Hamiltonian

Hamiltonian for a scalar field

January 3, 2016 phy2403 No comments , , ,

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In [1] it is left as an exersize to expand the scalar field Hamiltonian in terms of the raising and lowering operators. Let’s do that.

The field operator expanded in terms of the raising and lowering operators is

\begin{equation}\label{eqn:scalarFieldHamiltonian:20}
\phi(x) =
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i k \cdot x}
+ a_k^\dagger e^{i k \cdot x}
}.
\end{equation}

Note that \( x \) and \( k \) here are both four-vectors, so this field is dependent on a spacetime point, but the integration is over a spatial volume.

The Hamiltonian in terms of the fields was
\begin{equation}\label{eqn:scalarFieldHamiltonian:40}
H = \inv{2} \int d^3 x \lr{ \Pi^2 + \lr{ \spacegrad \phi }^2 + \mu^2 \phi^2 }.
\end{equation}

The field derivatives are

\begin{equation}\label{eqn:scalarFieldHamiltonian:60}
\Pi
= \partial_0 \phi
= \partial_0
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}
=
i
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \frac{\omega_k}{ 2 \omega_k } } \lr{
-a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
},
\end{equation}

and

\begin{equation}\label{eqn:scalarFieldHamiltonian:80}
\partial_n \phi
= \partial_n
\int \frac{ d^3 k}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}
=
i
\int \frac{ d^3 k k^n}{ (2 \pi)^{3/2} \sqrt{ 2 \omega_k } } \lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
-a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
}.
\end{equation}

Introducing a second set of momentum variables with \( j = \Abs{\Bj} \), the momentum portion of the Hamiltonian is

\begin{equation}\label{eqn:scalarFieldHamiltonian:100}
\begin{aligned}
\inv{2} \int d^3 x \Pi^2
&=
-\inv{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\omega_j
\omega_k
\lr{
-a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
+a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
-a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
-\inv{4}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\sqrt{
\omega_j
\omega_k}
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t – i (\Bk + \Bj) \cdot \Bx}
+ a_j a_k e^{-i (\omega_j + \omega_k) t + i (\Bj + \Bk) \cdot \Bx}
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t – i (\Bj – \Bk) \cdot \Bx}
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t – i (\Bk – \Bj) \cdot \Bx}
} \\
&=
-\inv{4}
\int
d^3 j
d^3 k
\sqrt{
\omega_j
\omega_k}
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t } \delta^3(\Bk + \Bj)
+ a_j a_k e^{-i (\omega_j + \omega_k) t } \delta^3(-\Bj – \Bk)
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t } \delta^3(\Bj – \Bk)
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t } \delta^3(\Bk – \Bj)
} \\
&=
-\inv{4}
\int
d^3 k
\omega_k
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}.
\end{aligned}
\end{equation}

For the gradient portion of the Hamiltonian we have

\begin{equation}\label{eqn:scalarFieldHamiltonian:120}
\begin{aligned}
\inv{2} \int d^3 x \lr{ \spacegrad \phi }^2
&=
-\inv{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{ \sum_{n=1}^3 j^n k^n }
\lr{
a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
-a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
-a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
-\inv{4}
\int
d^3 j
d^3 k
\frac{\Bj \cdot \Bk}{ \sqrt{ \omega_j \omega_k } }
\lr{
a_j^\dagger a_k^\dagger e^{i (\omega_k + \omega_j) t } \delta^3(\Bk + \Bj)
+ a_j a_k e^{-i (\omega_j + \omega_k) t } \delta^3(-\Bj – \Bk)
– a_j^\dagger a_k e^{-i (\omega_k -\omega_j) t } \delta^3(\Bj – \Bk)
– a_j a_k^\dagger e^{-i (\omega_j – \omega_k) t } \delta^3(\Bk – \Bj)
} \\
&=
-\inv{4}
\int
d^3 k
\frac{\Bk^2}{ \omega_k }
\lr{
– a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
– a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}.
\end{aligned}
\end{equation}

Finally, for the mass term, we have

\begin{equation}\label{eqn:scalarFieldHamiltonian:140}
\begin{aligned}
\inv{2} \int d^3 x \mu^2 \phi^2
&=
\frac{\mu^2}{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j e^{-i \omega_j t + i \Bj \cdot \Bx}
+a_j^\dagger e^{i \omega_j t – i \Bj \cdot \Bx}
}
\lr{
a_k e^{-i \omega_k t + i \Bk \cdot \Bx}
+a_k^\dagger e^{i \omega_k t – i \Bk \cdot \Bx}
} \\
&=
\frac{\mu^2}{2}
\inv{(2 \pi)^{3}}
\int d^3 x
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j a_k e^{-i (\omega_k + \omega_j) t + i (\Bk + \Bj) \cdot \Bx}
+a_j^\dagger a_k^\dagger e^{i (\omega_j + \omega_k) t – i (\Bk + \Bj) \cdot \Bx}
+a_j a_k^\dagger e^{i (\omega_k – \omega_j) t – i (\Bk – \Bj) \cdot \Bx}
+a_j^\dagger a_k e^{-i (\omega_k + \omega_j) t – i (\Bj – \Bk) \cdot \Bx}
} \\
&=
\frac{\mu^2}{2}
\int
d^3 j
d^3 k
\inv{ \sqrt{ 4 \omega_j \omega_k } }
\lr{
a_j a_k e^{-i (\omega_k + \omega_j) t } \delta^3(- \Bk – \Bj)
+a_j^\dagger a_k^\dagger e^{i (\omega_j + \omega_k) t } \delta^3( \Bk + \Bj)
+a_j a_k^\dagger e^{i (\omega_k – \omega_j) t } \delta^3 (\Bk – \Bj)
+a_j^\dagger a_k e^{-i (\omega_k + \omega_j) t } \delta^3 (\Bj – \Bk)
} \\
&=
\frac{\mu^2}{4}
\int
d^3 k
\inv{ \omega_k }
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}.
\end{aligned}
\end{equation}

Now all the pieces can be put back together again

\begin{equation}\label{eqn:scalarFieldHamiltonian:160}
\begin{aligned}
H
&=
\inv{4}
\int d^3 k
\inv{\omega_k}
\lr{
-\omega_k^2
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
– a_k^\dagger a_k
– a_k a_k^\dagger
}
+
\Bk^2
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+ a_{-k} a_k e^{- 2 i \omega_k t }
+ a_k^\dagger a_k
+ a_k a_k^\dagger
}
+
\mu^2
\lr{
a_{-k} a_k e^{- 2 i \omega_k t }
+a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
+a_k a_k^\dagger
+a_k^\dagger a_k
}
} \\
&=
\inv{4}
\int d^3 k
\inv{\omega_k}
\lr{
a_{-k}^\dagger a_k^\dagger e^{2 i \omega_k t }
\lr{
-\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_{-k} a_k e^{- 2 i \omega_k t }
\lr{
-\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_k a_k^\dagger
\lr{
\omega_k^2
+ \Bk^2
+
\mu^2
}
+ a_k^\dagger a_k
\lr{
\omega_k^2
+ \Bk^2
+
\mu^2
}
}.
\end{aligned}
\end{equation}

With \( \omega_k^2 = \Bk^2 + \mu^2 \), the time dependent terms are killed leaving
\begin{equation}\label{eqn:scalarFieldHamiltonian:180}
H
=
\inv{2}
\int d^3 k
\omega_k
\lr{
a_k a_k^\dagger
+ a_k^\dagger a_k
}.
\end{equation}

References

[1] Michael Luke. PHY2403F Lecture Notes: Quantum Field Theory, 2015. URL lecturenotes.pdf. [Online; accessed 02-Jan-2016].

Time reversal behavior of solutions to crystal spin Hamiltonian

December 15, 2015 phy1520 No comments , , , ,

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Q: [1] pr 4.12

Solve the spin 1 Hamiltonian
\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:20}
H = A S_z^2 + B(S_x^2 – S_y^2).
\end{equation}

Is this Hamiltonian invariant under time reversal?

How do the eigenkets change under time reversal?

A:

In spinMatrices.nb the matrix representation of the Hamiltonian is found to be
\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:40}
H =
\Hbar^2
\begin{bmatrix}
A+\frac{B}{2} & 0 & \frac{B}{2} \\
-\frac{i B}{\sqrt{2}} & B & -\frac{i B}{\sqrt{2}} \\
\frac{B}{2} & 0 & A+\frac{B}{2} \\
\end{bmatrix}.
\end{equation}

The eigenvalues are
\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:60}
\setlr{ \Hbar^2 A, \Hbar^2 B, \Hbar^2(A + B)},
\end{equation}

and the respective eigenvalues (unnormalized) are

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:80}
\setlr{
\begin{bmatrix}
-1 \\
0 \\
1
\end{bmatrix},
\begin{bmatrix}
0 \\
1 \\
0
\end{bmatrix},
\begin{bmatrix}
1 \\
-\frac{i \sqrt{2} B}{A} \\
1 \\
\end{bmatrix}
}.
\end{equation}

Under time reversal, the Hamiltonian is

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:100}
H \rightarrow A (-S_z)^2 + B ( (-S_x)^2 – (-S_y)^2 ) = H,
\end{equation}

so we expect the eigenkets for this Hamiltonian to vary by at most a phase factor. To check this, first recall that the time reversal action on a spin one state is

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:120}
\Theta \ket{1, m} = (-1)^m \ket{1, -m},
\end{equation}

or

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:140}
\begin{aligned}
\Theta \ket{1} &= -\ket{-1} \\
\Theta \ket{0} &= \ket{0} \\
\Theta \ket{-1} &= -\ket{1}.
\end{aligned}
\end{equation}

Let’s write the eigenkets respectively as

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:160}
\begin{aligned}
\ket{A} &= -\ket{1} + \ket{-1} \\
\ket{B} &= \ket{0} \\
\ket{A+B} &= \ket{1} + \ket{-1} – \frac{i \sqrt{2} B}{A} \ket{0}.
\end{aligned}
\end{equation}

Noting that the time reversal operator maps complex numbers onto their conjugates, the time reversed eigenkets are

\begin{equation}\label{eqn:crystalSpinHamiltonianTimeReversal:180}
\begin{aligned}
\ket{A} &\rightarrow \ket{-1} – \ket{-1} = -\ket{A} \\
\ket{B} &\rightarrow \ket{0} = \ket{B} \\
\ket{A+B} &\rightarrow -\ket{1} – \ket{-1} + \frac{i \sqrt{2} B}{A} \ket{0} = -\ket{A+B}.
\end{aligned}
\end{equation}

Up to a sign, the time reversed states match the unreversed states.

References

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

Energy estimate for an absolute value potential

December 4, 2015 phy1520 No comments , , ,

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Here’s a simple problem, a lot like the problem set 6 variational calculation.

Q: [1] 5.21

Estimate the lowest eigenvalue \( \lambda \) of the differential equation

\begin{equation}\label{eqn:absolutePotentialVariation:20}
\frac{d^2}{dx^2}\psi + \lr{ \lambda – \Abs{x} } \psi = 0.
\end{equation}

Using \( \alpha \) variation with the trial function

\begin{equation}\label{eqn:absolutePotentialVariation:40}
\psi =
\left\{
\begin{array}{l l}
c(\alpha – \Abs{x}) & \quad \mbox{\(\Abs{x} < \alpha \) } \\ 0 & \quad \mbox{\(\Abs{x} > \alpha \) }
\end{array}
\right.
\end{equation}

A:

First rewrite the differential equation in a Hamiltonian like fashion

\begin{equation}\label{eqn:absolutePotentialVariation:60}
H \psi = -\frac{d^2}{dx^2}\psi + \Abs{x} \psi = \lambda \psi.
\end{equation}

We need the derivatives of the trial distribution. The first derivative is

\begin{equation}\label{eqn:absolutePotentialVariation:80}
\begin{aligned}
\frac{d}{dx} \psi
&=
-c \frac{d}{dx} \Abs{x} \\
&=
-c \frac{d}{dx} \lr{ x \theta(x) – x \theta(-x) } \\
&=
-c \lr{
\theta(x) – \theta(-x)
+
x \delta(x) + x \delta(-x)
} \\
&=
-c \lr{
\theta(x) – \theta(-x)
+
2 x \delta(x)
}.
\end{aligned}
\end{equation}

The second derivative is
\begin{equation}\label{eqn:absolutePotentialVariation:100}
\begin{aligned}
\frac{d^2}{dx^2} \psi
&=
-c \frac{d}{dx} \lr{
\theta(x) – \theta(-x)
+
2 x \delta(x)
} \\
&=
-c \lr{
\delta(x) + \delta(-x)
+
2 \delta(x)
+
2 x \delta'(x)
} \\
&=
-c \lr{
4 \delta(x)
+
2 x \frac{-\delta(x) }{x}
} \\
&=
-2 c \delta(x).
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:absolutePotentialVariation:120}
H \psi = -2 c \delta(x) + \Abs{x} c \lr{ \alpha – \Abs{x} }.
\end{equation}

We are now set to compute some of the inner products. The normalization is the simplest

\begin{equation}\label{eqn:absolutePotentialVariation:140}
\begin{aligned}
\braket{\psi}{\psi}
&= c^2 \int_{-\alpha}^\alpha ( \alpha – \Abs{x} )^2 dx \\
&= 2 c^2 \int_{0}^\alpha ( x – \alpha )^2 dx \\
&= 2 c^2 \int_{-\alpha}^0 u^2 du \\
&= 2 c^2 \lr{ -\frac{(-\alpha)^3}{3} } \\
&= \frac{2}{3} c^2 \alpha^3.
\end{aligned}
\end{equation}

For the energy
\begin{equation}\label{eqn:absolutePotentialVariation:160}
\begin{aligned}
\braket{\psi}{H \psi}
&=
c^2 \int dx \lr{ \alpha – \Abs{x} } \lr{ -2 \delta(x) + \Abs{x} \lr{ \alpha – \Abs{x} } } \\
&=
c^2 \lr{ – 2 \alpha + \int_{-\alpha}^\alpha dx \lr{ \alpha – \Abs{x} }^2 \Abs{x} } \\
&=
c^2 \lr{ – 2 \alpha + 2 \int_{-\alpha}^0 du u^2 \lr{ u + \alpha } } \\
&=
c^2 \lr{ – 2 \alpha + 2 \evalrange{\lr{ \frac{u^4}{4} + \alpha \frac{u^3}{3} }}{-\alpha}{0} } \\
&=
c^2 \lr{ – 2 \alpha – 2 \lr{ \frac{\alpha^4}{4} – \frac{\alpha^4}{3} } } \\
&=
c^2 \lr{ – 2 \alpha + \inv{6} \alpha^4 }.
\end{aligned}
\end{equation}

The energy estimate is

\begin{equation}\label{eqn:absolutePotentialVariation:180}
\begin{aligned}
\overline{{E}}
&=
\frac{\braket{\psi}{H \psi}}{\braket{\psi}{\psi}} \\
&=
\frac{ – 2 \alpha + \inv{6} \alpha^4 }{ \frac{2}{3} \alpha^3} \\
&=
– \frac{3}{\alpha^2} + \inv{4} \alpha.
\end{aligned}
\end{equation}

This has its minimum at
\begin{equation}\label{eqn:absolutePotentialVariation:200}
0 = -\frac{6}{\alpha^3} + \inv{4},
\end{equation}

or
\begin{equation}\label{eqn:absolutePotentialVariation:220}
\alpha = 2 \times 3^{1/3}.
\end{equation}

Back subst into the energy gives

\begin{equation}\label{eqn:absolutePotentialVariation:240}
\begin{aligned}
\overline{{E}}
&=
– \frac{3}{4 \times 3^{2/3}} + \inv{2} 3^{1/3} \\
&= \frac{3^{4/3}}{4} \\
&\approx 1.08.
\end{aligned}
\end{equation}

The problem says the exact answer is 1.019, so the variation gets within 6 %.

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 20: Perturbation theory. Taught by Prof. Arun Paramekanti

December 3, 2015 phy1520 No comments , , , , , , , , , ,

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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] ch. 5 content.

Perturbation theory

Given a \( 2 \times 2 \) Hamiltonian \( H = H_0 + V \), where

\begin{equation}\label{eqn:qmLecture20:20}
H =
\begin{bmatrix}
a & c \\
c^\conj & b
\end{bmatrix}
\end{equation}

which has eigenvalues

\begin{equation}\label{eqn:qmLecture20:40}
\lambda_\pm = \frac{a + b}{2} \pm \sqrt{ \lr{ \frac{a – b}{2}}^2 + \Abs{c}^2 }.
\end{equation}

If \( c = 0 \),

\begin{equation}\label{eqn:qmLecture20:60}
H_0 =
\begin{bmatrix}
a & 0 \\
0 & b
\end{bmatrix},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:80}
V =
\begin{bmatrix}
0 & c \\
c^\conj & 0
\end{bmatrix}.
\end{equation}

Suppose that \( \Abs{c} \ll \Abs{a – b} \), then

\begin{equation}\label{eqn:qmLecture20:100}
\lambda_\pm \approx \frac{a + b}{2} \pm \Abs{ \frac{a – b}{2} } \lr{ 1 + 2 \frac{\Abs{c}^2}{\Abs{a – b}^2} }.
\end{equation}

If \( a > b \), then

\begin{equation}\label{eqn:qmLecture20:120}
\lambda_\pm \approx \frac{a + b}{2} \pm \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} }.
\end{equation}

\begin{equation}\label{eqn:qmLecture20:140}
\begin{aligned}
\lambda_{+}
&= \frac{a + b}{2} + \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&= a + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= a + \frac{\Abs{c}^2}{a – b},
\end{aligned}
\end{equation}

and
\begin{equation}\label{eqn:qmLecture20:680}
\begin{aligned}
\lambda_{-}
&= \frac{a + b}{2} – \frac{a – b}{2} \lr{ 1 + 2 \frac{\Abs{c}^2}{\lr{a – b}^2} } \\
&=
b + \lr{a – b} \frac{\Abs{c}^2}{\lr{a – b}^2} \\
&= b + \frac{\Abs{c}^2}{a – b}.
\end{aligned}
\end{equation}

This adiabatic evolution displays a “level repulsion”, quadradic in \( \Abs{c} \) as sketched in fig. 1, and is described as a non-degenerate perbutation.

fig. 1.  Adiabatic (non-degenerate) perturbation

fig. 1. Adiabatic (non-degenerate) perturbation

If \( \Abs{c} \gg \Abs{a -b} \), then

\begin{equation}\label{eqn:qmLecture20:160}
\begin{aligned}
\lambda_\pm
&= \frac{a + b}{2} \pm \Abs{c} \sqrt{ 1 + \inv{\Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&\approx \frac{a + b}{2} \pm \Abs{c} \lr{ 1 + \inv{2 \Abs{c}^2} \lr{ \frac{a – b}{2}}^2 } \\
&= \frac{a + b}{2} \pm \Abs{c} \pm \frac{\lr{a – b}^2}{8 \Abs{c}}.
\end{aligned}
\end{equation}

Here we loose the adiabaticity, and have “level repulsion” that is linear in \( \Abs{c} \), as sketched in fig. 2. We no longer have the sign of \( a – b \) in the expansion. This is described as a degenerate perbutation.

fig. 2.  Degenerate perbutation

fig. 2. Degenerate perbutation

General non-degenerate perturbation

Given an unperturbed system with solutions of the form

\begin{equation}\label{eqn:qmLecture20:180}
H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}},
\end{equation}

we want to solve the perturbed Hamiltonian equation

\begin{equation}\label{eqn:qmLecture20:200}
\lr{ H_0 + \lambda V } \ket{ n } = \lr{ E_n^{(0)} + \Delta n } \ket{n}.
\end{equation}

Here \( \Delta n \) is an energy shift as that goes to zero as \( \lambda \rightarrow 0 \). We can write this as

\begin{equation}\label{eqn:qmLecture20:220}
\lr{ E_n^{(0)} – H_0 } \ket{ n } = \lr{ \lambda V – \Delta_n } \ket{n}.
\end{equation}

We are hoping to iterate with application of the inverse to an initial estimate of \( \ket{n} \)

\begin{equation}\label{eqn:qmLecture20:240}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n}.
\end{equation}

This gets us into trouble if \( \lambda \rightarrow 0 \), which can be fixed by using

\begin{equation}\label{eqn:qmLecture20:260}
\ket{n} = \lr{ E_n^{(0)} – H_0 }^{-1} \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

which can be seen to be a solution to \ref{eqn:qmLecture20:220}. We want to ask if

\begin{equation}\label{eqn:qmLecture20:280}
\lr{ \lambda V – \Delta_n } \ket{n} ,
\end{equation}

contains a bit of \( \ket{ n^{(0)} } \)? To determine this act with \( \bra{n^{(0)}} \) on the left

\begin{equation}\label{eqn:qmLecture20:300}
\begin{aligned}
\bra{ n^{(0)} } \lr{ \lambda V – \Delta_n } \ket{n}
&=
\bra{ n^{(0)} } \lr{ E_n^{(0)} – H_0 } \ket{n} \\
&=
\lr{ E_n^{(0)} – E_n^{(0)} } \braket{n^{(0)}}{n} \\
&=
0.
\end{aligned}
\end{equation}

This shows that \( \ket{n} \) is entirely orthogonal to \( \ket{n^{(0)}} \).

Define a projection operator

\begin{equation}\label{eqn:qmLecture20:320}
P_n = \ket{n^{(0)}}\bra{n^{(0)}},
\end{equation}

which has the idempotent property \( P_n^2 = P_n \) that we expect of a projection operator.

Define a rejection operator
\begin{equation}\label{eqn:qmLecture20:340}
\overline{{P}}_n
= 1 –
\ket{n^{(0)}}\bra{n^{(0)}}
= \sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}.
\end{equation}

Because \( \ket{n} \) has no component in the direction \( \ket{n^{(0)}} \), the rejection operator can be inserted much like we normally do with the identity operator, yielding

\begin{equation}\label{eqn:qmLecture20:360}
\ket{n}’ = \lr{ E_n^{(0)} – H_0 }^{-1} \overline{{P}}_n \lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

valid for any initial \( \ket{n} \).

Power series perturbation expansion

Instead of iterating, suppose that the unknown state and unknown energy difference operator can be expanded in a \( \lambda \) power series, say

\begin{equation}\label{eqn:qmLecture20:380}
\ket{n}
=
\ket{n_0}
+ \lambda \ket{n_1}
+ \lambda^2 \ket{n_2}
+ \lambda^3 \ket{n_3} + \cdots
\end{equation}

and

\begin{equation}\label{eqn:qmLecture20:400}
\Delta_{n} = \Delta_{n_0}
+ \lambda \Delta_{n_1}
+ \lambda^2 \Delta_{n_2}
+ \lambda^3 \Delta_{n_3} + \cdots
\end{equation}

We usually interpret functions of operators in terms of power series expansions. In the case of \( \lr{ E_n^{(0)} – H_0 }^{-1} \), we have a concrete interpretation when acting on one of the unpertubed eigenstates

\begin{equation}\label{eqn:qmLecture20:420}
\inv{ E_n^{(0)} – H_0 } \ket{m^{(0)}} =
\inv{ E_n^{(0)} – E_m^0 } \ket{m^{(0)}}.
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture20:440}
\ket{n}
=
\inv{ E_n^{(0)} – H_0 }
\sum_{m \ne n}
\ket{m^{(0)}}\bra{m^{(0)}}
\lr{ \lambda V – \Delta_n } \ket{n} + \ket{ n^{(0)} },
\end{equation}

or

\begin{equation}\label{eqn:qmLecture20:460}
\boxed{
\ket{n}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_n } \ket{n}.
}
\end{equation}

From \ref{eqn:qmLecture20:220}, note that

\begin{equation}\label{eqn:qmLecture20:500}
\Delta_n =
\frac{\bra{n^{(0)}} \lambda V \ket{n}}{\braket{n^0}{n}},
\end{equation}

however, we will normalize by setting \( \braket{n^0}{n} = 1 \), so

\begin{equation}\label{eqn:qmLecture20:521}
\boxed{
\Delta_n =
\bra{n^{(0)}} \lambda V \ket{n}.
}
\end{equation}

to \( O(\lambda^0) \)

If all \( \lambda^n, n > 0 \) are zero, then we have

\label{eqn:qmLecture20:780}
\begin{equation}\label{eqn:qmLecture20:740}
\ket{n_0}
=
\ket{ n^{(0)} }
+
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ – \Delta_{n_0} } \ket{n_0}
\end{equation}
\begin{equation}\label{eqn:qmLecture20:800}
\Delta_{n_0} \braket{n^{(0)}}{n_0} = 0
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:540}
\begin{aligned}
\ket{n_0} &= \ket{n^{(0)}} \\
\Delta_{n_0} &= 0.
\end{aligned}
\end{equation}

to \( O(\lambda^1) \)

Requiring identity for all \( \lambda^1 \) terms means

\begin{equation}\label{eqn:qmLecture20:760}
\ket{n_1} \lambda
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – \Delta_{n_1} \lambda } \ket{n_0},
\end{equation}

so

\begin{equation}\label{eqn:qmLecture20:560}
\ket{n_1}
=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}} \bra{ m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} } \ket{n_0}.
\end{equation}

With the assumption that \( \ket{n^{(0)}} \) is normalized, and with the shorthand

\begin{equation}\label{eqn:qmLecture20:600}
V_{m n} = \bra{ m^{(0)}} V \ket{n^{(0)}},
\end{equation}

that is

\begin{equation}\label{eqn:qmLecture20:580}
\begin{aligned}
\ket{n_1}
&=
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}
\\
\Delta_{n_1} &= \bra{ n^{(0)} } V \ket{ n^0} = V_{nn}.
\end{aligned}
\end{equation}

to \( O(\lambda^2) \)

The second order perturbation states are found by selecting only the \( \lambda^2 \) contributions to

\begin{equation}\label{eqn:qmLecture20:820}
\lambda^2 \ket{n_2}
=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ \lambda V – (\lambda \Delta_{n_1} + \lambda^2 \Delta_{n_2}) }
\lr{
\ket{n_0}
+ \lambda \ket{n_1}
}.
\end{equation}

Because \( \ket{n_0} = \ket{n^{(0)}} \), the \( \lambda^2 \Delta_{n_2} \) is killed, leaving

\begin{equation}\label{eqn:qmLecture20:840}
\begin{aligned}
\ket{n_2}
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\ket{n_1} \\
&=
\sum_{m \ne n}
\frac{\ket{m^{(0)}}\bra{m^{(0)}}}
{
E_n^{(0)} – E_m^{(0)}
}
\lr{ V – \Delta_{n_1} }
\sum_{l \ne n}
\frac{
\ket{l^{(0)}}
}
{
E_n^{(0)} – E_l^{(0)}
}
V_{l n},
\end{aligned}
\end{equation}

which can be written as

\begin{equation}\label{eqn:qmLecture20:620}
\ket{n_2}
=
\sum_{l,m \ne n}
\ket{m^{(0)}}
\frac{V_{m l} V_{l n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}

\sum_{m \ne n}
\ket{m^{(0)}}
\frac{V_{n n} V_{m n}}
{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.
\end{equation}

For the second energy perturbation we have

\begin{equation}\label{eqn:qmLecture20:860}
\lambda^2 \Delta_{n_2} =
\bra{n^{(0)}} \lambda V \lr{ \lambda \ket{n_1} },
\end{equation}

or

\begin{equation}\label{eqn:qmLecture20:880}
\begin{aligned}
\Delta_{n_2}
&=
\bra{n^{(0)}} V \ket{n_1} \\
&=
\bra{n^{(0)}} V
\sum_{m \ne n}
\frac{
\ket{m^{(0)}}
}
{
E_n^{(0)} – E_m^{(0)}
}
V_{m n}.
\end{aligned}
\end{equation}

That is

\begin{equation}\label{eqn:qmLecture20:900}
\Delta_{n_2}
=
\sum_{m \ne n} \frac{V_{n m} V_{m n} }{E_n^{(0)} – E_m^{(0)}}.
\end{equation}

to \( O(\lambda^3) \)

Similarily, it can be shown that

\begin{equation}\label{eqn:qmLecture20:640}
\Delta_{n_3} =
\sum_{l, m \ne n} \frac{V_{n m} V_{m l} V_{l n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }
\lr{ E_n^{(0)} – E_l^{(0)} }
}

\sum_{ m \ne n} \frac{V_{n m} V_{n n} V_{m n} }{
\lr{ E_n^{(0)} – E_m^{(0)} }^2
}.
\end{equation}

In general, the energy perturbation is given by

\begin{equation}\label{eqn:qmLecture20:660}
\Delta_n^{(l)} = \bra{n^{(0)}} V \ket{n^{(l-1)}}.
\end{equation}

References

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

Alternate Dirac equation representation

November 27, 2015 phy1520 No comments , ,

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Given an alternate representation of the Dirac equation

\begin{equation}\label{eqn:diracAlternate:20}
H =
\begin{bmatrix}
m c^2 + V_0 & c \hat{p} \\
c \hat{p} & – m c^2 + V_0
\end{bmatrix},
\end{equation}

calculate the constant momentum solutions, the Heisenberg velocity operator \( \hat{v} \), and find the form of the probability density current.

Plane wave solutions

The action of the Hamiltonian on

\begin{equation}\label{eqn:diracAlternate:40}
\psi =
e^{i k x – i E t/\Hbar}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
\end{equation}

is
\begin{equation}\label{eqn:diracAlternate:60}
\begin{aligned}
H \psi
&=
\begin{bmatrix}
m c^2 + V_0 & c (-i \Hbar) i k \\
c (-i \Hbar) i k & – m c^2 + V_0
\end{bmatrix}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
e^{i k x – i E t/\Hbar} \\
&=
\begin{bmatrix}
m c^2 + V_0 & c \Hbar k \\
c \Hbar k & – m c^2 + V_0
\end{bmatrix}
\psi.
\end{aligned}
\end{equation}

Writing

\begin{equation}\label{eqn:diracAlternate:80}
H_k
=
\begin{bmatrix}
m c^2 + V_0 & c \Hbar k \\
c \Hbar k & – m c^2 + V_0
\end{bmatrix}
\end{equation}

the characteristic equation is

\begin{equation}\label{eqn:diracAlternate:100}
0
=
(m c^2 + V_0 – \lambda)
(-m c^2 + V_0 – \lambda)
– (c \Hbar k)^2
=
\lr{ (\lambda – V_0)^2 – (m c^2)^2 } – (c \Hbar k)^2,
\end{equation}

so

\begin{equation}\label{eqn:diracAlternate:120}
\lambda = V_0 \pm \epsilon,
\end{equation}

where
\begin{equation}\label{eqn:diracAlternate:140}
\epsilon^2 = (m c^2)^2 + (c \Hbar k)^2.
\end{equation}

We’ve got

\begin{equation}\label{eqn:diracAlternate:160}
\begin{aligned}
H – ( V_0 + \epsilon )
&=
\begin{bmatrix}
m c^2 – \epsilon & c \Hbar k \\
c \Hbar k & – m c^2 – \epsilon
\end{bmatrix} \\
H – ( V_0 – \epsilon )
&=
\begin{bmatrix}
m c^2 + \epsilon & c \Hbar k \\
c \Hbar k & – m c^2 + \epsilon
\end{bmatrix},
\end{aligned}
\end{equation}

so the eigenkets are

\begin{equation}\label{eqn:diracAlternate:180}
\begin{aligned}
\ket{V_0+\epsilon}
&\propto
\begin{bmatrix}
-c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&\propto
\begin{bmatrix}
-c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix}.
\end{aligned}
\end{equation}

Up to an arbitrary phase for each, these are

\begin{equation}\label{eqn:diracAlternate:200}
\begin{aligned}
\ket{V_0 + \epsilon}
&=
\inv{\sqrt{ 2 \epsilon ( \epsilon – m c^2) }}
\begin{bmatrix}
c \Hbar k \\
\epsilon -m c^2
\end{bmatrix} \\
\ket{V_0 – \epsilon}
&=
\inv{\sqrt{ 2 \epsilon ( \epsilon + m c^2) }}
\begin{bmatrix}
-c \Hbar k \\
\epsilon + m c^2
\end{bmatrix} \\
\end{aligned}
\end{equation}

We can now write

\begin{equation}\label{eqn:diracAlternate:220}
H_k =
E
\begin{bmatrix}
V_0 + \epsilon & 0 \\
0 & V_0 – \epsilon
\end{bmatrix}
E^{-1},
\end{equation}

where
\begin{equation}\label{eqn:diracAlternate:240}
\begin{aligned}
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
\sqrt{ \epsilon – m c^2 } & \sqrt{ \epsilon + m c^2 }
\end{bmatrix}, \qquad k > 0 \\
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
-\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
-\sqrt{ \epsilon – m c^2 } & \sqrt{ \epsilon + m c^2 }
\end{bmatrix}, \qquad k < 0. \end{aligned} \end{equation} Here the signs have been adjusted to ensure the transformation matrix has a unit determinant. Observe that there's redundancy in this matrix since \( \ifrac{c \Hbar \Abs{k}}{ \sqrt{ \epsilon - m c^2 } } = \sqrt{ \epsilon + m c^2 } \), and \( \ifrac{c \Hbar \Abs{k}}{ \sqrt{ \epsilon + m c^2 } } = \sqrt{ \epsilon - m c^2 } \), which allows the transformation matrix to be written in the form of a rotation matrix \begin{equation}\label{eqn:diracAlternate:260} \begin{aligned} E &= \inv{\sqrt{2 \epsilon} } \begin{bmatrix} \frac{c \Hbar k}{ \sqrt{ \epsilon - m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\ \frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } & \frac{c \Hbar k}{ \sqrt{ \epsilon - m c^2 } } \end{bmatrix}, \qquad k > 0 \\
E &=
\inv{\sqrt{2 \epsilon} }
\begin{bmatrix}
-\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } \\
\frac{c \Hbar k}{ \sqrt{ \epsilon + m c^2 } } & -\frac{c \Hbar k}{ \sqrt{ \epsilon – m c^2 } }
\end{bmatrix}, \qquad k < 0 \\ \end{aligned} \end{equation} With \begin{equation}\label{eqn:diracAlternate:280} \begin{aligned} \cos\theta &= \frac{c \Hbar \Abs{k}}{ \sqrt{ 2 \epsilon( \epsilon - m c^2) } } = \frac{\sqrt{ \epsilon + m c^2} }{ \sqrt{ 2 \epsilon}}\\ \sin\theta &= \frac{c \Hbar k}{ \sqrt{ 2 \epsilon( \epsilon + m c^2) } } = \frac{\textrm{sgn}(k) \sqrt{ \epsilon - m c^2}}{ \sqrt{ 2 \epsilon } }, \end{aligned} \end{equation} the transformation matrix (and eigenkets) is \begin{equation}\label{eqn:diracAlternate:300} \boxed{ E = \begin{bmatrix} \ket{V_0 + \epsilon} & \ket{V_0 - \epsilon} \end{bmatrix} = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}. } \end{equation} Observe that \ref{eqn:diracAlternate:280} can be simplified by using double angle formulas \begin{equation}\label{eqn:diracAlternate:320} \begin{aligned} \cos(2 \theta) &= \frac{\lr{ \epsilon + m c^2} }{ 2 \epsilon } - \frac{\lr{ \epsilon - m c^2}}{ 2 \epsilon } \\ &= \frac{1}{ 2 \epsilon } \lr{ \epsilon + m c^2 - \epsilon + m c^2 } \\ &= \frac{m c^2 }{ \epsilon }, \end{aligned} \end{equation} and \begin{equation}\label{eqn:diracAlternate:340} \sin(2\theta) = 2 \frac{1}{2 \epsilon} \textrm{sgn}(k ) \sqrt{ \epsilon^2 - (m c^2)^2 } = \frac{\Hbar k c}{\epsilon}. \end{equation} This allows all the \( \theta \) dependence on \( \Hbar k c \) and \( m c^2 \) to be expressed as a ratio of momenta \begin{equation}\label{eqn:diracAlternate:360} \boxed{ \tan(2\theta) = \frac{\Hbar k}{m c}. } \end{equation}

Hyperbolic solutions

For a wave function of the form

\begin{equation}\label{eqn:diracAlternate:380}
\psi =
e^{k x – i E t/\Hbar}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix},
\end{equation}

some of the work above can be recycled if we substitute \( k \rightarrow -i k \), which yields unnormalized eigenfunctions

\begin{equation}\label{eqn:diracAlternate:400}
\begin{aligned}
\ket{V_0+\epsilon}
&\propto
\begin{bmatrix}
i c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&\propto
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix},
\end{aligned}
\end{equation}

where

\begin{equation}\label{eqn:diracAlternate:420}
\epsilon^2 = (m c^2)^2 – (c \Hbar k)^2.
\end{equation}

The squared magnitude of these wavefunctions are

\begin{equation}\label{eqn:diracAlternate:440}
\begin{aligned}
(c \Hbar k)^2 + (m c^2 \mp \epsilon)^2
&=
(c \Hbar k)^2 + (m c^2)^2 + \epsilon^2 \mp 2 m c^2 \epsilon \\
&=
(c \Hbar k)^2 + (m c^2)^2 + (m c^2)^2 \mp (c \Hbar k)^2 – 2 m c^2 \epsilon \\
&= 2 (m c^2)^2 \mp 2 m c^2 \epsilon \\
&= 2 m c^2 ( m c^2 \mp \epsilon ),
\end{aligned}
\end{equation}

so, up to a constant phase for each, the normalized kets are

\begin{equation}\label{eqn:diracAlternate:460}
\begin{aligned}
\ket{V_0+\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 – \epsilon ) }}
\begin{bmatrix}
i c \Hbar k \\
m c^2 – \epsilon
\end{bmatrix} \\
\ket{V_0-\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 + \epsilon ) }}
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix},
\end{aligned}
\end{equation}

After the \( k \rightarrow -i k \) substitution, \( H_k \) is not Hermitian, so these kets aren’t expected to be orthonormal, which is readily verified

\begin{equation}\label{eqn:diracAlternate:480}
\begin{aligned}
\braket{V_0+\epsilon}{V_0-\epsilon}
&=
\inv{\sqrt{ 2 m c^2 ( m c^2 – \epsilon ) }}
\inv{\sqrt{ 2 m c^2 ( m c^2 + \epsilon ) }}
\begin{bmatrix}
-i c \Hbar k &
m c^2 – \epsilon
\end{bmatrix}
\begin{bmatrix}
i c \Hbar k \\
m c^2 + \epsilon
\end{bmatrix} \\
&=
\frac{ 2 ( c \Hbar k )^2 }{2 m c^2 \sqrt{(\Hbar k c)^2} } \\
&=
\textrm{sgn}(k)
\frac{
\Hbar k }{m c } .
\end{aligned}
\end{equation}

Heisenberg velocity operator

\begin{equation}\label{eqn:diracAlternate:500}
\begin{aligned}
\hat{v}
&= \inv{i \Hbar} \antisymmetric{ \hat{x} }{ H} \\
&= \inv{i \Hbar} \antisymmetric{ \hat{x} }{ m c^2 \sigma_z + V_0 + c \hat{p} \sigma_x } \\
&= \frac{c \sigma_x}{i \Hbar} \antisymmetric{ \hat{x} }{ \hat{p} } \\
&= c \sigma_x.
\end{aligned}
\end{equation}

Probability current

Acting against a completely general wavefunction the Hamiltonian action \( H \psi \) is

\begin{equation}\label{eqn:diracAlternate:520}
\begin{aligned}
i \Hbar \PD{t}{\psi}
&= m c^2 \sigma_z \psi + V_0 \psi + c \hat{p} \sigma_x \psi \\
&= m c^2 \sigma_z \psi + V_0 \psi -i \Hbar c \sigma_x \PD{x}{\psi}.
\end{aligned}
\end{equation}

Conversely, the conjugate \( (H \psi)^\dagger \) is

\begin{equation}\label{eqn:diracAlternate:540}
-i \Hbar \PD{t}{\psi^\dagger}
= m c^2 \psi^\dagger \sigma_z + V_0 \psi^\dagger +i \Hbar c \PD{x}{\psi^\dagger} \sigma_x.
\end{equation}

These give

\begin{equation}\label{eqn:diracAlternate:560}
\begin{aligned}
i \Hbar \psi^\dagger \PD{t}{\psi}
&=
m c^2 \psi^\dagger \sigma_z \psi + V_0 \psi^\dagger \psi -i \Hbar c \psi^\dagger \sigma_x \PD{x}{\psi} \\
-i \Hbar \PD{t}{\psi^\dagger} \psi
&= m c^2 \psi^\dagger \sigma_z \psi + V_0 \psi^\dagger \psi +i \Hbar c \PD{x}{\psi^\dagger} \sigma_x \psi.
\end{aligned}
\end{equation}

Taking differences
\begin{equation}\label{eqn:diracAlternate:580}
\psi^\dagger \PD{t}{\psi} + \PD{t}{\psi^\dagger} \psi
=
– c \psi^\dagger \sigma_x \PD{x}{\psi} – c \PD{x}{\psi^\dagger} \sigma_x \psi,
\end{equation}

or

\begin{equation}\label{eqn:diracAlternate:600}
0
=
\PD{t}{}
\lr{
\psi^\dagger \psi
}
+
\PD{x}{}
\lr{
c \psi^\dagger \sigma_x \psi
}.
\end{equation}

The probability current still has the usual form \( \rho = \psi^\dagger \psi = \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2 \), but the probability current with this representation of the Dirac Hamiltonian is

\begin{equation}\label{eqn:diracAlternate:620}
\begin{aligned}
j
&= c \psi^\dagger \sigma_x \psi \\
&= c
\begin{bmatrix}
\psi_1^\conj &
\psi_2^\conj
\end{bmatrix}
\begin{bmatrix}
\psi_2 \\
\psi_1
\end{bmatrix} \\
&= c \lr{ \psi_1^\conj \psi_2 + \psi_2^\conj \psi_1 }.
\end{aligned}
\end{equation}

PHY1520H Graduate Quantum Mechanics. Lecture 18: Approximation methods. Taught by Prof. Arun Paramekanti

November 26, 2015 phy1520 No comments , ,

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Disclaimer

Peeter’s lecture notes from class. These may be incoherent and rough, especially since I didn’t attend this class myself, and am doing a walkthrough of notes provided by Nishant.

These are notes for the UofT course PHY1520, Graduate Quantum Mechanics, taught by Prof. Paramekanti, covering [1] chap. 5 content.

Approximation methods

Suppose we have a perturbed Hamiltonian

\begin{equation}\label{eqn:qmLecture18:20}
H = H_0 + \lambda V,
\end{equation}

where \( \lambda = 0 \) represents a solvable (perhaps known) system, and \( \lambda = 1 \) is the case of interest. There are two approaches of interest

  1. Direct solution of \( H \) with \( \lambda = 1 \).
  2. Take \( \lambda \) small, and do a series expansion. This is perturbation theory.

Variational methods

Given

\begin{equation}\label{eqn:qmLecture18:40}
H \ket{\phi_n} = E_n \ket{\phi_n},
\end{equation}

where we don’t know \( \ket{\phi_n} \), we can compute the expectation with respect to an arbitrary state \( \ket{\psi} \)

\begin{equation}\label{eqn:qmLecture18:60}
\bra{\psi} H \ket{\psi}
=
\bra{\psi} H \lr{ \sum_n \ket{\phi_n} \bra{\phi_n} } \ket{\psi}
=
\sum_n E_n \braket{\psi}{\phi_n} \braket{\phi_n}{\psi}
=
\sum_n E_n \Abs{\braket{\psi}{\phi_n}}^2.
\end{equation}

Define

\begin{equation}\label{eqn:qmLecture18:80}
\overline{{E}}
= \frac{\bra{\psi} H \ket{\psi}}{\braket{\psi}{\psi}}.
\end{equation}

Assuming that it is possible to express the state in the Hamiltonian energy basis

\begin{equation}\label{eqn:qmLecture18:100}
\ket{\psi}
=
\sum_n a_n \ket{\phi_n},
\end{equation}

this average energy is
\begin{equation}\label{eqn:qmLecture18:120}
\overline{{E}}
= \frac{ \sum_{m,n}\bra{\phi_m} a_m^\conj H a_n \ket{\phi_n}}{ \sum_n \Abs{a_n}^2 }
= \frac{ \sum_{n} \Abs{a_n}^2 E_n }{ \sum_n \Abs{a_n}^2 }.
= \sum_{n}
\frac{\Abs{a_n}^2 }{ \sum_n \Abs{a_n}^2 }
E_n
= \sum_n \frac{P_n}{\sum_m P_m} E_n,
\end{equation}

where \( P_m = \Abs{a_m}^2 \), which has the structure of a probability coefficient once divided by \( \sum_m P_m \), as sketched in fig. 1.

fig. 1.  A decreasing probability distribution

fig. 1. A decreasing probability distribution

This average energy is a probability weighted average of the individual energy basis states. One of those energies is the ground state energy \( E_1 \), so we necessarily have

\begin{equation}\label{eqn:qmLecture18:140}
\boxed{
\overline{{E}} \ge E_1.
}
\end{equation}

Example: particle in a \( [0,L] \) box.

For the infinite potential box sketched in fig. 2.

fig. 2.  Infinite potential  [0,L]  box.

fig. 2. Infinite potential [0,L] box.

The exact solutions for such a system are found to be

\begin{equation}\label{eqn:qmLecture18:220}
\psi(x) = \sqrt{\frac{2}{L}} \sin\lr{ \frac{n \pi}{L} x },
\end{equation}

where the energies are

\begin{equation}\label{eqn:qmLecture18:240}
E = \frac{\Hbar^2}{2m} \frac{n^2 \pi^2}{L^2}.
\end{equation}

The function \( \psi’ = x (L-x) \) also satisfies the boundary value constraints? How close in energy is that function to the ground state?

\begin{equation}\label{eqn:qmLecture18:260}
\overline{{E}}
=
-\frac{\Hbar^2}{2m} \frac{\int_0^L dx x (L-x) \frac{d^2}{dx^2} \lr{ x (L-x) }}{
\int_0^L dx x^2 (L-x)^2
}
=
\frac{\Hbar^2}{2m} \frac{\frac{2 L^3}{6}}{
\frac{L^5}{30}
}
=
\frac{\Hbar^2}{2m} \frac{10}{L^2}.
\end{equation}

This average energy is quite close to the ground state energy

\begin{equation}\label{eqn:qmLecture18:280}
\frac{\overline{{E}} }{E_1} = \frac{10}{\pi^2} = 1.014.
\end{equation}

Example II: particle in a \( [-L/2,L/2] \) box.

fig. 3.  Infinite potential  [-L/2,L/2]  box.

fig. 3. Infinite potential [-L/2,L/2] box.

Shifting the boundaries, as sketched in fig. 3 doesn’t change the energy levels. For this potential let’s try a shifted trial function

\begin{equation}\label{eqn:qmLecture18:300}
\psi(x) = \lr{ x – \frac{L}{2} } \lr{ x + \frac{L}{2} } = x^2 – \frac{L^2}{4},
\end{equation}

without worrying about the form of the exact solution. This produces the same result as above

\begin{equation}\label{eqn:qmLecture18:270}
\overline{{E}}
=
-\frac{\Hbar^2}{2m} \frac{\int_0^L dx \lr{ x^2 – \frac{L^2}{4} } \frac{d^2}{dx^2} \lr{ x^2 – \frac{L^2}{4} }}{
\int_0^L dx \lr{x^2 – \frac{L^2}{4} }^2
}
=
-\frac{\Hbar^2}{2m} \frac{- 2 L^3/6}{
\frac{L^5}{30}
}
=
\frac{\Hbar^2}{2m} \frac{10}{L^2}.
\end{equation}

Summary (Nishant)

The above example is that of a particle in a box. The actual wave function is a sin as shown. But we can
come up with a guess wave function that meets the boundary conditions and ask how accurate it is
compared to the actual one.

Basically we are assuming a wave function form and then seeing how it differs from the exact form.
We cannot do this if we have nothing to compare it against. But, we note that the variance of the
number operator in the systems eigenstate is zero. So we can still calculate the variance and try to
minimize it. This is one way of coming up with an approximate wave function. This does not necessarily
give the ground state wave function though. For this we need to minimize the energy itself.

References

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

Two spin time evolution

November 14, 2015 phy1520 No comments , , , ,

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Motivation

Our midterm posed a (low mark “quick question”) that I didn’t complete (or at least not properly). This shouldn’t have been a difficult question, but I spend way too much time on it, costing me time that I needed for other questions.

It turns out that there isn’t anything fancy required for this question, just perseverance and careful work.

Guts

The question asked for the time evolution of a two particle state

\begin{equation}\label{eqn:twoSpinHamiltonian:20}
\psi = \inv{\sqrt{2}} \lr{ \ket{\uparrow \downarrow} – \ket{\downarrow \uparrow} }
\end{equation}

under the action of the Hamiltonian

\begin{equation}\label{eqn:twoSpinHamiltonian:40}
H = – B S_{z,1} + 2 B S_{x,2} = \frac{\Hbar B}{2}\lr{ -\sigma_{z,1} + 2 \sigma_{x,2} } .
\end{equation}

We have to know the action of the Hamiltonian on all the states

\begin{equation}\label{eqn:twoSpinHamiltonian:60}
\begin{aligned}
H \ket{\uparrow \uparrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \uparrow} + 2 \ket{\uparrow \downarrow} } \\
H \ket{\uparrow \downarrow} &= \frac{B \Hbar}{2} \lr{ -\ket{\uparrow \downarrow} + 2 \ket{\uparrow \uparrow} } \\
H \ket{\downarrow \uparrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \uparrow} + 2 \ket{\downarrow \downarrow} } \\
H \ket{\downarrow \downarrow} &= \frac{B \Hbar}{2} \lr{ \ket{\downarrow \downarrow} + 2 \ket{\downarrow \uparrow} } \\
\end{aligned}
\end{equation}

With respect to the basis \( \setlr{ \ket{\uparrow \uparrow}, \ket{\uparrow \downarrow}, \ket{\downarrow \uparrow}, \ket{\downarrow \downarrow} } \), the matrix of the Hamiltonian is

\begin{equation}\label{eqn:twoSpinHamiltonian:80}
H =
\frac{ \Hbar B }{2}
\begin{bmatrix}
-1 & 2 & 0 & 0 \\
2 & -1 & 0 & 0 \\
0 & 0 & 1 & 2 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}
\end{equation}

Utilizing the block diagonal form (and ignoring the \( \Hbar B/2 \) factor for now), the characteristic equation is

\begin{equation}\label{eqn:twoSpinHamiltonian:100}
0
=
\begin{vmatrix}
-1 -\lambda & 2 \\
2 & -1 – \lambda
\end{vmatrix}
\begin{vmatrix}
1 -\lambda & 2 \\
2 & 1 – \lambda
\end{vmatrix}
=
\lr{(1 + \lambda)^2 – 4}
\lr{(1 – \lambda)^2 – 4}.
\end{equation}

This has solutions

\begin{equation}\label{eqn:twoSpinHamiltonian:120}
1 \pm \lambda = \pm 2,
\end{equation}

or, with the \( \Hbar B/2 \) factors put back in

\begin{equation}\label{eqn:twoSpinHamiltonian:140}
\lambda = \pm \Hbar B/2 , \pm 3 \Hbar B/2.
\end{equation}

I was thinking that we needed to compute the time evolution operator

\begin{equation}\label{eqn:twoSpinHamiltonian:160}
U = e^{-i H t/\Hbar},
\end{equation}

but we actually only need the eigenvectors, and the inverse relations. We can find the eigenvectors by inspection in each case from

\begin{equation}\label{eqn:twoSpinHamiltonian:180}
\begin{aligned}
H – (1) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
-2 & 2 & 0 & 0 \\
2 & -2 & 0 & 0 \\
0 & 0 & 0 & 2 \\
0 & 0 & 2 & 0 \\
\end{bmatrix} \\
H – (-1) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
0 & 2 & 0 & 0 \\
2 & 0 & 0 & 0 \\
0 & 0 & 2 & 2 \\
0 & 0 & 2 & 2 \\
\end{bmatrix} \\
H – (3) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
-4 & 2 & 0 & 0 \\
2 & -4 & 0 & 0 \\
0 & 0 &-2 & 2 \\
0 & 0 & 2 &-2 \\
\end{bmatrix} \\
H – (-3) \frac{ \Hbar B }{2}
&=
\frac{ \Hbar B }{2}
\begin{bmatrix}
2 & 2 & 0 & 0 \\
2 & 2 & 0 & 0 \\
0 & 0 & 4 & 2 \\
0 & 0 & 2 & 1 \\
\end{bmatrix}.
\end{aligned}
\end{equation}

The eigenkets are

\begin{equation}\label{eqn:twoSpinHamiltonian:280}
\begin{aligned}
\ket{1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
1 \\
0 \\
0 \\
\end{bmatrix} \\
\ket{-1} &=
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
0 \\
1 \\
-1 \\
\end{bmatrix} \\
\ket{3} &=
\inv{\sqrt{2}}
\begin{bmatrix}
0 \\
0 \\
1 \\
1 \\
\end{bmatrix} \\
\ket{-3} &=
\inv{\sqrt{2}}
\begin{bmatrix}
1 \\
-1 \\
0 \\
0 \\
\end{bmatrix},
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:twoSpinHamiltonian:300}
\begin{aligned}
\sqrt{2} \ket{1} &= \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} \\
\sqrt{2} \ket{-1} &= \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} \\
\sqrt{2} \ket{3} &= \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} \\
\sqrt{2} \ket{-3} &= \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow}.
\end{aligned}
\end{equation}

We can invert these

\begin{equation}\label{eqn:twoSpinHamiltonian:220}
\begin{aligned}
\ket{\uparrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{1} + \ket{-3} } \\
\ket{\uparrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{1} – \ket{-3} } \\
\ket{\downarrow \uparrow} &= \inv{\sqrt{2}} \lr{ \ket{3} + \ket{-1} } \\
\ket{\downarrow \downarrow} &= \inv{\sqrt{2}} \lr{ \ket{3} – \ket{-1} } \\
\end{aligned}
\end{equation}

The original state of interest can now be expressed in terms of the eigenkets

\begin{equation}\label{eqn:twoSpinHamiltonian:240}
\psi
=
\inv{2} \lr{
\ket{1} – \ket{-3} –
\ket{3} – \ket{-1}
}
\end{equation}

The time evolution of this ket is

\begin{equation}\label{eqn:twoSpinHamiltonian:260}
\begin{aligned}
\psi(t)
&=
\inv{2}
\lr{
e^{-i B t/2} \ket{1}
– e^{3 i B t/2} \ket{-3}
– e^{-3 i B t/2} \ket{3}
– e^{i B t/2} \ket{-1}
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
e^{-i B t/2} \lr{ \ket{\uparrow \uparrow} + \ket{\uparrow \downarrow} }
– e^{3 i B t/2} \lr{ \ket{\uparrow \uparrow} – \ket{\uparrow \downarrow} }
– e^{-3 i B t/2} \lr{ \ket{\downarrow \uparrow} + \ket{\downarrow \downarrow} }
– e^{i B t/2} \lr{ \ket{\downarrow \uparrow} – \ket{\downarrow \downarrow} }
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
\lr{ e^{-i B t/2} – e^{3 i B t/2} } \ket{\uparrow \uparrow}
+ \lr{ e^{-i B t/2} + e^{3 i B t/2} } \ket{\uparrow \downarrow}
– \lr{ e^{-3 i B t/2} + e^{i B t/2} } \ket{\downarrow \uparrow}
+ \lr{ e^{i B t/2} – e^{-3 i B t/2} } \ket{\downarrow \downarrow}
} \\
&=
\inv{2 \sqrt{2}}
\Biglr{
e^{i B t/2} \lr{ e^{-2 i B t/2} – e^{2 i B t/2} } \ket{\uparrow \uparrow}
+ e^{i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\uparrow \downarrow}
– e^{- i B t/2} \lr{ e^{-2 i B t/2} + e^{2 i B t/2} } \ket{\downarrow \uparrow}
+ e^{- i B t/2} \lr{ e^{2 i B t/2} – e^{-2 i B t/2} } \ket{\downarrow \downarrow}
} \\
&=
\inv{\sqrt{2}}
\lr{
i \sin( B t )
\lr{
e^{- i B t/2} \ket{\downarrow \downarrow} – e^{i B t/2} \ket{\uparrow \uparrow}
}
+ \cos( B t ) \lr{
e^{i B t/2} \ket{\uparrow \downarrow}
– e^{- i B t/2} \ket{\downarrow \uparrow}
}
}
\end{aligned}
\end{equation}

Note that this returns to the original state when \( t = \frac{2 \pi n}{B}, n \in \mathbb{Z} \). I think I’ve got it right this time (although I got a slightly different answer on paper before typing it up.)

This doesn’t exactly seem like a quick answer question, at least to me. Is there some easier way to do it?

PHY1520H Graduate Quantum Mechanics. Lecture 12: Symmetry (cont.). Taught by Prof. Arun Paramekanti

November 5, 2015 phy1520 No comments , , , , , , ,

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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. 4 content from [1].

Parity (review)

\begin{equation}\label{eqn:qmLecture12:20}
\hat{\Pi} \hat{x} \hat{\Pi} = – \hat{x}
\end{equation}
\begin{equation}\label{eqn:qmLecture12:40}
\hat{\Pi} \hat{p} \hat{\Pi} = – \hat{p}
\end{equation}

These are polar vectors, in contrast to an axial vector such as \( \BL = \Br \cross \Bp \).

\begin{equation}\label{eqn:qmLecture12:60}
\hat{\Pi}^2 = 1
\end{equation}

\begin{equation}\label{eqn:qmLecture12:80}
\Psi(x) \rightarrow \Psi(-x)
\end{equation}

If \( \antisymmetric{\hat{\Pi}}{\hat{H}} = 0 \) then all the eigenstates are either

  • even: \( \hat{\Pi} \) eigenvalue is \( + 1 \).
  • odd: \( \hat{\Pi} \) eigenvalue is \( – 1 \).

We are done with discrete symmetry operators for now.

Translations

Define a (continuous) translation operator

\begin{equation}\label{eqn:qmLecture12:100}
\hat{T}_\epsilon \ket{x} = \ket{x + \epsilon}
\end{equation}

The action of this operator is sketched in fig. 1.

lecture12Fig1

fig. 1. Translation operation.

 

This is a unitary operator

\begin{equation}\label{eqn:qmLecture12:120}
\hat{T}_{-\epsilon} = \hat{T}_{\epsilon}^\dagger = \hat{T}_{\epsilon}^{-1}
\end{equation}

In a position basis, the action of this operator is

\begin{equation}\label{eqn:qmLecture12:140}
\bra{x} \hat{T}_{\epsilon} \ket{\psi} = \braket{x-\epsilon}{\psi} = \psi(x – \epsilon)
\end{equation}

\begin{equation}\label{eqn:qmLecture12:160}
\Psi(x – \epsilon) \approx \Psi(x) – \epsilon \PD{x}{\Psi}
\end{equation}

\begin{equation}\label{eqn:qmLecture12:180}
\bra{x} \hat{T}_{\epsilon} \ket{\Psi}
= \braket{x}{\Psi} – \frac{\epsilon}{\Hbar} \bra{ x} i \hat{p} \ket{\Psi}
\end{equation}

\begin{equation}\label{eqn:qmLecture12:200}
\hat{T}_{\epsilon} \approx \lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{p} }
\end{equation}

A non-infinitesimal translation can be composed of many small translations, as sketched in fig. 2.

fig. 2. Composition of small translations

fig. 2. Composition of small translations

For \( \epsilon \rightarrow 0, N \rightarrow \infty, N \epsilon = a \), the total translation operator is

\begin{equation}\label{eqn:qmLecture12:220}
\begin{aligned}
\hat{T}_{a}
&= \hat{T}_{\epsilon}^N \\
&= \lim_{\epsilon \rightarrow 0, N \rightarrow \infty, N \epsilon = a }
\lr{ 1 – \frac{\epsilon}{\Hbar} \hat{p} }^N \\
&= e^{-i a \hat{p}/\Hbar}
\end{aligned}
\end{equation}

The momentum \( \hat{p} \) is called a “Generator” generator of translations. If a Hamiltonian \( H \) is translationally invariant, then

\begin{equation}\label{eqn:qmLecture12:240}
\antisymmetric{\hat{T}_{a}}{H} = 0, \qquad \forall a.
\end{equation}

This means that momentum will be a good quantum number

\begin{equation}\label{eqn:qmLecture12:260}
\antisymmetric{\hat{p}}{H} = 0.
\end{equation}

Rotations

Rotations form a non-Abelian group, since the order of rotations \( \hatR_1 \hatR_2 \ne \hatR_2 \hatR_1 \).

Given a rotation acting on a ket

\begin{equation}\label{eqn:qmLecture12:280}
\hatR \ket{\Br} = \ket{R \Br},
\end{equation}

observe that the action of the rotation operator on a wave function is inverted

\begin{equation}\label{eqn:qmLecture12:300}
\bra{\Br} \hatR \ket{\Psi}
=
\bra{R^{-1} \Br} \ket{\Psi}
= \Psi(R^{-1} \Br).
\end{equation}

Example: Z axis normal rotation

Consider an infinitesimal rotation about the z-axis as sketched in fig. 3(a),(b)

lecture12Fig3

fig 3(a). Rotation about z-axis.

fig 3(b). Rotation about z-axis.

fig 3(b). Rotation about z-axis.

\begin{equation}\label{eqn:qmLecture12:320}
\begin{aligned}
x’ &= x – \epsilon y \\
y’ &= y + \epsilon y \\
z’ &= z
\end{aligned}
\end{equation}

The rotated wave function is

\begin{equation}\label{eqn:qmLecture12:340}
\tilde{\Psi}(x,y,z)
= \Psi( x + \epsilon y, y – \epsilon x, z )
=
\Psi( x, y, z )
+
\epsilon y \underbrace{\PD{x}{\Psi}}_{i \hat{p}_x/\Hbar}

\epsilon x \underbrace{\PD{y}{\Psi}}_{i \hat{p}_y/\Hbar}.
\end{equation}

The state must then transform as

\begin{equation}\label{eqn:qmLecture12:360}
\ket{\tilde{\Psi}}
=
\lr{
1
+ i \frac{\epsilon}{\Hbar} \hat{y} \hat{p}_x
– i \frac{\epsilon}{\Hbar} \hat{x} \hat{p}_y
}
\ket{\Psi}.
\end{equation}

Observe that the combination \( \hat{x} \hat{p}_y – \hat{y} \hat{p}_x \) is the \( \hat{L}_z \) component of angular momentum \( \hat{\BL} = \hat{\Br} \cross \hat{\Bp} \), so the infinitesimal rotation can be written

\begin{equation}\label{eqn:qmLecture12:380}
\boxed{
\hatR_z(\epsilon) \ket{\Psi}
=
\lr{ 1 – i \frac{\epsilon}{\Hbar} \hat{L}_z } \ket{\Psi}.
}
\end{equation}

For a finite rotation \( \epsilon \rightarrow 0, N \rightarrow \infty, \phi = \epsilon N \), the total rotation is

\begin{equation}\label{eqn:qmLecture12:420}
\hatR_z(\phi)
=
\lr{ 1 – \frac{i \epsilon}{\Hbar} \hat{L}_z }^N,
\end{equation}

or
\begin{equation}\label{eqn:qmLecture12:440}
\boxed{
\hatR_z(\phi)
=
e^{-i \frac{\phi}{\Hbar} \hat{L}_z}.
}
\end{equation}

Note that \( \antisymmetric{\hat{L}_x}{\hat{L}_y} \ne 0 \).

By construction using Euler angles or any other method, a general rotation will include contributions from components of all the angular momentum operator, and will have the structure

\begin{equation}\label{eqn:qmLecture12:480}
\boxed{
\hatR_\ncap(\phi)
=
e^{-i \frac{\phi}{\Hbar} \lr{ \hat{\BL} \cdot \ncap }}.
}
\end{equation}

Rotationally invariant \( \hat{H} \).

Given a rotationally invariant Hamiltonian

\begin{equation}\label{eqn:qmLecture12:520}
\antisymmetric{\hat{R}_\ncap(\phi)}{\hat{H}} = 0 \qquad \forall \ncap, \phi,
\end{equation}

then every

\begin{equation}\label{eqn:qmLecture12:540}
\antisymmetric{\BL \cdot \ncap}{\hat{H}} = 0,
\end{equation}

or
\begin{equation}\label{eqn:qmLecture12:560}
\antisymmetric{L_i}{\hat{H}} = 0,
\end{equation}

Non-Abelian implies degeneracies in the spectrum.

Time-reversal

Imagine that we have something moving along a curve at time \( t = 0 \), and ending up at the final position at time \( t = t_f \).

fig. 4. Time reversal trajectory.

fig. 4. Time reversal trajectory.

Imagine that we flip the direction of motion (i.e. flipping the velocity) and run time backwards so the final-time state becomes the initial state.

If the time reversal operator is designated \( \hat{\Theta} \), with operation

\begin{equation}\label{eqn:qmLecture12:580}
\hat{\Theta} \ket{\Psi} = \ket{\tilde{\Psi}},
\end{equation}

so that

\begin{equation}\label{eqn:qmLecture12:600}
\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(t)} = \ket{\Psi(0)},
\end{equation}

or

\begin{equation}\label{eqn:qmLecture12:620}
\hat{\Theta}^{-1} e^{-i \hat{H} t/\Hbar} \hat{\Theta} \ket{\Psi(0)} = \ket{\Psi(-t)}.
\end{equation}

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.

Commutators of angular momentum and a central force Hamiltonian

September 30, 2015 phy1520 No comments , , , ,

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

In problem 1.17 of [1] we are to show that non-commuting operators that both commute with the Hamiltonian, have, in general, degenerate energy eigenvalues. It suggests considering \( L_x, L_z \) and a central force Hamiltonian \( H = \Bp^2/2m + V(r) \) as examples.

Let’s just demonstrate these commutators act as expected in these cases.

With \( \BL = \Bx \cross \Bp \), we have

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:20}
\begin{aligned}
L_x &= y p_z – z p_y \\
L_y &= z p_x – x p_z \\
L_z &= x p_y – y p_x.
\end{aligned}
\end{equation}

The \( L_x, L_z \) commutator is

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:40}
\begin{aligned}
\antisymmetric{L_x}{L_z}
&=
\antisymmetric{y p_z – z p_y }{x p_y – y p_x} \\
&=
\antisymmetric{y p_z}{x p_y}
-\antisymmetric{y p_z}{y p_x}
-\antisymmetric{z p_y }{x p_y}
+\antisymmetric{z p_y }{y p_x} \\
&=
x p_z \antisymmetric{y}{p_y}
+ z p_x \antisymmetric{p_y }{y} \\
&=
i \Hbar \lr{ x p_z – z p_x } \\
&=
– i \Hbar L_y
\end{aligned}
\end{equation}

cyclicly permuting the indexes shows that no pairs of different \( \BL \) components commute. For \( L_y, L_x \) that is

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:60}
\begin{aligned}
\antisymmetric{L_y}{L_x}
&=
\antisymmetric{z p_x – x p_z }{y p_z – z p_y} \\
&=
\antisymmetric{z p_x}{y p_z}
-\antisymmetric{z p_x}{z p_y}
-\antisymmetric{x p_z }{y p_z}
+\antisymmetric{x p_z }{z p_y} \\
&=
y p_x \antisymmetric{z}{p_z}
+ x p_y \antisymmetric{p_z }{z} \\
&=
i \Hbar \lr{ y p_x – x p_y } \\
&=
– i \Hbar L_z,
\end{aligned}
\end{equation}

and for \( L_z, L_y \)

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:80}
\begin{aligned}
\antisymmetric{L_z}{L_y}
&=
\antisymmetric{x p_y – y p_x }{z p_x – x p_z} \\
&=
\antisymmetric{x p_y}{z p_x}
-\antisymmetric{x p_y}{x p_z}
-\antisymmetric{y p_x }{z p_x}
+\antisymmetric{y p_x }{x p_z} \\
&=
z p_y \antisymmetric{x}{p_x}
+ y p_z \antisymmetric{p_x }{x} \\
&=
i \Hbar \lr{ z p_y – y p_z } \\
&=
– i \Hbar L_x.
\end{aligned}
\end{equation}

If these angular momentum components are also shown to commute with themselves (which they do), the commutator relations above can be summarized as

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:100}
\antisymmetric{L_a}{L_b} = i \Hbar \epsilon_{a b c} L_c.
\end{equation}

In the example to consider, we’ll have to consider the commutators with \( \Bp^2 \) and \( V(r) \). Picking any one component of \( \BL \) is sufficent due to the symmetries of the problem. For example

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:120}
\begin{aligned}
\antisymmetric{L_x}{\Bp^2}
&=
\antisymmetric{y p_z – z p_y}{p_x^2 + p_y^2 + p_z^2} \\
&=
\antisymmetric{y p_z}{{p_x^2} + p_y^2 + {p_z^2}}
-\antisymmetric{z p_y}{{p_x^2} + {p_y^2} + p_z^2} \\
&=
p_z \antisymmetric{y}{p_y^2}
-p_y \antisymmetric{z}{p_z^2} \\
&=
p_z 2 i \Hbar p_y
2 i \Hbar p_y
-p_y 2 i \Hbar p_z \\
&=
0.
\end{aligned}
\end{equation}

How about the commutator of \( \BL \) with the potential? It is sufficient to consider one component again, for example

\begin{equation}\label{eqn:angularMomentumAndCentralForceCommutators:140}
\begin{aligned}
\antisymmetric{L_x}{V}
&=
\antisymmetric{y p_z – z p_y}{V} \\
&=
y \antisymmetric{p_z}{V} – z \antisymmetric{p_y}{V} \\
&=
-i \Hbar y \PD{z}{V(r)} + i \Hbar z \PD{y}{V(r)} \\
&=
-i \Hbar y \PD{r}{V}\PD{z}{r} + i \Hbar z \PD{r}{V}\PD{y}{r} \\
&=
-i \Hbar y \PD{r}{V} \frac{z}{r} + i \Hbar z \PD{r}{V}\frac{y}{r} \\
&=
0.
\end{aligned}
\end{equation}

We’ve shown that all the components of \( \BL \) commute with a central force Hamiltonian, and each different component of \( \BL \) do not commute.

The next step will be figuring out how to use this to show that there are energy degeneracies.

References

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