## PHY1520H Graduate Quantum Mechanics. Lecture 2: Basic concepts, time evolution, and density operators. Taught by Prof. Arun Paramekanti

September 22, 2015 phy1520 , , ,

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering chap 1 (basic concepts),3 (density operator). content from [1].

### Basic concepts

We’ve reviewed the basic concepts that we will encounter in Quantum Mechanics.

1. Abstract state vector. $$\ket{ \psi}$$
2. Basis states. $$\ket{ x }$$
3. Observables, special Hermitian operators. We’ll only deal with linear observables.
4. Measurement.

We can either express the wave functions $$\psi(x) = \braket{x}{\psi}$$ in terms of a basis for the observable, or can express the observable in terms of the basis of the wave function (position or momentum for example).

We saw that the position space representation of a momentum operator (also an observable) was

\label{eqn:lecture2:20}
\hat{p} \rightarrow -i \Hbar \PD{x}{}.

In general we can find the matrix element representation of any operator by considering its representation in a given basis. For example, in a position basis, that would be

\label{eqn:lecture2:40}
\bra{x’} \hat{A} \ket{x} \leftrightarrow A_{x x’}

The Hermitian property of the observable means that $$A_{x x’} = A_{x’ x}^\conj$$

\label{eqn:lecture2:60}
\int dx \bra{x’} \hat{A} \ket{x} \braket{x }{\psi} = \braket{x’}{\phi}
\leftrightarrow
A_{x’ x} \psi_x = \phi_{x’}.

## Example: Measurement example

fig. 1. Polarizer apparatus

Consider a polarization apparatus as sketched in fig. 1, where the output is of the form $$I_{\textrm{out}} = I_{\textrm{in}} \cos^2 \theta$$.

A general input state can be written in terms of each of the possible polarizations

\label{eqn:lecture2:80}
\alpha \ket{ \updownarrow } + \beta \ket{ \leftrightarrow } \sim
\cos\theta \ket{ \updownarrow } + \sin\theta \ket{ \leftrightarrow }

Here $$\abs{\alpha}^2$$ is the probability that the input state is in the upwards polarization state, and $$\abs{\beta}^2$$ is the probability that the input state is in the downwards polarization state.

The measurement of the polarization results in an output state that has a specific polarization. That measurement is said to collapse the wavefunction.

When attempting a measurement, looking for a specific value, effects the state of the system, and is call a strong or projective measurement. Such a measurement is

• (i) Probabilistic.
• (ii) Requires many measurements.

This measurement process results a determination of the eigenvalue of the operator. The eigenvalue production of measurement is why we demand that operators be Hermitian.

It is also possible to try to do a weaker (perturbative) measurement, where some information is extracted from the input state without completely altering it.

### Time evolution

1. Schrodinger picture.
The time evolution process is governed by a Schrodinger equation of the following form\label{eqn:lecture2:100}
i \Hbar \PD{t}{} \ket{\Psi(t)} = \hat{H} \ket{\Psi(t)}.

This Hamiltonian could be, for example,

\label{eqn:lecture2:120}
\hat{H} = \frac{\hat{p}^2}{2m} + V(x),

Such a representation of time evolution is expressed in terms of operators $$\hat{x}, \hat{p}, \hat{H}, \cdots$$ that are independent of time.

2. Heisenberg picture.Suppose we have a state $$\ket{\Psi(t)}$$ and operate on this with an operator

\label{eqn:lecture2:140}
\hat{A} \ket{\Psi(t)}.

This will have time evolution of the form

\label{eqn:lecture2:160}
\hat{A} e^{-i \hat{H} t/\Hbar} \ket{\Psi(0)},

or in matrix element form

\label{eqn:lecture2:180}
\bra{\phi(t)} \hat{A} \ket{\Psi(t)}
=
\bra{\phi(0)}
e^{i \hat{H} t/\Hbar}
\hat{A} e^{-i \hat{H} t/\Hbar} \ket{\Psi(0)}.

We work with states that do not evolve in time $$\ket{\phi(0)}, \ket{\Psi(0)}, \cdots$$, but operators do evolve in time according to

\label{eqn:lecture2:200}
\hat{A}(t) =
e^{i \hat{H} t/\Hbar}
\hat{A} e^{-i \hat{H} t/\Hbar}.

### Density operator

We can have situations where it is impossible to determine a single state that describes the system. For example, given the gas in the room that you are sitting in, there are things that we can measure, but it is impossible to describe the state that describes all the particles and also impossible to construct a Hamiltonian that governs all the interactions of those many many particles.

We need a probabilistic description to even describe such a complex system.

Suppose we have a complex system that can be partitioned into two subsets, left and right, as sketched in fig. 2.

fig. 2. System partitioned into separate set of states

If the states in each partition can be enumerated separately, we can write the state of the system as sums over the probability amplitudes that for the combined states.

\label{eqn:lecture2:220}
\ket{\Psi}
=
\sum_{m, n} C_{m,n} \ket{m} \ket{n}

Here $$C_{m, n}$$ is the probability amplitude to find the state in the combined state $$\ket{m} \ket{n}$$.

As an example of such a system, we could investigate a two particle configuration where spin up or spin down can be separately measured for each particle.

\label{eqn:lecture2:240}
\ket{\psi} = \inv{\sqrt{2}} \lr{
\ket{\uparrow}\ket{\downarrow}
+
\ket{\downarrow}\ket{\rightarrow}
}

Considering such a system we could ask questions such as

• What is the probability that the left half is in state $$m$$? This would be\label{eqn:lecture2:260}
\sum_n \Abs{C_{m, n}}^2
• Probability that the left half is in state $$m$$, and the
probability that the right half is in state $$n$$? That is\label{eqn:lecture2:280}
\Abs{C_{m, n}}^2

We define the density operator

\label{eqn:lecture2:300}
\hat{\rho} = \ket{\Psi} \bra{\Psi}.

This is idempotent

\label{eqn:lecture2:320}
\hat{\rho}^2 =
\lr{ \ket{\Psi} \bra{\Psi} }
\lr{ \ket{\Psi} \bra{\Psi} }
=
\ket{\Psi} \bra{\Psi}

# References

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

## PHY1520H Graduate Quantum Mechanics. Lecture 1: Lighting review. Taught by Prof. Arun Paramekanti

### Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 1 content.

### Classical mechanics

We’ll be talking about one body physics for most of this course. In classical mechanics we can figure out the particle trajectories using both of $$(\Br, \Bp$$, where

\label{eqn:qmLecture1:20}
\begin{aligned}
\ddt{\Br} &= \inv{m} \Bp \\
\end{aligned}

A two dimensional phase space as sketched in fig. 1 shows the trajectory of a point particle subject to some equations of motion

fig. 1. One dimensional classical phase space example

### Quantum mechanics

For this lecture, we’ll work with natural units, setting

\label{eqn:qmLecture1:480}
\boxed{
\Hbar = 1.
}

In QM we are no longer allowed to think of position and momentum, but have to start asking about state vectors $$\ket{\Psi}$$.

We’ll consider the state vector with respect to some basis, for example, in a position basis, we write

\label{eqn:qmLecture1:40}
\braket{ x }{\Psi } = \Psi(x),

a complex numbered “wave function”, the probability amplitude for a particle in $$\ket{\Psi}$$ to be in the vicinity of $$x$$.

We could also consider the state in a momentum basis

\label{eqn:qmLecture1:60}
\braket{ p }{\Psi } = \Psi(p),

a probability amplitude with respect to momentum $$p$$.

More precisely,

\label{eqn:qmLecture1:80}
\Abs{\Psi(x)}^2 dx \ge 0

is the probability of finding the particle in the range $$(x, x + dx )$$. To have meaning as a probability, we require

\label{eqn:qmLecture1:100}
\int_{-\infty}^\infty \Abs{\Psi(x)}^2 dx = 1.

The average position can be calculated using this probability density function. For example

\label{eqn:qmLecture1:120}
\expectation{x} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 x dx,

or
\label{eqn:qmLecture1:140}
\expectation{f(x)} = \int_{-\infty}^\infty \Abs{\Psi(x)}^2 f(x) dx.

Similarly, calculation of an average of a function of momentum can be expressed as

\label{eqn:qmLecture1:160}
\expectation{f(p)} = \int_{-\infty}^\infty \Abs{\Psi(p)}^2 f(p) dp.

### Transformation from a position to momentum basis

We have a problem, if we which to compute an average in momentum space such as $$\expectation{p}$$, when given a wavefunction $$\Psi(x)$$.

How do we convert

\label{eqn:qmLecture1:180}
\Psi(p)
\stackrel{?}{\leftrightarrow}
\Psi(x),

or equivalently
\label{eqn:qmLecture1:200}
\braket{p}{\Psi}
\stackrel{?}{\leftrightarrow}
\braket{x}{\Psi}.

Such a conversion can be performed by virtue of an the assumption that we have a complete orthonormal basis, for which we can introduce identity operations such as

\label{eqn:qmLecture1:220}
\int_{-\infty}^\infty dp \ket{p}\bra{p} = 1,

or
\label{eqn:qmLecture1:240}
\int_{-\infty}^\infty dx \ket{x}\bra{x} = 1

Some interpretations:

1. $$\ket{x_0} \leftrightarrow \text{sits at} x = x_0$$
2. $$\braket{x}{x’} \leftrightarrow \delta(x – x’)$$
3. $$\braket{p}{p’} \leftrightarrow \delta(p – p’)$$
4. $$\braket{x}{p’} = \frac{e^{i p x}}{\sqrt{V}}$$, where $$V$$ is the volume of the box containing the particle. We’ll define the appropriate normalization for an infinite box volume later.

The delta function interpretation of the braket $$\braket{p}{p’}$$ justifies the identity operator, since we recover any state in the basis when operating with it. For example, in momentum space

\label{eqn:qmLecture1:260}
\begin{aligned}
1 \ket{p}
&=
\lr{ \int_{-\infty}^\infty dp’
\ket{p’}\bra{p’} }
\ket{p} \\
&=
\int_{-\infty}^\infty dp’
\ket{p’}
\braket{p’}{p} \\
&=
\int_{-\infty}^\infty dp’
\ket{p’}
\delta(p – p’) \\
&=
\ket{p}.
\end{aligned}

This also the determination of an integral operator representation for the delta function

\label{eqn:qmLecture1:500}
\begin{aligned}
\delta(x – x’)
&=
\braket{x}{x’} \\
&=
\int dp \braket{x}{p} \braket{p}{x’} \\
&=
\inv{V} \int dp e^{i p x} e^{-i p x’},
\end{aligned}

or

\label{eqn:qmLecture1:520}
\delta(x – x’)
=
\inv{V} \int dp e^{i p (x- x’)}.

Here we used the fact that $$\braket{p}{x} = \braket{x}{p}^\conj$$.

FIXME: do we have a justification for that conjugation with what was defined here so far?

The conversion from a position basis to momentum space is now possible

\label{eqn:qmLecture1:280}
\begin{aligned}
\braket{p}{\Psi}
&= \Psi(p) \\
&= \int_{-\infty}^\infty \braket{p}{x} \braket{x}{\Psi} dx \\
&= \int_{-\infty}^\infty \frac{e^{-ip x}}{\sqrt{V}} \Psi(x) dx.
\end{aligned}

The momentum space to position space conversion can be written as

\label{eqn:qmLecture1:300}
\Psi(x)
= \int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi(p) dp.

Now we can go back and figure out the an expectation

\label{eqn:qmLecture1:320}
\begin{aligned}
\expectation{p}
&=
\int \Psi^\conj(p) \Psi(p) p d p \\
&=
\int dp
\lr{
\int_{-\infty}^\infty \frac{e^{ip x}}{\sqrt{V}} \Psi^\conj(x) dx
}
\lr{
\int_{-\infty}^\infty \frac{e^{-ip x’}}{\sqrt{V}} \Psi(x’) dx’
}
p \\
&=\int dp dx dx’
\Psi^\conj(x)
\inv{V} e^{ip (x-x’)} \Psi(x’) p \\
&=
\int dp dx dx’
\Psi^\conj(x)
\inv{V} \lr{ -i\PD{x}{e^{ip (x-x’)}} }\Psi(x’) \\
&=
\int dp dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\inv{V} \int dx’ e^{ip (x-x’)} \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\int dx’ \lr{ \inv{V} \int dp e^{ip (x-x’)} } \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\int dx’ \delta(x – x’) \Psi(x’) \\
&=
\int dx
\Psi^\conj(x) \lr{ -i \PD{x}{} }
\Psi(x)
\end{aligned}

Here we’ve essentially calculated the position space representation of the momentum operator, allowing identifications of the following form

\label{eqn:qmLecture1:380}
p \leftrightarrow -i \PD{x}{}

\label{eqn:qmLecture1:400}
p^2 \leftrightarrow – \PDSq{x}{}.

### Alternate starting point.

Most of the above results followed from the claim that $$\braket{x}{p} = e^{i p x}$$. Note that this position space representation of the momentum operator can also be taken as the starting point. Given that, the exponential representation of the position-momentum braket follows

\label{eqn:qmLecture1:540}
\bra{x} P \ket{p}
=
-i \Hbar \PD{x}{} \braket{x}{p},

but $$\bra{x} P \ket{p} = p \braket{x}{p}$$, providing a differential equation for $$\braket{x}{p}$$

\label{eqn:qmLecture1:560}
p \braket{x}{p} = -i \Hbar \PD{x}{} \braket{x}{p},

with solution

\label{eqn:qmLecture1:580}
i p x/\Hbar = \ln \braket{x}{p} + \text{const},

or
\label{eqn:qmLecture1:600}
\braket{x}{p} \propto e^{i p x/\Hbar}.

### Matrix interpretation

1. Ket’s $$\ket{\Psi} \leftrightarrow \text{column vector}$$
2. Bra’s $$\bra{\Psi} \leftrightarrow {(\text{row vector})}^\conj$$
3. Operators $$\leftrightarrow$$ matrices that act on vectors.

\label{eqn:qmLecture1:420}
\hat{p} \ket{\Psi} \rightarrow \ket{\Psi’}

### Time evolution

For a state subject to the equations of motion given by the Hamiltonian operator $$\hat{H}$$

\label{eqn:qmLecture1:440}
i \PD{t}{} \ket{\Psi} = \hat{H} \ket{\Psi},

the time evolution is given by
\label{eqn:qmLecture1:460}
\ket{\Psi(t)} = e^{-i \hat{H} t} \ket{\Psi(0)}.

### Incomplete information

We’ll need to introduce the concept of Density matrices. This will bring us to concepts like entanglement.

# References

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

## Question: Quantum Virial Theorem ([1] pr. 2.7)

Consider a particle with Hamiltonian

\label{eqn:qmVirialTheorem:20}
H = \frac{\Bp^2}{2 m} + V(\Bx),

By calculating the time evolution of $$\antisymmetric{\Bx \cdot \Bp}{H}$$, identify the quantum virial theorem and show the conditions where it is satisfied.

\label{eqn:qmVirialTheorem:40}
\begin{aligned}
\antisymmetric{\Bx \cdot \Bp}{H}
&=
\inv{2 m} \antisymmetric{\Bx \cdot \Bp}{\Bp^2} + \antisymmetric{\Bx \cdot \Bp}{V(\Bx)} \\
&=
\inv{2 m} \lr{ x_r p_r \Bp^2 – \Bp^2 x_r p_r}
+
\lr{ x_r p_r V(\Bx) – V(\Bx) x_r p_r } \\
&=
\inv{2 m} \antisymmetric{ x_r }{\Bp^2} p_r
+
x_r \antisymmetric{ p_r}{ V(\Bx)},
\end{aligned}

Evaluating those commutators separately, gives

\label{eqn:qmVirialTheorem:60}
\begin{aligned}
\antisymmetric{ x_r }{\Bp^2}
&=
\antisymmetric{ x_r }{p_r^2}\qquad \text{no sum} \\
&=
2 i \Hbar p_r,
\end{aligned}

and

\label{eqn:qmVirialTheorem:80}
\antisymmetric{ p_r}{ V(\Bx)}
= -i \Hbar \PD{x_r}{V(\Bx)},

so
\label{eqn:qmVirialTheorem:100}
\begin{aligned}
\ddt{}\lr{\Bx \cdot \Bp}
&=
\inv{i \Hbar}
\antisymmetric{\Bx \cdot \Bp}{H} \\
&=
\inv{2 m} 2 p_r p_r – x_r \PD{x_r}{V(\Bx)} \\
&=
\frac{\Bp^2}{m} – \Bx \cdot \spacegrad V(\Bx).
\end{aligned}

Taking expectation values, assuming that the states are independent of time, we have

\label{eqn:qmVirialTheorem:120}
\begin{aligned}
0
&= \ddt{} \expectation{ \Bx \cdot \Bp } \\
&= \expectation{\frac{\Bp^2}{m}} – \expectation{\Bx \cdot \spacegrad V(\Bx)}.
\end{aligned}

Note that taking the expectation with respect to stationary states was required to reverse the order of the time derivative with the expectation operation.

The right hand side is the quantum equivalent of the virial theorem, relating the average kinetic energy to the potential

\label{eqn:qmVirialTheorem:140}
2 \expectation{T} = \expectation{\Bx \cdot \spacegrad V(\Bx)}

# References

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

## A symmetric real Hamiltonian

August 31, 2015 phy1520 , ,

## Question: A symmetric real Hamiltonian ([1] pr. 2.9)

Find the time evolution for the state $$\ket{a’}$$ for a Hamiltian of the form

\label{eqn:symmetricHamiltonianEvolution:20}
H = \delta \lr{ \ket{a’}\bra{a’} + \ket{a”}\bra{a”} }

This Hamiltonian has the matrix representation

\label{eqn:symmetricHamiltonianEvolution:40}
H =
\begin{bmatrix}
0 & \delta \\
\delta & 0
\end{bmatrix},

which has a characteristic equation of

\label{eqn:symmetricHamiltonianEvolution:60}
\lambda^2 -\delta^2 = 0,

so the energy eigenvalues are $$\pm \delta$$.

The diagonal basis states are respectively

\label{eqn:symmetricHamiltonianEvolution:80}
\ket{\pm\delta} =
\inv{\sqrt{2}}
\begin{bmatrix}
\pm 1 \\
1
\end{bmatrix}.

The time evolution operator is

\label{eqn:symmetricHamiltonianEvolution:100}
\begin{aligned}
U
&= e^{-i H t/\Hbar} \\
&=
e^{-i \delta t/\Hbar} \ket{+\delta}\bra{+\delta}
+ e^{i \delta t/\Hbar} \ket{-\delta}\bra{-\delta} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1
\end{bmatrix}
\begin{bmatrix}
1 \\
1
\end{bmatrix}
+ \frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
-1 & 1
\end{bmatrix}
\begin{bmatrix}
-1 \\
1
\end{bmatrix} \\
&=
\frac{e^{-i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & 1 \\
1 & 1
\end{bmatrix}
+\frac{e^{i \delta t/\Hbar} }{2}
\begin{bmatrix}
1 & -1 \\
-1 & 1
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}.
\end{aligned}

The desired time evolution in the original basis is

\label{eqn:symmetricHamiltonianEvolution:140}
\begin{aligned}
\ket{a’, t}
&=
e^{-i H t/\Hbar}
\ket{a’, 0} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) & -i\sin(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar) & \cos(\delta t/\Hbar) \\
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
\cos(\delta t/\Hbar) \\
-i \sin(\delta t/\Hbar)
\end{bmatrix} \\
&=
\cos(\delta t/\Hbar) \ket{a’,0} -i \sin(\delta t/\Hbar) \ket{a”,0}.
\end{aligned}

This evolution has the same structure as left circularly polarized light.

The probability of finding the system in state $$\ket{a”}$$ given an initial state of $$\ket{a’,0}$$ is

\label{eqn:symmetricHamiltonianEvolution:160}
P
=
\Abs{\braket{a”}{a’,t}}^2
=
\sin^2 \lr{ \delta t/\Hbar }.

# References

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

## Quantum SHO ladder operators as a diagonal change of basis for the Heisenberg EOMs

Many authors pull the definitions of the raising and lowering (or ladder) operators out of their butt with no attempt at motivation. This is pointed out nicely in [1] by Eli along with one justification based on factoring the Hamiltonian.

In [2] is a small exception to the usual presentation. In that text, these operators are defined as usual with no motivation. However, after the utility of these operators has been shown, the raising and lowering operators show up in a context that does provide that missing motivation as a side effect.
It doesn’t look like the author was trying to provide a motivation, but it can be interpreted that way.

When seeking the time evolution of Heisenberg-picture position and momentum operators, we will see that those solutions can be trivially expressed using the raising and lowering operators. No special tools nor black magic is required to find the structure of these operators. Unfortunately, we must first switch to both the Heisenberg picture representation of the position and momentum operators, and also employ the Heisenberg equations of motion. Neither of these last two fit into standard narrative of most introductory quantum mechanics treatments. We will also see that these raising and lowering “operators” could also be introduced in classical mechanics, provided we were attempting to solve the SHO system using the Hamiltonian equations of motion.

I’ll outline this route to finding the structure of the ladder operators below. Because these are encountered trying to solve the time evolution problem, I’ll first show a simpler way to solve that problem. Because that simpler method depends a bit on lucky observation and is somewhat unstructured, I’ll then outline a more structured procedure that leads to the ladder operators directly, also providing the solution to the time evolution problem as a side effect.

The starting point is the Heisenberg equations of motion. For a time independent Hamiltonian $$H$$, and a Heisenberg operator $$A^{(H)}$$, those equations are

\label{eqn:harmonicOscDiagonalize:20}
\ddt{A^{(H)}} = \inv{i \Hbar} \antisymmetric{A^{(H)}}{H}.

Here the Heisenberg operator $$A^{(H)}$$ is related to the Schrodinger operator $$A^{(S)}$$ by

\label{eqn:harmonicOscDiagonalize:60}
A^{(H)} = U^\dagger A^{(S)} U,

where $$U$$ is the time evolution operator. For this discussion, we need only know that $$U$$ commutes with $$H$$, and do not need to know the specific structure of that operator. In particular, the Heisenberg equations of motion take the form

\label{eqn:harmonicOscDiagonalize:80}
\begin{aligned}
\ddt{A^{(H)}}
&= \inv{i \Hbar}
\antisymmetric{A^{(H)}}{H} \\
&= \inv{i \Hbar}
\antisymmetric{U^\dagger A^{(S)} U}{H} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} U H
– H U^\dagger A^{(S)} U
} \\
&= \inv{i \Hbar}
\lr{
U^\dagger A^{(S)} H U
– U^\dagger H A^{(S)} U
} \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{A^{(S)}}{H} U.
\end{aligned}

The Hamiltonian for the harmonic oscillator, with Schrodinger-picture position and momentum operators $$x, p$$ is

\label{eqn:harmonicOscDiagonalize:40}
H = \frac{p^2}{2m} + \inv{2} m \omega^2 x^2,

so the equations of motions are

\label{eqn:harmonicOscDiagonalize:100}
\begin{aligned}
\ddt{x^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{x}{\frac{p^2}{2m}} U \\
&= \inv{2 m i \Hbar} U^\dagger \lr{ i \Hbar \PD{p}{p^2} } U \\
&= \inv{m } U^\dagger p U \\
&= \inv{m } p^{(H)},
\end{aligned}

and
\label{eqn:harmonicOscDiagonalize:120}
\begin{aligned}
\ddt{p^{(H)}}
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{H} U \\
&= \inv{i \Hbar} U^\dagger \antisymmetric{p}{\inv{2} m \omega^2 x^2 } U \\
&= \frac{m \omega^2}{2 i \Hbar} U^\dagger \lr{ -i \Hbar \PD{x}{x^2} } U \\
&= -m \omega^2 U^\dagger x U \\
&= -m \omega^2 x^{(H)}.
\end{aligned}

In the Heisenberg picture the equations of motion are precisely those of classical Hamiltonian mechanics, except that we are dealing with operators instead of scalars

\label{eqn:harmonicOscDiagonalize:140}
\begin{aligned}
\ddt{p^{(H)}} &= -m \omega^2 x^{(H)} \\
\ddt{x^{(H)}} &= \inv{m } p^{(H)}.
\end{aligned}

In the text the ladder operators are used to simplify the solution of these coupled equations, since they can decouple them. That’s not really required since we can solve them directly in matrix form with little work

\label{eqn:harmonicOscDiagonalize:160}
\ddt{}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix}
=
\begin{bmatrix}
0 & -m \omega^2 \\
\inv{m} & 0
\end{bmatrix}
\begin{bmatrix}
p^{(H)} \\
x^{(H)}
\end{bmatrix},

or, with length scaled variables

\label{eqn:harmonicOscDiagonalize:180}
\begin{aligned}
\ddt{}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}
&=
\begin{bmatrix}
0 & -\omega \\
\omega & 0
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix} \\
&=
-i \omega
\sigma_y
\begin{bmatrix}
\frac{p^{(H)}}{m \omega} \\
x^{(H)}
\end{bmatrix}.
\end{aligned}

Writing $$y = \begin{bmatrix} \frac{p^{(H)}}{m \omega} \\ x^{(H)} \end{bmatrix}$$, the solution can then be written immediately as

\label{eqn:harmonicOscDiagonalize:200}
\begin{aligned}
y(t)
&=
\exp\lr{ -i \omega \sigma_y t } y(0) \\
&=
\lr{ \cos \lr{ \omega t } I – i \sigma_y \sin\lr{ \omega t } } y(0) \\
&=
\begin{bmatrix}
\cos\lr{ \omega t } & \sin\lr{ \omega t } \\
-\sin\lr{ \omega t } & \cos\lr{ \omega t }
\end{bmatrix}
y(0),
\end{aligned}

or

\label{eqn:harmonicOscDiagonalize:220}
\begin{aligned}
\frac{p^{(H)}(t)}{m \omega} &= \cos\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \sin\lr{ \omega t } x^{(H)}(0) \\
x^{(H)}(t) &= -\sin\lr{ \omega t } \frac{p^{(H)}(0)}{m \omega} + \cos\lr{ \omega t } x^{(H)}(0).
\end{aligned}

This solution depends on being lucky enough to recognize that the matrix has a Pauli matrix as a factor (which squares to unity, and allows the exponential to be evaluated easily.)

If we hadn’t been that observant, then the first tool we’d have used instead would have been to diagonalize the matrix. For such diagonalization, it’s natural to work in completely dimensionless variables. Such a non-dimensionalisation can be had by defining

\label{eqn:harmonicOscDiagonalize:240}
x_0 = \sqrt{\frac{\Hbar}{m \omega}},

and dividing the working (operator) variables through by those values. Let $$z = \inv{x_0} y$$, and $$\tau = \omega t$$ so that the equations of motion are

\label{eqn:harmonicOscDiagonalize:260}
\frac{dz}{d\tau}
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
z.

This matrix can be diagonalized as

\label{eqn:harmonicOscDiagonalize:280}
A
=
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
=
V
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
V^{-1},

where

\label{eqn:harmonicOscDiagonalize:300}
V =
\inv{\sqrt{2}}
\begin{bmatrix}
i & -i \\
1 & 1
\end{bmatrix}.

The equations of motion can now be written

\label{eqn:harmonicOscDiagonalize:320}
\frac{d}{d\tau} \lr{ V^{-1} z } =
\begin{bmatrix}
i & 0 \\
0 & -i
\end{bmatrix}
\lr{ V^{-1} z }.

This final change of variables $$V^{-1} z$$ decouples the system as desired. Expanding that gives

\label{eqn:harmonicOscDiagonalize:340}
\begin{aligned}
V^{-1} z
&=
\inv{\sqrt{2}}
\begin{bmatrix}
-i & 1 \\
i & 1
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i \frac{p^{(H)}}{m \omega} + x^{(H)} \\
i \frac{p^{(H)}}{m \omega} + x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
a^\dagger \\
a
\end{bmatrix},
\end{aligned}

where
\label{eqn:harmonicOscDiagonalize:n}
\begin{aligned}
a^\dagger &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ -i \frac{p^{(H)}}{m \omega} + x^{(H)} } \\
a &= \sqrt{\frac{m \omega}{2 \Hbar}} \lr{ i \frac{p^{(H)}}{m \omega} + x^{(H)} }.
\end{aligned}

Lo and behold, we have the standard form of the raising and lowering operators, and can write the system equations as

\label{eqn:harmonicOscDiagonalize:360}
\begin{aligned}
\ddt{a^\dagger} &= i \omega a^\dagger \\
\ddt{a} &= -i \omega a.
\end{aligned}

It is actually a bit fluky that this matched exactly, since we could have chosen eigenvectors that differ by constant phase factors, like

\label{eqn:harmonicOscDiagonalize:380}
V = \inv{\sqrt{2}}
\begin{bmatrix}
i e^{i\phi} & -i e^{i \psi} \\
1 e^{i\phi} & e^{i \psi}
\end{bmatrix},

so

\label{eqn:harmonicOscDiagonalize:341}
\begin{aligned}
V^{-1} z
&=
\frac{e^{-i(\phi + \psi)}}{\sqrt{2}}
\begin{bmatrix}
-i e^{i\psi} & e^{i \psi} \\
i e^{i\phi} & e^{i \phi}
\end{bmatrix}
\begin{bmatrix}
\frac{p^{(H)}}{x_0 m \omega} \\
\frac{x^{(H)}}{x_0}
\end{bmatrix} \\
&=
\inv{\sqrt{2} x_0}
\begin{bmatrix}
-i e^{i\phi} \frac{p^{(H)}}{m \omega} + e^{i\phi} x^{(H)} \\
i e^{i\psi} \frac{p^{(H)}}{m \omega} + e^{i\psi} x^{(H)}
\end{bmatrix} \\
&=
\begin{bmatrix}
e^{i\phi} a^\dagger \\
e^{i\psi} a
\end{bmatrix}.
\end{aligned}

To make the resulting pairs of operators Hermitian conjugates, we’d want to constrain those constant phase factors by setting $$\phi = -\psi$$. If we were only interested in solving the time evolution problem no such additional constraints are required.

The raising and lowering operators are seen to naturally occur when seeking the solution of the Heisenberg equations of motion. This is found using the standard technique of non-dimensionalisation and then seeking a change of basis that diagonalizes the system matrix. Because the Heisenberg equations of motion are identical to the classical Hamiltonian equations of motion in this case, what we call the raising and lowering operators in quantum mechanics could also be utilized in the classical simple harmonic oscillator problem. However, in a classical context we wouldn’t have a justification to call this more than a change of basis.

# References

[1] Eli Lansey. The Quantum Harmonic Oscillator Ladder Operators, 2009. URL http://behindtheguesses.blogspot.ca/2009/03/quantum-harmonic-oscillator-ladder.html. [Online; accessed 18-August-2015].

[2] Jun John Sakurai and Jim J Napolitano. Modern quantum mechanics, chapter {Time Development of the Oscillator}. Pearson Higher Ed, 2014.