Harmonic oscillator

2D SHO xy perturbation

December 7, 2015 phy1520 No comments , , , , ,

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Q: [1] pr. 5.4

Given a 2D SHO with Hamiltonian

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:20}
H_0 = \inv{2m} \lr{ p_x^2 + p_y^2 } + \frac{m \omega^2}{2} \lr{ x^2 + y^2 },
\end{equation}

  • (a)
    What are the energies and degeneracies of the three lowest states?

  • (b)
    With perturbation

    \begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:40}
    V = m \omega^2 x y,
    \end{equation}

    calculate the first order energy perturbations and the zeroth order perturbed states.

  • (c)
    Solve the \( H_0 + \delta V \) problem exactly, and compare.

A: part (a)

Recall that we have

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:60}
H \ket{n_1, n_2} =
\Hbar\omega
\lr{
n_1 + n_2 + 1
}
\ket{n_1, n_2},
\end{equation}

So the three lowest energy states are \( \ket{0,0}, \ket{1,0}, \ket{0,1} \) with energies \( \Hbar \omega, 2 \Hbar \omega, 2 \Hbar \omega \) respectively (with a two fold degeneracy for the second two energy eigenkets).

A: part (b)

Consider the action of \( x y \) on the \( \beta = \setlr{ \ket{0,0}, \ket{1,0}, \ket{0,1} } \) subspace. Those are

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:200}
\begin{aligned}
x y \ket{0,0}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{0,0} \\
&=
\frac{x_0^2}{2} \lr{ b + b^\dagger } \ket{1,0} \\
&=
\frac{x_0^2}{2} \ket{1,1}.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:220}
\begin{aligned}
x y \ket{1, 0}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{1,0} \\
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \ket{1,1} \\
&=
\frac{x_0^2}{2} \lr{ \ket{0,1} + \sqrt{2} \ket{2,1} } .
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:240}
\begin{aligned}
x y \ket{0, 1}
&=
\frac{x_0^2}{2} \lr{ a + a^\dagger } \lr{ b + b^\dagger } \ket{0,1} \\
&=
\frac{x_0^2}{2} \lr{ b + b^\dagger } \ket{1,1} \\
&=
\frac{x_0^2}{2} \lr{ \ket{1,0} + \sqrt{2} \ket{1,2} }.
\end{aligned}
\end{equation}

The matrix representation of \( m \omega^2 x y \) with respect to the subspace spanned by basis \( \beta \) above is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:260}
x y
\sim
\inv{2} \Hbar \omega
\begin{bmatrix}
0 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0 \\
\end{bmatrix}.
\end{equation}

This diagonalizes with

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:300}
U
=
\begin{bmatrix}
1 & 0 \\
0 & \tilde{U}
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:320}
\tilde{U}
=
\inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1 \\
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:340}
D =
\inv{2} \Hbar \omega
\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1 \\
\end{bmatrix}
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:360}
x y = U D U^\dagger = U D U.
\end{equation}

The unperturbed Hamiltonian in the original basis is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:380}
H_0
=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & 2 I
\end{bmatrix},
\end{equation}

So the transformation to the diagonal \( x y \) basis leaves the initial Hamiltonian unaltered

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:400}
\begin{aligned}
H_0′
&= U^\dagger H_0 U \\
&=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & \tilde{U} 2 I \tilde{U}
\end{bmatrix} \\
&=
\Hbar \omega
\begin{bmatrix}
1 & 0 \\
0 & 2 I
\end{bmatrix}.
\end{aligned}
\end{equation}

Now we can compute the first order energy shifts almost by inspection. Writing the new basis as \( \beta’ = \setlr{ \ket{0}, \ket{1}, \ket{2} } \) those energy shifts are just the diagonal elements from the \( x y \) operators matrix representation

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:420}
\begin{aligned}
E^{{(1)}}_0 &= \bra{0} V \ket{0} = 0 \\
E^{{(1)}}_1 &= \bra{1} V \ket{1} = \inv{2} \Hbar \omega \\
E^{{(1)}}_2 &= \bra{2} V \ket{2} = -\inv{2} \Hbar \omega.
\end{aligned}
\end{equation}

The new energies are

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:440}
\begin{aligned}
E_0 &\rightarrow \Hbar \omega \\
E_1 &\rightarrow \Hbar \omega \lr{ 2 + \delta/2 } \\
E_2 &\rightarrow \Hbar \omega \lr{ 2 – \delta/2 }.
\end{aligned}
\end{equation}

A: part (c)

For the exact solution, it’s possible to rotate the coordinate system in a way that kills the explicit \( x y \) term of the perturbation. That we could do this for \( x, y \) operators wasn’t obvious to me, but after doing so (and rotating the momentum operators the same way) the new operators still have the required commutators. Let

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:80}
\begin{aligned}
\begin{bmatrix}
u \\
v
\end{bmatrix}
&=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix} \\
&=
\begin{bmatrix}
x \cos\theta + y \sin\theta \\
-x \sin\theta + y \cos\theta
\end{bmatrix}.
\end{aligned}
\end{equation}

Similarly, for the momentum operators, let
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:100}
\begin{aligned}
\begin{bmatrix}
p_u \\
p_v
\end{bmatrix}
&=
\begin{bmatrix}
\cos\theta & \sin\theta \\
-\sin\theta & \cos\theta
\end{bmatrix}
\begin{bmatrix}
p_x \\
p_y
\end{bmatrix} \\
&=
\begin{bmatrix}
p_x \cos\theta + p_y \sin\theta \\
-p_x \sin\theta + p_y \cos\theta
\end{bmatrix}.
\end{aligned}
\end{equation}

For the commutators of the new operators we have

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:120}
\begin{aligned}
\antisymmetric{u}{p_u}
&=
\antisymmetric{x \cos\theta + y \sin\theta}{p_x \cos\theta + p_y \sin\theta} \\
&=
\antisymmetric{x}{p_x} \cos^2\theta + \antisymmetric{y}{p_y} \sin^2\theta \\
&=
i \Hbar \lr{ \cos^2\theta + \sin^2\theta } \\
&=
i\Hbar.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:140}
\begin{aligned}
\antisymmetric{v}{p_v}
&=
\antisymmetric{-x \sin\theta + y \cos\theta}{-p_x \sin\theta + p_y \cos\theta} \\
&=
\antisymmetric{x}{p_x} \sin^2\theta + \antisymmetric{y}{p_y} \cos^2\theta \\
&=
i \Hbar.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:160}
\begin{aligned}
\antisymmetric{u}{p_v}
&=
\antisymmetric{x \cos\theta + y \sin\theta}{-p_x \sin\theta + p_y \cos\theta} \\
&= \cos\theta \sin\theta \lr{ -\antisymmetric{x}{p_x} + \antisymmetric{y}{p_p} } \\
&=
0.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:180}
\begin{aligned}
\antisymmetric{v}{p_u}
&=
\antisymmetric{-x \sin\theta + y \cos\theta}{p_x \cos\theta + p_y \sin\theta} \\
&= \cos\theta \sin\theta \lr{ -\antisymmetric{x}{p_x} + \antisymmetric{y}{p_p} } \\
&=
0.
\end{aligned}
\end{equation}

We see that the new operators are canonical conjugate as required. For this problem, we just want a 45 degree rotation, with

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:460}
\begin{aligned}
x &= \inv{\sqrt{2}} \lr{ u + v } \\
y &= \inv{\sqrt{2}} \lr{ u – v }.
\end{aligned}
\end{equation}

We have
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:480}
\begin{aligned}
x^2 + y^2
&=
\inv{2} \lr{ (u+v)^2 + (u-v)^2 } \\
&=
\inv{2} \lr{ 2 u^2 + 2 v^2 + 2 u v – 2 u v } \\
&=
u^2 + v^2,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:500}
\begin{aligned}
p_x^2 + p_y^2
&=
\inv{2} \lr{ (p_u+p_v)^2 + (p_u-p_v)^2 } \\
&=
\inv{2} \lr{ 2 p_u^2 + 2 p_v^2 + 2 p_u p_v – 2 p_u p_v } \\
&=
p_u^2 + p_v^2,
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:520}
\begin{aligned}
x y
&=
\inv{2} \lr{ (u+v)(u-v) } \\
&=
\inv{2} \lr{ u^2 – v^2 }.
\end{aligned}
\end{equation}

The perturbed Hamiltonian is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:540}
\begin{aligned}
H_0 + \delta V
&=
\inv{2m} \lr{ p_u^2 + p_v^2 }
+ \inv{2} m \omega^2 \lr{ u^2 + v^2 + \delta u^2 – \delta v^2 } \\
&=
\inv{2m} \lr{ p_u^2 + p_v^2 }
+ \inv{2} m \omega^2 \lr{ u^2(1 + \delta) + v^2 (1 – \delta) }.
\end{aligned}
\end{equation}

In this coordinate system, the corresponding eigensystem is

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:560}
H \ket{n_1, n_2}
= \Hbar \omega \lr{ 1 + n_1 \sqrt{1 + \delta} + n_2 \sqrt{ 1 – \delta } } \ket{n_1, n_2}.
\end{equation}

For small \( \delta \)

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:580}
n_1 \sqrt{1 + \delta} + n_2 \sqrt{ 1 – \delta }
\approx
n_1 + n_2
+ \inv{2} n_1 \delta
– \inv{2} n_2 \delta,
\end{equation}

so
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:600}
H \ket{n_1, n_2}
\approx \Hbar \omega \lr{ 1 + n_1 + n_2 + \inv{2} n_1 \delta – \inv{2} n_2 \delta
} \ket{n_1, n_2}.
\end{equation}

The lowest order perturbed energy levels are

\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:620}
\ket{0,0} \rightarrow \Hbar \omega
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:640}
\ket{1,0} \rightarrow \Hbar \omega \lr{ 2 + \inv{2} \delta }
\end{equation}
\begin{equation}\label{eqn:2dHarmonicOscillatorXYPerturbation:660}
\ket{0,1} \rightarrow \Hbar \omega \lr{ 2 – \inv{2} \delta }
\end{equation}

The degeneracy of the \( \ket{0,1}, \ket{1,0} \) states has been split, and to first order match the zeroth order perturbation result.

References

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

Harmonic oscillator with energy shift

December 5, 2015 phy1520 No comments , ,

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Q: [1] pr 5.1

Given a perturbed 1D SHO Hamiltonian

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:20}
H = \inv{2m} p^2 + \inv{2} m \omega^2 x^2 + \lambda b x,
\end{equation}

calculate the first non-zero perturbation to the ground state energy. Then solve for that energy directly and compare.

A:

The first order energy shift is seen to be zero

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:40}
\begin{aligned}
\Delta_0^{(0)}
&= V_{00} \\
&= \bra{0} b x \ket{0} \\
&= \frac{x_0}{\sqrt{2}} \bra{0} a + a^\dagger \ket{0} \\
&= \frac{x_0}{\sqrt{2}} \braket{0}{1} \\
&= 0.
\end{aligned}
\end{equation}

The first order perturbation to the ground state is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:60}
\begin{aligned}
\ket{0^{(1)}}
&= \sum_{m \ne 0} \frac{ \ket{m} \bra{m} b x \ket{0} }{ \Hbar \omega/2 – \Hbar
\omega (m – 1/2) } \\
&= -b \frac{x_0}{\sqrt{2} \Hbar \omega} \sum_{m \ne 0} \frac{ \ket{m}
\braket{m}{1} }{ m } \\
&= -b \frac{x_0}{\sqrt{2} \Hbar \omega} \ket{1}.
\end{aligned}
\end{equation}

The second order ground state energy perturbation is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:80}
\begin{aligned}
\Delta_0^{(2)}
&=
\bra{0} b x \ket{0^{(1)}} \\
&=
\frac{b x_0}{\sqrt{2}} \bra{0} a + a^\dagger \lr{ -b \frac{x_0}{\sqrt{2} \Hbar \omega} \ket{1} } \\
&=
\frac{b x_0}{\sqrt{2}} \lr{ -b \frac{x_0}{\sqrt{2} \Hbar \omega} } \\
&=
-\frac{b^2 x_0^2}{ 2 \Hbar \omega } \\
&=
-\frac{b^2 }{ 2 \Hbar \omega } \frac{\Hbar}{m \omega} \\
&=
-\frac{b^2 }{ 2 m \omega^2 },
\end{aligned}
\end{equation}

so the total energy perturbation up to second order is

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:100}
\Delta_0 = -\lambda^2 \frac{b^2 }{ 2 m \omega^2 }.
\end{equation}

To compare to the exact result, rewrite the Hamiltonian as

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:120}
\begin{aligned}
H
&= \inv{2m} p^2 + \inv{2} m \omega^2 \lr{ x^2 + \frac{2 \lambda b x}{m \omega^2} } \\
&= \inv{2m} p^2 + \inv{2} m \omega^2 \lr{ x + \frac{\lambda b }{m \omega^2} }^2 – \inv{2} m \omega^2 \lr{ \frac{\lambda b }{m \omega^2} }^2.
\end{aligned}
\end{equation}

The Hamiltonian is subject to a constant energy shift

\begin{equation}\label{eqn:harmonicOscillatorEnergyShiftPertubation:140}
\begin{aligned}
\Delta E
&=
– \inv{2} m \omega^2 \frac{\lambda^2 b^2 }{m^2 \omega^4} \\
&=
– \frac{\lambda^2 b^2 }{2 m \omega^2}.
\end{aligned}
\end{equation}

This is an exact match with the second order perturbation result of \ref{eqn:harmonicOscillatorEnergyShiftPertubation:100}.

References

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

Plane wave ground state expectation for SHO

October 18, 2015 phy1520 No comments , , , , , , , , , , ,

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Problem [1] 2.18 is, for a 1D SHO, show that

\begin{equation}\label{eqn:exponentialExpectationGroundState:20}
\bra{0} e^{i k x} \ket{0} = \exp\lr{ -k^2 \bra{0} x^2 \ket{0}/2 }.
\end{equation}

Despite the simple appearance of this problem, I found this quite involved to show. To do so, start with a series expansion of the expectation

\begin{equation}\label{eqn:exponentialExpectationGroundState:40}
\bra{0} e^{i k x} \ket{0}
=
\sum_{m=0}^\infty \frac{(i k)^m}{m!} \bra{0} x^m \ket{0}.
\end{equation}

Let

\begin{equation}\label{eqn:exponentialExpectationGroundState:60}
X = \lr{ a + a^\dagger },
\end{equation}

so that

\begin{equation}\label{eqn:exponentialExpectationGroundState:80}
x
= \sqrt{\frac{\Hbar}{2 \omega m}} X
= \frac{x_0}{\sqrt{2}} X.
\end{equation}

Consider the first few values of \( \bra{0} X^n \ket{0} \)

\begin{equation}\label{eqn:exponentialExpectationGroundState:100}
\begin{aligned}
\bra{0} X \ket{0}
&=
\bra{0} \lr{ a + a^\dagger } \ket{0} \\
&=
\braket{0}{1} \\
&=
0,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:exponentialExpectationGroundState:120}
\begin{aligned}
\bra{0} X^2 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^2 \ket{0} \\
&=
\braket{1}{1} \\
&=
1,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:exponentialExpectationGroundState:140}
\begin{aligned}
\bra{0} X^3 \ket{0}
&=
\bra{0} \lr{ a + a^\dagger }^3 \ket{0} \\
&=
\bra{1} \lr{ \sqrt{2} \ket{2} + \ket{0} } \\
&=
0.
\end{aligned}
\end{equation}

Whenever the power \( n \) in \( X^n \) is even, the braket can be split into a bra that has only contributions from odd eigenstates and a ket with even eigenstates. We conclude that \( \bra{0} X^n \ket{0} = 0 \) when \( n \) is odd.

Noting that \( \bra{0} x^2 \ket{0} = \ifrac{x_0^2}{2} \), this leaves

\begin{equation}\label{eqn:exponentialExpectationGroundState:160}
\begin{aligned}
\bra{0} e^{i k x} \ket{0}
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \bra{0} x^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{(i k)^{2 m}}{(2 m)!} \lr{ \frac{x_0^2}{2} }^m \bra{0} X^{2m} \ket{0} \\
&=
\sum_{m=0}^\infty \frac{1}{(2 m)!} \lr{ -k^2 \bra{0} x^2 \ket{0} }^m \bra{0} X^{2m} \ket{0}.
\end{aligned}
\end{equation}

This problem is now reduced to showing that

\begin{equation}\label{eqn:exponentialExpectationGroundState:180}
\frac{1}{(2 m)!} \bra{0} X^{2m} \ket{0} = \inv{m! 2^m},
\end{equation}

or

\begin{equation}\label{eqn:exponentialExpectationGroundState:200}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&= \frac{(2m)!}{m! 2^m} \\
&= \frac{ (2m)(2m-1)(2m-2) \cdots (2)(1) }{2^m m!} \\
&= \frac{ 2^m (m)(2m-1)(m-1)(2m-3)(m-2) \cdots (2)(3)(1)(1) }{2^m m!} \\
&= (2m-1)!!,
\end{aligned}
\end{equation}

where \( n!! = n(n-2)(n-4)\cdots \).

It looks like \( \bra{0} X^{2m} \ket{0} \) can be expanded by inserting an identity operator and proceeding recursively, like

\begin{equation}\label{eqn:exponentialExpectationGroundState:220}
\begin{aligned}
\bra{0} X^{2m} \ket{0}
&=
\bra{0} X^2 \lr{ \sum_{n=0}^\infty \ket{n}\bra{n} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^2 \lr{ \ket{0}\bra{0} + \ket{2}\bra{2} } X^{2m-2} \ket{0} \\
&=
\bra{0} X^{2m-2} \ket{0} + \bra{0} X^2 \ket{2} \bra{2} X^{2m-2} \ket{0}.
\end{aligned}
\end{equation}

This has made use of the observation that \( \bra{0} X^2 \ket{n} = 0 \) for all \( n \ne 0,2 \). The remaining term includes the factor

\begin{equation}\label{eqn:exponentialExpectationGroundState:240}
\begin{aligned}
\bra{0} X^2 \ket{2}
&=
\bra{0} \lr{a + a^\dagger}^2 \ket{2} \\
&=
\lr{ \bra{0} + \sqrt{2} \bra{2} } \ket{2} \\
&=
\sqrt{2},
\end{aligned}
\end{equation}

Since \( \sqrt{2} \ket{2} = \lr{a^\dagger}^2 \ket{0} \), the expectation of interest can be written

\begin{equation}\label{eqn:exponentialExpectationGroundState:260}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + \bra{0} a^2 X^{2m-2} \ket{0}.
\end{equation}

How do we expand the second term. Let’s look at how \( a \) and \( X \) commute

\begin{equation}\label{eqn:exponentialExpectationGroundState:280}
\begin{aligned}
a X
&=
\antisymmetric{a}{X} + X a \\
&=
\antisymmetric{a}{a + a^\dagger} + X a \\
&=
\antisymmetric{a}{a^\dagger} + X a \\
&=
1 + X a,
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:exponentialExpectationGroundState:300}
\begin{aligned}
a^2 X
&=
a \lr{ a X } \\
&=
a \lr{ 1 + X a } \\
&=
a + a X a \\
&=
a + \lr{ 1 + X a } a \\
&=
2 a + X a^2.
\end{aligned}
\end{equation}

Proceeding to expand \( a^2 X^n \) we find
\begin{equation}\label{eqn:exponentialExpectationGroundState:320}
\begin{aligned}
a^2 X^3 &= 6 X + 6 X^2 a + X^3 a^2 \\
a^2 X^4 &= 12 X^2 + 8 X^3 a + X^4 a^2 \\
a^2 X^5 &= 20 X^3 + 10 X^4 a + X^5 a^2 \\
a^2 X^6 &= 30 X^4 + 12 X^5 a + X^6 a^2.
\end{aligned}
\end{equation}

It appears that we have
\begin{equation}\label{eqn:exponentialExpectationGroundState:340}
\antisymmetric{a^2 X^n}{X^n a^2} = \beta_n X^{n-2} + 2 n X^{n-1} a,
\end{equation}

where

\begin{equation}\label{eqn:exponentialExpectationGroundState:360}
\beta_n = \beta_{n-1} + 2 (n-1),
\end{equation}

and \( \beta_2 = 2 \). Some goofing around shows that \( \beta_n = n(n-1) \), so the induction hypothesis is

\begin{equation}\label{eqn:exponentialExpectationGroundState:380}
\antisymmetric{a^2 X^n}{X^n a^2} = n(n-1) X^{n-2} + 2 n X^{n-1} a.
\end{equation}

Let’s check the induction
\begin{equation}\label{eqn:exponentialExpectationGroundState:400}
\begin{aligned}
a^2 X^{n+1}
&=
a^2 X^{n} X \\
&=
\lr{ n(n-1) X^{n-2} + 2 n X^{n-1} a + X^n a^2 } X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} a X + X^n a^2 X \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} \lr{ 1 + X a } + X^n \lr{ 2 a + X a^2 } \\
&=
n(n-1) X^{n-1} + 2 n X^{n-1} + 2 n X^{n} a
+ 2 X^n a
+ X^{n+1} a^2 \\
&=
X^{n+1} a^2 + (2 + 2 n) X^{n} a + \lr{ 2 n + n(n-1) } X^{n-1} \\
&=
X^{n+1} a^2 + 2(n + 1) X^{n} a + (n+1) n X^{n-1},
\end{aligned}
\end{equation}

which concludes the induction, giving

\begin{equation}\label{eqn:exponentialExpectationGroundState:420}
\bra{ 0 } a^2 X^{n} \ket{0 } = n(n-1) \bra{0} X^{n-2} \ket{0},
\end{equation}

and

\begin{equation}\label{eqn:exponentialExpectationGroundState:440}
\bra{0} X^{2m} \ket{0}
=
\bra{0} X^{2m-2} \ket{0} + (2m-2)(2m-3) \bra{0} X^{2m-4} \ket{0}.
\end{equation}

Let

\begin{equation}\label{eqn:exponentialExpectationGroundState:460}
\sigma_{n} = \bra{0} X^n \ket{0},
\end{equation}

so that the recurrence relation, for \( 2n \ge 4 \) is

\begin{equation}\label{eqn:exponentialExpectationGroundState:480}
\sigma_{2n} = \sigma_{2n -2} + (2n-2)(2n-3) \sigma_{2n -4}
\end{equation}

We want to show that this simplifies to

\begin{equation}\label{eqn:exponentialExpectationGroundState:500}
\sigma_{2n} = (2n-1)!!
\end{equation}

The first values are

\begin{equation}\label{eqn:exponentialExpectationGroundState:540}
\sigma_0 = \bra{0} X^0 \ket{0} = 1
\end{equation}
\begin{equation}\label{eqn:exponentialExpectationGroundState:560}
\sigma_2 = \bra{0} X^2 \ket{0} = 1
\end{equation}

which gives us the right result for the first term in the induction

\begin{equation}\label{eqn:exponentialExpectationGroundState:580}
\begin{aligned}
\sigma_4
&= \sigma_2 + 2 \times 1 \times \sigma_0 \\
&= 1 + 2 \\
&= 3!!
\end{aligned}
\end{equation}

For the general induction term, consider

\begin{equation}\label{eqn:exponentialExpectationGroundState:600}
\begin{aligned}
\sigma_{2n + 2}
&= \sigma_{2n} + 2 n (2n – 1) \sigma_{2n -2} \\
&= (2n-1)!! + 2n ( 2n – 1) (2n -3)!! \\
&= (2n + 1) (2n -1)!! \\
&= (2n + 1)!!,
\end{aligned}
\end{equation}

which completes the final induction. That was also the last thing required to complete the proof, so we are done!

References

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

Expectations for SHO Hamiltonian, and virial theorem.

October 15, 2015 phy1520 No comments , , , , ,

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Question: Expectations for SHO Hamiltonian, and virial theorem. ([1] pr. 2.3)

(a)

For a 1D SHO, compute \(
\bra{m} x \ket{n},
\bra{m} x^2 \ket{n},
\bra{m} p \ket{n},
\bra{m} p^2 \ket{n} \) and \( \bra{m} \symmetric{x}{p} \ket{n} \).

(b)

Verify the virial theorem is satisfied for energy eigenstates.

Answer

(a)

Using

\begin{equation}\label{eqn:shoExpectations:20}
\begin{aligned}
x &= \frac{x_0}{\sqrt{2}} \lr{ a + a^\dagger } \\
p &= \frac{i\Hbar}{x_0 \sqrt{2}} \lr{ a^\dagger – a} \\
a(t) &= a(0) e^{-i \omega t} \\
a(0) \ket{n} &= \sqrt{n} \ket{n-1} \\
a^\dagger(0) \ket{n} &= \sqrt{n+1} \ket{n+1} \\
x_0^2 &= \frac{\Hbar}{\omega m},
\end{aligned}
\end{equation}

we have

\begin{equation}\label{eqn:shoExpectations:40}
\begin{aligned}
\bra{m} x \ket{n}
&=
\frac{x_0}{\sqrt{2}} \bra{m} \lr{ a + a^\dagger } \ket{n} \\
&=
\frac{x_0}{\sqrt{2}} \bra{m}
\lr{
e^{-i \omega t} \sqrt{n} \ket{n-1}
+
e^{i \omega t} \sqrt{n+1} \ket{n+1}
} \\
&=
\frac{x_0}{\sqrt{2}} \lr{
\delta_{m, n-1} e^{-i \omega t} \sqrt{n}
+
\delta_{m, n+1} e^{i \omega t} \sqrt{n+1}
},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:shoExpectations:60}
\begin{aligned}
\bra{m} x^2 \ket{n}
&=
\frac{x_0^2}{2} \bra{m} \lr{ a + a^\dagger }^2 \ket{n} \\
&=
\frac{x_0^2}{2}
\lr{
e^{i \omega t} \sqrt{m} \bra{m-1}
+
e^{-i \omega t} \sqrt{m+1} \bra{m+1}
}
\lr{
e^{-i \omega t} \sqrt{n} \ket{n-1}
+
e^{i \omega t} \sqrt{n+1} \ket{n+1}
} \\
&=
\frac{x_0^2}{2}
\lr{
\delta_{m+1,n+1} \sqrt{(m+1)(n+1)}
+\delta_{m+1,n-1} \sqrt{(m+1)n} e^{-2 i \omega t}
+\delta_{m-1,n+1} \sqrt{m(n+1)} e^{2 i \omega t}
+\delta_{m-1,n-1} \sqrt{m n}
},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:shoExpectations:80}
\begin{aligned}
\bra{m} p \ket{n}
&=
\frac{i\Hbar}{\sqrt{2} x_0} \bra{m} \lr{ a^\dagger – a} \ket{n} \\
&=
\frac{i\Hbar}{\sqrt{2} x_0} \bra{m} \lr{
e^{i \omega t} \sqrt{n+1} \ket{n+1}

e^{-i \omega t} \sqrt{n} \ket{n-1}
} \\
&=
\frac{i\Hbar}{\sqrt{2} x_0} \lr{
\delta_{m,n+1} e^{i \omega t} \sqrt{n+1}

\delta_{m,n-1} e^{-i \omega t} \sqrt{n}
},
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:shoExpectations:100}
\begin{aligned}
\bra{m} p^2 \ket{n}
&=
\frac{\Hbar^2}{2 x_0^2} \ket{m} \lr{ a – a^\dagger } \lr{ a^\dagger – a}
\ket{n} \\
&=
\frac{\Hbar^2}{2 x_0^2}
\lr{
-e^{-i \omega t} \sqrt{m+1} \bra{m+1}
+
e^{i \omega t} \sqrt{m} \bra{m-1}
}
\lr{
e^{i \omega t} \sqrt{n+1} \ket{n+1}

e^{-i \omega t} \sqrt{n} \ket{n-1}
} \\
&=
\frac{\Hbar^2}{2 x_0^2}
\lr{
\delta_{m+1,n+1} \sqrt{(m+1)(n+1)}
+\delta_{m+1,n-1} \sqrt{(m+1)n} e^{-2 i \omega t}
+\delta_{m-1,n+1} \sqrt{m(n+1)} e^{2 i \omega t}
+\delta_{m-1,n-1} \sqrt{m n}
}.
\end{aligned}
\end{equation}

For the anticommutator \( \symmetric{x}{p} \), we have

\begin{equation}\label{eqn:shoExpectations:120}
\begin{aligned}
\symmetric{x}{p}
&=
\frac{i\Hbar}{2}
\lr{
\lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} } \lr{ a^\dagger e^{i \omega t} – a e^{-i \omega t} }

\lr{ a^\dagger e^{i \omega t} – a e^{-i \omega t} }
\lr{ a e^{-i \omega t} + a^\dagger e^{i \omega t} }
} \\
&=
\frac{i\Hbar}{2}
\lr{
– a^2 e^{- 2 i \omega t}
+ (a^\dagger)^2 e^{ 2 i \omega t}
+ a a^\dagger
– a^\dagger a
+ a^2 e^{- 2 i \omega t}
– (a^\dagger)^2 e^{ 2 i \omega t}
– a^\dagger a
+ a a^\dagger
} \\
&=
i\Hbar
\lr{
a a^\dagger – a^\dagger a
},
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:shoExpectations:140}
\begin{aligned}
\bra{m} \symmetric{x}{p} \ket{n}
&=
i\Hbar
\bra{m}
\lr{
a a^\dagger – a^\dagger a
}
\ket{n} \\
&=
i\Hbar
\bra{m}
\lr{
\sqrt{(n+1)^2}\ket{n}
-\sqrt{n^2}\ket{n}
} \\
&=
i\Hbar
\bra{m}
\lr{
2 n + 1
}
\ket{n}.
\end{aligned}
\end{equation}

(b)

For the SHO, the virial theorem requires \( \expectation{p^2/m} = \expectation{m \omega x^2} \). That momentum expectation with respect to the eigenstate \( \ket{n} \) is

\begin{equation}\label{eqn:shoExpectations:160}
\begin{aligned}
\expectation{p^2/m}
&=
\frac{\Hbar^2}{2 x_0^2 m}
\lr{
\sqrt{(n+1)(n+1)}
+
\sqrt{n n}
} \\
&=
\frac{\Hbar^2 m \omega}{2 \Hbar m} \lr{ 2 n + 1 } \\
&=
\Hbar \omega \lr{ n + \inv{2} }.
\end{aligned}
\end{equation}

For the position expectation we’ve got

\begin{equation}\label{eqn:shoExpectations:180}
\begin{aligned}
\expectation{m \omega x^2}
&=
\frac{m \omega^2 x_0^2}{2}
\lr{
\sqrt{(n+1)(n+1)}
+ \sqrt{n n}
} \\
&=
\frac{m \omega^2 \Hbar}{2 m \omega}
\lr{
\sqrt{(n+1)(n+1)}
+ \sqrt{n n}
} \\
&=
\frac{\omega \Hbar}{2 }
\lr{ 2 n + 1 } \\
&=
\omega \Hbar
\lr{ n + \inv{2} }.
\end{aligned}
\end{equation}

This shows that the virial theorem holds for the SHO Hamiltonian for eigenstates.

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 5: time evolution of coherent states, and charged particles in a magnetic field. Taught by Prof. Arun Paramekanti

October 1, 2015 phy1520 No comments , , , , , , , , , , , ,

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

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 \textchapref{{1}} [1] content.

Coherent states (cont.)

A coherent state for the SHO \( H = \lr{ N + \inv{2} } \Hbar \omega \) was given by

\begin{equation}\label{eqn:qmLecture5:20}
a \ket{z} = z \ket{z},
\end{equation}

where we showed that

\begin{equation}\label{eqn:qmLecture5:40}
\ket{z} = c_0 e^{ z a^\dagger } \ket{0}.
\end{equation}

In the Heisenberg picture we found

\begin{equation}\label{eqn:qmLecture5:60}
\begin{aligned}
a_{\textrm{H}}(t) &= e^{i H t/\Hbar} a e^{-i H t/\Hbar} = a e^{-i\omega t} \\
a_{\textrm{H}}^\dagger(t) &= e^{i H t/\Hbar} a^\dagger e^{-i H t/\Hbar} = a^\dagger e^{i\omega t}.
\end{aligned}
\end{equation}

Recall that the position and momentum representation of the ladder operators was

\begin{equation}\label{eqn:qmLecture5:80}
\begin{aligned}
a &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} + i \hat{p} \sqrt{\inv{m \Hbar \omega}} } \\
a^\dagger &= \inv{\sqrt{2}} \lr{ \hat{x} \sqrt{\frac{m \omega}{\Hbar}} – i \hat{p} \sqrt{\inv{m \Hbar \omega}} },
\end{aligned}
\end{equation}

or equivalently
\begin{equation}\label{eqn:qmLecture5:100}
\begin{aligned}
\hat{x} &= \lr{ a + a^\dagger } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
\hat{p} &= i \lr{ a^\dagger – a } \sqrt{\frac{m \Hbar \omega}{2}}.
\end{aligned}
\end{equation}

Given this we can compute expectation value of position operator

\begin{equation}\label{eqn:qmLecture5:120}
\begin{aligned}
\bra{z} \hat{x} \ket{z}
&=
\sqrt{\frac{\Hbar}{ 2 m \omega}}
\bra{z}
\lr{ a + a^\dagger }
\ket{z} \\
&=
\lr{ z + z^\conj } \sqrt{\frac{\Hbar}{ 2 m \omega}} \\
&=
2 \textrm{Re} z \sqrt{\frac{\Hbar}{ 2 m \omega}} .
\end{aligned}
\end{equation}

Similarly

\begin{equation}\label{eqn:qmLecture5:140}
\begin{aligned}
\bra{z} \hat{p} \ket{z}
&=
i \sqrt{\frac{m \Hbar \omega}{2}}
\bra{z}
\lr{ a^\dagger – a }
\ket{z} \\
&=
\sqrt{\frac{m \Hbar \omega}{2}}
2 \textrm{Im} z.
\end{aligned}
\end{equation}

How about the expectation of the Heisenberg position operator? That is

\begin{equation}\label{eqn:qmLecture5:160}
\begin{aligned}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z}
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \bra{z} \lr{ a + a^\dagger } \ket{z} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ z e^{-i \omega t} + z^\conj e^{i \omega t}} \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \lr{z + z^\conj} \cos( \omega t ) -i \lr{ z – z^\conj } \sin( \omega t) } \\
&=
\sqrt{\frac{\Hbar}{2 m \omega}} \lr{ \expectation{x(0)} \sqrt{ \frac{2 m \omega}{\Hbar}} \cos( \omega t ) -i \expectation{p(0)} i \sqrt{\frac{2 m \omega}{\Hbar} } \sin( \omega t) } \\
&=
\expectation{x(0)} \cos( \omega t ) + \frac{\expectation{p(0)}}{m \omega} \sin( \omega t) .
\end{aligned}
\end{equation}

We find that the average of the Heisenberg position operator evolves in time in exactly the same fashion as position in the classical Harmonic oscillator. This phase space like trajectory is sketched in fig. 1.

fig. 1.  phase space like trajectory

fig. 1. phase space like trajectory

In the text it is shown that we have the same structure for the Heisenberg operator itself, before taking expectations

\begin{equation}\label{eqn:qmLecture5:220}
\hat{x}_{\textrm{H}}(t)
=
{x(0)} \cos( \omega t ) + \frac{{p(0)}}{m \omega} \sin( \omega t).
\end{equation}

Where the coherent states become useful is that we will see that the second moments of position and momentum are not time dependent with respect to the coherent states. Such states remain localized.

Uncertainty

First note that using the commutator relationship we have

\begin{equation}\label{eqn:qmLecture5:180}
\begin{aligned}
\bra{z} a a^\dagger \ket{z}
&=
\bra{z} \lr{ \antisymmetric{a}{a^\dagger} + a^\dagger a } \ket{z} \\
&=
\bra{z} \lr{ 1 + a^\dagger a } \ket{z}.
\end{aligned}
\end{equation}

For the second moment we have

\begin{equation}\label{eqn:qmLecture5:200}
\begin{aligned}
\bra{z} \hat{x}^2 \ket{z}
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{a + a^\dagger } \lr{a + a^\dagger } \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + a a^\dagger + a^\dagger a
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\bra{z} \lr{
a^2 + {(a^\dagger)}^2 + 2 a^\dagger a + 1
} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z^2 + {(z^\conj)}^2 + 2 z^\conj z + 1} \ket{z} \\
&=
\frac{\Hbar}{ 2 m \omega}
\lr{ z + z^\conj }^2
+
\frac{\Hbar}{ 2 m \omega}.
\end{aligned}
\end{equation}

We find

\begin{equation}\label{eqn:qmLecture5:240}
\sigma_x^2 = \frac{\Hbar}{ 2 m \omega},
\end{equation}

and

\begin{equation}\label{eqn:qmLecture5:260}
\sigma_p^2 = \frac{m \Hbar \omega}{2}
\end{equation}

so

\begin{equation}\label{eqn:qmLecture5:280}
\sigma_x^2 \sigma_p^2 = \frac{\Hbar^2}{4},
\end{equation}

or

\begin{equation}\label{eqn:qmLecture5:300}
\sigma_x \sigma_p = \frac{\Hbar}{2}.
\end{equation}

This is the minimum uncertainty.

Quantum Field theory

In Quantum Field theory the ideas of isolated oscillators is used to model particle creation. The lowest energy state (a no particle, vacuum state) is given the lowest energy level, with each additional quantum level modeling a new particle creation state as sketched in fig. 2.

fig. 2.  QFT energy levels

fig. 2. QFT energy levels

We have to imagine many oscillators, each with a distinct vacuum energy \( \sim \Bk^2 \) . The Harmonic oscillator can be used to model the creation of particles with \( \Hbar \omega \) energy differences from that “vacuum energy”.

Charged particle in a magnetic field

In the classical case ( with SI units or \( c = 1 \) ) we have

\begin{equation}\label{eqn:qmLecture5:320}
\BF = q \BE + q \Bv \cross \BB.
\end{equation}

Alternately, we can look at the Hamiltonian view of the system, written in terms of potentials

\begin{equation}\label{eqn:qmLecture5:340}
\BB = \spacegrad \cross \BA,
\end{equation}
\begin{equation}\label{eqn:qmLecture5:360}
\BE = – \spacegrad \phi – \PD{t}{\BA}.
\end{equation}

Note that the curl form for the magnetic field implies one of the required Maxwell’s equations \( \spacegrad \cdot \BB = 0 \).

Ignoring time dependence of the potentials, the Hamiltonian can be expressed as

\begin{equation}\label{eqn:qmLecture5:380}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.
\end{equation}

In this Hamiltonian the vector \( \Bp \) is called the canonical momentum, the momentum conjugate to position in phase space.

It is left as an exercise to show that the Lorentz force equation results from application of the Hamiltonian equations of motion, and that the velocity is given by \( \Bv = (\Bp – q \BA)/m \).

For quantum mechanics, we use the same Hamiltonian, but promote our position, momentum and potentials to operators.

\begin{equation}\label{eqn:qmLecture5:400}
\hat{H} = \inv{2 m} \lr{ \hat{\Bp} – q \hat{\BA}(\Br, t) }^2 + q \hat{\phi}(\Br, t).
\end{equation}

Gauge invariance

Can we say anything about this before looking at the question of a particle in a magnetic field?

Recall that the we can make a gauge transformation of the form

\label{eqn:qmLecture5:420a}
\begin{equation}\label{eqn:qmLecture5:420}
\BA \rightarrow \BA + \spacegrad \chi
\end{equation}
\begin{equation}\label{eqn:qmLecture5:440}
\phi \rightarrow \phi – \PD{t}{\chi}
\end{equation}

Does this notion of gauge invariance also carry over to the Quantum Hamiltonian. After gauge transformation we have

\begin{equation}\label{eqn:qmLecture5:460}
\hat{H}’
= \inv{2 m} \lr{ \hat{\Bp} – q \BA – q \spacegrad \chi }^2 + q \lr{ \phi – \PD{t}{\chi} }
\end{equation}

Now we are in a mess, since this function \( \chi \) can make the Hamiltonian horribly complicated. We don’t see how gauge invariance can easily be applied to the quantum problem. Next time we will introduce a transformation that resolves some of this mess.

Question: Lorentz force from classical electrodynamic Hamiltonian

Given the classical Hamiltonian

\begin{equation}\label{eqn:qmLecture5:381}
H = \inv{2 m} \lr{ \Bp – q \BA }^2 + q \phi.
\end{equation}

apply the Hamiltonian equations of motion

\begin{equation}\label{eqn:qmLecture5:480}
\begin{aligned}
\ddt{\Bp} &= – \PD{\Bq}{H} \\
\ddt{\Bq} &= \PD{\Bp}{H},
\end{aligned}
\end{equation}

to show that this is the Hamiltonian that describes the Lorentz force equation, and to find the velocity in terms of the canonical momentum and vector potential.

Answer

The particle velocity follows easily

\begin{equation}\label{eqn:qmLecture5:500}
\begin{aligned}
\Bv
&= \ddt{\Br} \\
&= \PD{\Bp}{H} \\
&= \inv{m} \lr{ \Bp – a \BA }.
\end{aligned}
\end{equation}

For the Lorentz force we can proceed in the coordinate representation

\begin{equation}\label{eqn:qmLecture5:520}
\begin{aligned}
\ddt{p_k}
&= – \PD{x_k}{H} \\
&= – \frac{2}{2m} \lr{ p_m – q A_m } \PD{x_k}{}\lr{ p_m – q A_m } – q \PD{x_k}{\phi} \\
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
\end{aligned}
\end{equation}

We also have

\begin{equation}\label{eqn:qmLecture5:540}
\begin{aligned}
\ddt{p_k}
&=
\ddt{} \lr{m x_k + q A_k } \\
&=
m \frac{d^2 x_k}{dt^2} + q \PD{x_m}{A_k} \frac{d x_m}{dt} + q \PD{t}{A_k}.
\end{aligned}
\end{equation}

Putting these together we’ve got

\begin{equation}\label{eqn:qmLecture5:560}
\begin{aligned}
m \frac{d^2 x_k}{dt^2}
&= q v_m \PD{x_k}{A_m} – q \PD{x_k}{\phi},
– q \PD{x_m}{A_k} \frac{d x_m}{dt} – q \PD{t}{A_k} \\
&=
q v_m \lr{ \PD{x_k}{A_m} – \PD{x_m}{A_k} } + q E_k \\
&=
q v_m \epsilon_{k m s} B_s + q E_k,
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:qmLecture5:580}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&=
q \Be_k v_m \epsilon_{k m s} B_s + q E_k \\
&= q \Bv \cross \BB + q \BE.
\end{aligned}
\end{equation}

Question: Show gauge invariance of the magnetic and electric fields

After the gauge transformation of \ref{eqn:qmLecture5:420} show that the electric and magnetic fields are unaltered.

Answer

For the magnetic field the transformed field is

\begin{equation}\label{eqn:qmLecture5:600}
\begin{aligned}
\BB’
&= \spacegrad \cross \lr{ \BA + \spacegrad \chi } \\
&= \spacegrad \cross \BA + \spacegrad \cross \lr{ \spacegrad \chi } \\
&= \spacegrad \cross \BA \\
&= \BB.
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:qmLecture5:620}
\begin{aligned}
\BE’
&=
– \PD{t}{\BA’} – \spacegrad \phi’ \\
&=
– \PD{t}{}\lr{\BA + \spacegrad \chi} – \spacegrad \lr{ \phi – \PD{t}{\chi}} \\
&=
– \PD{t}{\BA} – \spacegrad \phi \\
&=
\BE.
\end{aligned}
\end{equation}

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 4: Quantum Harmonic oscillator and coherent states. Taught by Prof. Arun Paramekanti

September 29, 2015 phy1520 No comments , , , ,

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

Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough. This lecture reviewed a lot of quantum harmonic oscillator theory, and wouldn’t make sense without having seen raising and lowering operators (ladder operators), number operators, and the like.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 2 content.

Classical Harmonic Oscillator

Recall the classical Harmonic oscillator equations in their Hamiltonian form

\begin{equation}\label{eqn:qmLecture4:40}
\ddt{x} = \frac{p}{m}
\end{equation}
\begin{equation}\label{eqn:qmLecture4:60}
\ddt{p} = -k x.
\end{equation}

With

\begin{equation}\label{eqn:qmLecture4:140}
\begin{aligned}
x(t = 0) &= x_0 \\
p(t = 0) &= p_0 \\
k &= m \omega^2,
\end{aligned}
\end{equation}

the solutions are ellipses in phase space

\begin{equation}\label{eqn:qmLecture4:100}
x(t) = x_0 \cos(\omega t) + \frac{p_0}{m \omega} \sin(\omega t)
\end{equation}
\begin{equation}\label{eqn:qmLecture4:120}
p(t) = p_0 \cos(\omega t) – m \omega x_0 \sin(\omega t).
\end{equation}

After a suitable scaling of the variables, these elliptical orbits can be transformed into circular trajectories.

Quantum Harmonic Oscillator

\begin{equation}\label{eqn:qmLecture4:160}
\hat{H} = \frac{\hat{p}^2}{2 m} + \inv{2} k \hat{x}^2
\end{equation}

Set

\begin{equation}\label{eqn:qmLecture4:200}
\hat{X} = \sqrt{\frac{m \omega}{\Hbar}} \hat{x}
\end{equation}
\begin{equation}\label{eqn:qmLecture4:220}
\hat{P} = \sqrt{\inv{m \omega \Hbar}} \hat{p}
\end{equation}

The commutators after this change of variables goes from

\begin{equation}\label{eqn:qmLecture4:240}
\antisymmetric{ \hat{x}}{\hat{p}} = i \Hbar,
\end{equation}

to
\begin{equation}\label{eqn:qmLecture4:260}
\antisymmetric{ \hat{X}}{\hat{P}} = i.
\end{equation}

The Hamiltonian takes the form

\begin{equation}\label{eqn:qmLecture4:280}
\begin{aligned}
\hat{H}
&= \frac{\Hbar \omega}{2} \lr{ \hat{X}^2 + \hat{P}^2 } \\
&= \Hbar \omega \lr{ \lr{ \frac{\hat{X} -i \hat{P}}{\sqrt{2}} } \lr{ \frac{\hat{X} +i \hat{P}}{\sqrt{2}}} + \inv{2} }.
\end{aligned}
\end{equation}

Define ladder operators (raising and lowering operators respectively)

\begin{equation}\label{eqn:qmLecture4:320}
\hat{a}^\dagger = \frac{\hat{X} -i \hat{P}}{\sqrt{2}}
\end{equation}
\begin{equation}\label{eqn:qmLecture4:340}
\hat{a} = \frac{\hat{X} +i \hat{P}}{\sqrt{2}}
\end{equation}

so

\begin{equation}\label{eqn:qmLecture4:360}
\hat{H} = \Hbar \omega \lr{ \hat{a}^\dagger \hat{a} + \inv{2} }.
\end{equation}

We can show

\begin{equation}\label{eqn:qmLecture4:380}
\antisymmetric{\hat{a}}{\hat{a}^\dagger} = 1,
\end{equation}

and

\begin{equation}\label{eqn:qmLecture4:400}
N \ket{n} \equiv \hat{a}^\dagger a = n \ket{n},
\end{equation}

where \( n \ge 0 \) is an integer. Recall that

\begin{equation}\label{eqn:qmLecture4:420}
\hat{a} \ket{0} = 0,
\end{equation}

and

\begin{equation}\label{eqn:qmLecture4:440}
\bra{X} X + i P \ket{0} = 0.
\end{equation}

With

\begin{equation}\label{eqn:qmLecture4:460}
\braket{x}{0} = \Psi_0(x),
\end{equation}

we can show

\begin{equation}\label{eqn:qmLecture4:480}
\inv{\sqrt{2}} \lr{ X + \PD{X}{} } \Psi_0(X) = 0.
\end{equation}

Also recall that

\begin{equation}\label{eqn:qmLecture4:520}
\hat{a} \ket{n} = \sqrt{n} \ket{n-1}
\end{equation}
\begin{equation}\label{eqn:qmLecture4:540}
\hat{a}^\dagger \ket{n} = \sqrt{n + 1} \ket{n+1}
\end{equation}

Coherent states

Coherent states for the quantum harmonic oscillator are the eigenkets for the creation and annihilation operators

\begin{equation}\label{eqn:qmLecture4:580}
\hat{a} \ket{z} = z \ket{z}
\end{equation}
\begin{equation}\label{eqn:qmLecture4:600}
\hat{a}^\dagger \ket{\tilde{z}} = \tilde{z} \ket{\tilde{z}} ,
\end{equation}

where

\begin{equation}\label{eqn:qmLecture4:620}
\ket{z} = \sum_{n = 0}^\infty c_n \ket{n},
\end{equation}

and \( z \) is allowed to be a complex number.

Looking for such a state, we compute

\begin{equation}\label{eqn:qmLecture4:640}
\begin{aligned}
\hat{a} \ket{z}
&= \sum_{n=1}^\infty c_n \hat{a} \ket{n} \\
&= \sum_{n=1}^\infty c_n \sqrt{n} \ket{n-1}
\end{aligned}
\end{equation}

compare this to

\begin{equation}\label{eqn:qmLecture4:660}
\begin{aligned}
z \ket{z}
&=
z \sum_{n=0}^\infty c_n \ket{n} \\
&=
\sum_{n=1}^\infty c_n \sqrt{n} \ket{n-1} \\
&=
\sum_{n=0}^\infty c_{n+1} \sqrt{n+1} \ket{n},
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:qmLecture4:680}
c_{n+1} \sqrt{n+1} = z c_n
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture4:700}
c_{n+1} = \frac{z c_n}{\sqrt{n+1}}
\end{equation}

\begin{equation}\label{eqn:qmLecture4:720}
\begin{aligned}
c_1 &= c_0 z \\
c_2 &= \frac{z c_1}{\sqrt{2}} = \frac{z^2 c_0}{\sqrt{2}} \\
\vdots &
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:qmLecture4:740}
c_n = \frac{z^n}{\sqrt{n!}}.
\end{equation}

So the desired state is

\begin{equation}\label{eqn:qmLecture4:760}
\ket{z} = c_0 \sum_{n=0}^\infty \frac{z^n}{\sqrt{n!}} \ket{n}.
\end{equation}

Also recall that

\begin{equation}\label{eqn:qmLecture4:780}
\ket{n} = \frac{\lr{ \hat{a}^\dagger }^n}{\sqrt{n!}} \ket{0},
\end{equation}

which gives

\begin{equation}\label{eqn:qmLecture4:800}
\begin{aligned}
\ket{z}
&= c_0 \sum_{n=0}^\infty \frac{\lr{z \hat{a}^\dagger}^n }{n!} \ket{0} \\
&= c_0 e^{z \hat{a}^\dagger} \ket{0}.
\end{aligned}
\end{equation}

The normalization is

\begin{equation}\label{eqn:qmLecture4:820}
c_0 = e^{-\Abs{z}^2/2}.
\end{equation}

While we have \( \braket{n_1}{n_2} = \delta_{n_1, n_2} \), these \( \ket{z} \) states are not orthonormal. Figuring out that this overlap

\begin{equation}\label{eqn:qmLecture4:840}
\braket{z_1}{z_2} \ne 0,
\end{equation}

will be left for homework.

Dynamics

We don’t know much about these coherent states. For example does a coherent state at time zero evolve to a coherent state?

\begin{equation}\label{eqn:qmLecture4:860}
\ket{z} \stackrel{?}{\rightarrow} \ket{z(t)}
\end{equation}

It turns out that these questions are best tackled in the Heisenberg picture, considering

\begin{equation}\label{eqn:qmLecture4:880}
e^{-i \hat{H} t/\Hbar } \ket{z}.
\end{equation}

For example, what is the average of the position operator

\begin{equation}\label{eqn:qmLecture4:900}
\bra{z} e^{i \hat{H} t/\Hbar } \hat{x} e^{-i \hat{H} t/\Hbar } \ket{z}
=
\sum_{n, n’ = 0}^\infty
\bra{n} c_n^\conj e^{i E_n t/\Hbar}
\lr{ a + a^\dagger} \sqrt{ \frac{\Hbar}{m \omega} }
c_{n’} e^{i E_{n’} t/\Hbar}
\ket{n}.
\end{equation}

This is very messy to attempt. Instead if we know how the operator evolves we can calculate

\begin{equation}\label{eqn:qmLecture4:920}
\bra{z} \hat{x}_{\textrm{H}}(t) \ket{z},
\end{equation}

that is

\begin{equation}\label{eqn:qmLecture4:940}
\expectation{\hat{x}}(t) = \bra{z} \hat{x}_{\textrm{H}}(t) \ket{z},
\end{equation}

and for momentum

\begin{equation}\label{eqn:qmLecture4:960}
\expectation{\hat{p}}(t) = \bra{z} \hat{p}_{\textrm{H}}(t) \ket{z}.
\end{equation}

The question to ask is what are the expansions of

\begin{equation}\label{eqn:qmLecture4:1000}
\hat{a}_{\textrm{H}}(t) = e^{i \hat{H} t/\Hbar} \hat{a} e^{-i \hat{H} t/\Hbar}.
\end{equation}
\begin{equation}\label{eqn:qmLecture4:1020}
\hat{a}^\dagger_{\textrm{H}}(t) = e^{i \hat{H} t/\Hbar} \hat{a}^\dagger e^{-i \hat{H} t/\Hbar}.
\end{equation}

The question to ask is how do these operators ask on the basis states

\begin{equation}\label{eqn:qmLecture4:1040}
\begin{aligned}
\hat{a}_{\textrm{H}}(t) \ket{n}
&= e^{i \hat{H} t/\Hbar} \hat{a} e^{-i \hat{H} t/\Hbar} \ket{n} \\
&= e^{i \hat{H} t/\Hbar} \hat{a} e^{-i t \omega (n + 1/2)} \ket{n} \\
&=
e^{-i t \omega (n + 1/2)}
e^{i \hat{H} t/\Hbar}
\sqrt{n} \ket{n-1} \\
&=
\sqrt{n}
e^{-i t \omega (n + 1/2)}
e^{i t \omega (n – 1/2)}
\ket{n-1} \\
&=
\sqrt{n} e^{-i \omega t} \ket{n-1} \\
&=
e^{-i \omega t} \ket{n}.
\end{aligned}
\end{equation}

So we have found

\begin{equation}\label{eqn:qmLecture4:1060}
\begin{aligned}
\hat{a}_{\textrm{H}}(t) &= a e^{-i\omega t} \\
\hat{a}^\dagger_{\textrm{H}}(t) &= a^\dagger e^{i\omega t}
\end{aligned}
\end{equation}

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

August 19, 2015 phy1520 No comments , , , , , , , , , , , , , , , , , , ,

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

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

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

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

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

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

\begin{equation}\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}
\end{equation}

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

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

so the equations of motions are

\begin{equation}\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}
\end{equation}

and
\begin{equation}\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}
\end{equation}

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

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

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

\begin{equation}\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},
\end{equation}

or, with length scaled variables

\begin{equation}\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}
\end{equation}

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

\begin{equation}\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}
\end{equation}

or

\begin{equation}\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}
\end{equation}

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

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

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

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

This matrix can be diagonalized as

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

where

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

The equations of motion can now be written

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

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

\begin{equation}\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}
\end{equation}

where
\begin{equation}\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}
\end{equation}

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

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

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

\begin{equation}\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},
\end{equation}

so

\begin{equation}\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}
\end{equation}

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.

Update to old phy356 (Quantum Mechanics I) notes.

February 12, 2015 math and physics play No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

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.