{-B} \\
&=
x^2.
\end{aligned}
\end{equation}

End proof.

In particular, observe that we did not need to construct the usual infinitesimal representations of rotation and boost transformation matrices or tensors in order to demonstrate that we have spacetime invariance for the transformations. The rough idea of such a transformation is that the exponential commutes with components of the four-vector that lie off the spacetime plane specified by the bivector \( B \), and anticommutes with components of the four-vector that lie in the plane. The end result is that the sandwich operation simplifies to
\begin{equation}\label{eqn:lorentzTransform:60}
x’ = x_\parallel e^{-B} + x_\perp,
\end{equation}
where \( x = x_\perp + x_\parallel \) and \( x_\perp \cdot B = 0 \), and \( x_\parallel \wedge B = 0 \). In particular, using \( x = x B B^{-1} = \lr{ x \cdot B + x \wedge B } B^{-1} \), we find that
\begin{equation}\label{eqn:lorentzTransform:80}
\begin{aligned}
x_\parallel &= \lr{ x \cdot B } B^{-1} \\
x_\perp &= \lr{ x \wedge B } B^{-1}.
\end{aligned}
\end{equation}
When \( B \) is a spacetime plane \( B = b \wedge \gamma_0 \), then this exponential has a hyperbolic nature, and we end up with a Lorentz boost. When \( B \) is a spatial bivector, we end up with a single complex exponential, encoding our plane old 3D rotation. More general \( B \)’s that encode composite boosts and rotations are also possible, but \( B \) must be invertible (it should have no lightlike factors.) The rough geometry of these projections is illustrated in fig 1, where the spacetime plane is represented by \( B \).

Projection and rejection geometry.

fig 1. Projection and rejection geometry.

 

What is not so obvious is how to pick \( B \)’s that correspond to specific rotation axes or boost directions. Let’s consider each of those cases in turn.

Theorem 1.2: Boost.

The boost along a direction vector \( \vcap \) and rapidity \( \alpha \) is given by
\begin{equation}\label{eqn:lorentzTransform:620}
x’ = e^{-\vcap \alpha/2} x e^{\vcap \alpha/2},
\end{equation}
where \( \vcap = \gamma_{k0} \cos\theta^k \) is an STA bivector representing a spatial direction with direction cosines \( \cos\theta^k \).

Start proof:

We want to demonstrate that this is equivalent to the usual boost formulation. We can start with decomposition of the four-vector \( x \) into components that lie in and off of the spacetime plane \( \vcap \).
\begin{equation}\label{eqn:lorentzTransform:100}
\begin{aligned}
x
&= \lr{ x^0 + \Bx } \gamma_0 \\
&= \lr{ x^0 + \Bx \vcap^2 } \gamma_0 \\
&= \lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap + \lr{ \Bx \wedge \vcap} \vcap } \gamma_0,
\end{aligned}
\end{equation}
where \( \Bx = x \wedge \gamma_0 \). The first two components lie in the boost plane, whereas the last is the spatial component of the vector that lies perpendicular to the boost plane. Observe that \( \vcap \) anticommutes with the dot product term and commutes with he wedge product term, so we have
\begin{equation}\label{eqn:lorentzTransform:120}
\begin{aligned}
x’
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap } \vcap } \gamma_0
e^{\vcap \alpha/2 }
e^{\vcap \alpha/2 }
+
\lr{ \Bx \wedge \vcap } \vcap \gamma_0
e^{-\vcap \alpha/2 }
e^{\vcap \alpha/2 } \\
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap } \vcap } \gamma_0
e^{\vcap \alpha }
+
\lr{ \Bx \wedge \vcap } \vcap \gamma_0.
\end{aligned}
\end{equation}
Noting that \( \vcap^2 = 1 \), we may expand the exponential in hyperbolic functions, and find that the boosted portion of the vector expands as
\begin{equation}\label{eqn:lorentzTransform:260}
\begin{aligned}
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0 e^{\vcap \alpha}
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0 \lr{ \cosh\alpha + \vcap \sinh \alpha} \\
&=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \lr{ \cosh\alpha – \vcap \sinh \alpha} \gamma_0 \\
&=
\lr{ x^0 \cosh\alpha – \lr{ \Bx \cdot \vcap} \sinh \alpha} \gamma_0
+
\lr{ -x^0 \sinh \alpha + \lr{ \Bx \cdot \vcap} \cosh \alpha } \vcap \gamma_0.
\end{aligned}
\end{equation}
We are left with
\begin{equation}\label{eqn:lorentzTransform:320}
\begin{aligned}
x’
&=
\lr{ x^0 \cosh\alpha – \lr{ \Bx \cdot \vcap} \sinh \alpha} \gamma_0
+
\lr{ \lr{ \Bx \cdot \vcap} \cosh \alpha -x^0 \sinh \alpha } \vcap \gamma_0
+
\lr{ \Bx \wedge \vcap} \vcap \gamma_0 \\
&=
\begin{bmatrix}
\gamma_0 & \vcap \gamma_0
\end{bmatrix}
\begin{bmatrix}
\cosh\alpha & – \sinh\alpha \\
-\sinh\alpha & \cosh\alpha
\end{bmatrix}
\begin{bmatrix}
x^0 \\
\Bx \cdot \vcap
\end{bmatrix}
+
\lr{ \Bx \wedge \vcap} \vcap \gamma_0,
\end{aligned}
\end{equation}
which has the desired Lorentz boost structure. Of course, this is usually seen with \( \vcap = \gamma_{10} \) so that the components in the coordinate column vector are \( (ct, x) \).

End proof.

Theorem 1.3: Spatial rotation.

Given two linearly independent spatial bivectors \( \Ba = a^k \gamma_{k0}, \Bb = b^k \gamma_{k0} \), a rotation of \(\theta\) radians in the plane of \( \Ba, \Bb \) from \( \Ba \) towards \( \Bb \), is given by
\begin{equation}\label{eqn:lorentzTransform:640}
x’ = e^{-i\theta} x e^{i\theta},
\end{equation}
where \( i = (\Ba \wedge \Bb)/\Abs{\Ba \wedge \Bb} \), is a unit (spatial) bivector.

Start proof:

Without loss of generality, we may pick \( i = \acap \bcap \), where \( \acap^2 = \bcap^2 = 1 \), and \( \acap \cdot \bcap = 0 \). With such an orthonormal basis for the plane, we can decompose our four vector into portions that lie in and off the plane
\begin{equation}\label{eqn:lorentzTransform:400}
\begin{aligned}
x
&= \lr{ x^0 + \Bx } \gamma_0 \\
&= \lr{ x^0 + \Bx i i^{-1} } \gamma_0 \\
&= \lr{ x^0 + \lr{ \Bx \cdot i } i^{-1} + \lr{ \Bx \wedge i } i^{-1} } \gamma_0.
\end{aligned}
\end{equation}
The projective term lies in the plane of rotation, whereas the timelike and spatial rejection term are perpendicular. That is
\begin{equation}\label{eqn:lorentzTransform:420}
\begin{aligned}
x_\parallel &= \lr{ \Bx \cdot i } i^{-1} \gamma_0 \\
x_\perp &= \lr{ x^0 + \lr{ \Bx \wedge i } i^{-1} } \gamma_0,
\end{aligned}
\end{equation}
where \( x_\parallel \wedge i = 0 \), and \( x_\perp \cdot i = 0 \). The plane pseudoscalar \( i \) anticommutes with \( x_\parallel \), and commutes with \( x_\perp \), so
\begin{equation}\label{eqn:lorentzTransform:440}
\begin{aligned}
x’
&= e^{-i\theta/2} \lr{ x_\parallel + x_\perp } e^{i\theta/2} \\
&= x_\parallel e^{i\theta} + x_\perp.
\end{aligned}
\end{equation}
However
\begin{equation}\label{eqn:lorentzTransform:460}
\begin{aligned}
\lr{ \Bx \cdot i } i^{-1}
&=
\lr{ \Bx \cdot \lr{ \acap \wedge \bcap } } \bcap \acap \\
&=
\lr{\Bx \cdot \acap} \bcap \bcap \acap
-\lr{\Bx \cdot \bcap} \acap \bcap \acap \\
&=
\lr{\Bx \cdot \acap} \acap
+\lr{\Bx \cdot \bcap} \bcap,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:lorentzTransform:480}
\begin{aligned}
x_\parallel e^{i\theta}
&=
\lr{
\lr{\Bx \cdot \acap} \acap
+
\lr{\Bx \cdot \bcap} \bcap
}
\gamma_0
\lr{
\cos\theta + \acap \bcap \sin\theta
} \\
&=
\acap \lr{
\lr{\Bx \cdot \acap} \cos\theta
–
\lr{\Bx \cdot \bcap} \sin\theta
}
\gamma_0
+
\bcap \lr{
\lr{\Bx \cdot \acap} \sin\theta
+
\lr{\Bx \cdot \bcap} \cos\theta
}
\gamma_0,
\end{aligned}
\end{equation}
so
\begin{equation}\label{eqn:lorentzTransform:500}
x’
=
\begin{bmatrix}
\acap & \bcap
\end{bmatrix}
\begin{bmatrix}
\cos\theta & – \sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
\Bx \cdot \acap \\
\Bx \cdot \bcap \\
\end{bmatrix}
\gamma_0
+
\lr{ x \wedge i} i^{-1} \gamma_0.
\end{equation}
Observe that this rejection term can be explicitly expanded to
\begin{equation}\label{eqn:lorentzTransform:520}
\lr{ \Bx \wedge i} i^{-1} \gamma_0 =
x –
\lr{ \Bx \cdot \acap } \acap \gamma_0
–
\lr{ \Bx \cdot \acap } \acap \gamma_0.
\end{equation}
This is the timelike component of the vector, plus the spatial component that is normal to the plane. This exponential sandwich transformation rotates only the portion of the vector that lies in the plane, and leaves the rest (timelike and normal) untouched.

End proof.

Problems.

Problem: Verify components relative to boost direction.

In the proof of thm. 1.2, the vector \( x \) was expanded in terms of the spacetime split. An alternate approach, is to expand as
\begin{equation}\label{eqn:lorentzTransform:340}
\begin{aligned}
x
&= x \vcap^2 \\
&= \lr{ x \cdot \vcap + x \wedge \vcap } \vcap \\
&= \lr{ x \cdot \vcap } \vcap + \lr{ x \wedge \vcap } \vcap.
\end{aligned}
\end{equation}
Show that
\begin{equation}\label{eqn:lorentzTransform:360}
\lr{ x \cdot \vcap } \vcap
=
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0,
\end{equation}
and
\begin{equation}\label{eqn:lorentzTransform:380}
\lr{ x \wedge \vcap } \vcap
=
\lr{ \Bx \wedge \vcap} \vcap \gamma_0.
\end{equation}

Answer

Let \( x = x^\mu \gamma_\mu \), so that
\begin{equation}\label{eqn:lorentzTransform:160}
\begin{aligned}
x \cdot \vcap
&=
\gpgradeone{ x^\mu \gamma_\mu \cos\theta^b \gamma_{b 0} } \\
&=
x^\mu \cos\theta^b \gpgradeone{ \gamma_\mu \gamma_{b 0} }
.
\end{aligned}
\end{equation}
The \( \mu = 0 \) component of this grade selection is
\begin{equation}\label{eqn:lorentzTransform:180}
\gpgradeone{ \gamma_0 \gamma_{b 0} }
=
-\gamma_b,
\end{equation}
and for \( \mu = a \ne 0 \), we have
\begin{equation}\label{eqn:lorentzTransform:200}
\gpgradeone{ \gamma_a \gamma_{b 0} }
=
-\delta_{a b} \gamma_0,
\end{equation}
so we have
\begin{equation}\label{eqn:lorentzTransform:220}
\begin{aligned}
x \cdot \vcap
&=
x^0 \cos\theta^b (-\gamma_b)
+
x^a \cos\theta^b (-\delta_{ab} \gamma_0 ) \\
&=
-x^0 \vcap \gamma_0
–
x^b \cos\theta^b \gamma_0 \\
&=
– \lr{ x^0 \vcap + \Bx \cdot \vcap } \gamma_0,
\end{aligned}
\end{equation}
where \( \Bx = x \wedge \gamma_0 \) is the spatial portion of the four vector \( x \) relative to the stationary observer frame. Since \( \vcap \) anticommutes with \( \gamma_0 \), the component of \( x \) in the spacetime plane \( \vcap \) is
\begin{equation}\label{eqn:lorentzTransform:240}
\lr{ x \cdot \vcap } \vcap =
\lr{ x^0 + \lr{ \Bx \cdot \vcap} \vcap } \gamma_0,
\end{equation}
as expected.

For the rejection term, we have
\begin{equation}\label{eqn:lorentzTransform:280}
x \wedge \vcap
=
x^\mu \cos\theta^s \gpgradethree{ \gamma_\mu \gamma_{s 0} }.
\end{equation}
The \( \mu = 0 \) term clearly contributes nothing, leaving us with:
\begin{equation}\label{eqn:lorentzTransform:300}
\begin{aligned}
\lr{ x \wedge \vcap } \vcap
&=
\lr{ x \wedge \vcap } \cdot \vcap \\
&=
x^r \cos\theta^s \cos\theta^t \lr{ \lr{ \gamma_r \wedge \gamma_{s}} \gamma_0 } \cdot \lr{ \gamma_{t0} } \\
&=
x^r \cos\theta^s \cos\theta^t \gpgradeone{
\lr{ \gamma_r \wedge \gamma_{s} } \gamma_0 \gamma_{t0}
} \\
&=
-x^r \cos\theta^s \cos\theta^t \lr{ \gamma_r \wedge \gamma_{s}} \cdot \gamma_t \\
&=
-x^r \cos\theta^s \cos\theta^t \lr{ -\gamma_r \delta_{st} + \gamma_s \delta_{rt} } \\
&=
x^r \cos\theta^t \cos\theta^t \gamma_r
–
x^t \cos\theta^s \cos\theta^t \gamma_s \\
&=
\Bx \gamma_0
– (\Bx \cdot \vcap) \vcap \gamma_0 \\
&=
\lr{ \Bx \wedge \vcap} \vcap \gamma_0,
\end{aligned}
\end{equation}
as expected. Is there a clever way to demonstrate this without resorting to coordinates?

Problem: Rotation transformation components.

Given a unit spatial bivector \( i = \acap \bcap \), where \( \acap \cdot \bcap = 0 \) and \( i^2 = -1 \), show that
\begin{equation}\label{eqn:lorentzTransform:540}
\lr{ x \cdot i } i^{-1}
=
\lr{ \Bx \cdot i } i^{-1} \gamma_0
=
\lr{\Bx \cdot \acap } \acap \gamma_0
+
\lr{\Bx \cdot \bcap } \bcap \gamma_0,
\end{equation}
and
\begin{equation}\label{eqn:lorentzTransform:560}
\lr{ x \wedge i } i^{-1}
=
\lr{ \Bx \wedge i } i^{-1} \gamma_0
=
x –
\lr{\Bx \cdot \acap } \acap \gamma_0
–
\lr{\Bx \cdot \bcap } \bcap \gamma_0.
\end{equation}
Also show that \( i \) anticommutes with \( \lr{ x \cdot i } i^{-1} \) and commutes with \( \lr{ x \wedge i } i^{-1} \).

Answer

This problem is left for the reader, as I don’t feel like typing out my solution.

The first part of this problem can be done in the tedious coordinate approach used above, but hopefully there is a better way.

For the last (commutation) part of the problem, here is a hint. Let \( x \wedge i = n i \), where \( n \cdot i = 0 \). The result then follows easily.

A curious proof of the Baker-Campbell-Hausdorff formula

November 4, 2015 phy1520 , , ,

[Click here for a PDF of this post with nicer formatting]

Equation (39) of [1] states the Baker-Campbell-Hausdorff formula for two operators \( a, b\) that commute with their commutator \( \antisymmetric{a}{b} \)

\begin{equation}\label{eqn:bakercambell:20}
e^a e^b = e^{a + b + \antisymmetric{a}{b}/2},
\end{equation}

and provides the outline of an interesting method of proof. That method is to consider the derivative of

\begin{equation}\label{eqn:bakercambell:40}
f(\lambda) = e^{\lambda a} e^{\lambda b} e^{-\lambda (a + b)},
\end{equation}

That derivative is
\begin{equation}\label{eqn:bakercambell:60}
\begin{aligned}
\frac{df}{d\lambda}
&=
e^{\lambda a} a e^{\lambda b} e^{-\lambda (a + b)}
+
e^{\lambda a} b e^{\lambda b} e^{-\lambda (a + b)}
–
e^{\lambda a} b e^{\lambda b} (a + b)e^{-\lambda (a + b)} \\
&=
e^{\lambda a} \lr{
a e^{\lambda b}
+
b e^{\lambda b}
–
e^{\lambda b} (a+b)
}
e^{-\lambda (a + b)} \\
&=
e^{\lambda a} \lr{
\antisymmetric{a}{e^{\lambda b}}
+
{\antisymmetric{b}{e^{\lambda b}}}
}
e^{-\lambda (a + b)} \\
&=
e^{\lambda a}
\antisymmetric{a}{e^{\lambda b}}
e^{-\lambda (a + b)}
.
\end{aligned}
\end{equation}

The commutator above is proportional to \( \antisymmetric{a}{b} \)

\begin{equation}\label{eqn:bakercambell:80}
\begin{aligned}
\antisymmetric{a}{e^{\lambda b}}
&=
\sum_{k=0}^\infty \frac{\lambda^k}{k!} \antisymmetric{a}{ b^k } \\
&=
\sum_{k=0}^\infty \frac{\lambda^k}{k!} k b^{k-1} \antisymmetric{a}{b} \\
&=
\lambda \sum_{k=1}^\infty \frac{\lambda^{k-1}}{(k-1)!} b^{k-1}
\antisymmetric{a}{b} \\
&=
\lambda e^{\lambda b} \antisymmetric{a}{b},
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:bakercambell:100}
\frac{df}{d\lambda} = \lambda \antisymmetric{a}{b} f.
\end{equation}

To get the above, we should also do the induction demonstration for \( \antisymmetric{a}{ b^k } = k b^{k-1} \antisymmetric{a}{b} \).

This clearly holds for \( k = 0,1 \). For any other \( k \) we have

\begin{equation}\label{eqn:bakercambell:120}
\begin{aligned}
\antisymmetric{a}{b^{k+1}}
&=
a b^{k+1} – b^{k+1} a \\
&=
\lr{ \antisymmetric{a}{b^{k}} + b^k a
} b – b^{k+1} a \\
&=
k b^{k-1} \antisymmetric{a}{b} b
+ b^k \lr{ \antisymmetric{a}{b} + {b a} }
– {b^{k+1} a} \\
&=
k b^{k} \antisymmetric{a}{b}
+ b^k \antisymmetric{a}{b} \\
&=
(k+1) b^k \antisymmetric{a}{b}.
\end{aligned}
\end{equation}

Observe that \ref{eqn:bakercambell:100} is solved by

\begin{equation}\label{eqn:bakercambell:140}
f = e^{\lambda^2\antisymmetric{a}{b}/2},
\end{equation}

which gives

\begin{equation}\label{eqn:bakercambell:160}
e^{\lambda^2 \antisymmetric{a}{b}/2} =
e^{\lambda a} e^{\lambda b} e^{-\lambda (a + b)}.
\end{equation}

Right multiplication by \( e^{\lambda (a + b)} \) which commutes with \( e^{\lambda^2 \antisymmetric{a}{b}/2} \) and setting \( \lambda = 1 \) recovers \ref{eqn:bakercambell:20} as desired.

What I wonder looking at this, is what thought process led to trying this in the first place? This is not what I would consider an obvious approach to demonstrating this identity.

References

[1] Roy J Glauber. Some notes on multiple-boson processes. Physical Review, 84 (3), 1951.

Update to old phy356 (Quantum Mechanics I) notes.

February 12, 2015 math and physics play , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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.