rotation

Lorentz boosts in Geometric Algebra paravector notation.

January 14, 2018 math and physics play , , , , , , , , , , , ,

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

Motivation.

The notation I prefer for relativistic geometric algebra uses Hestenes’ space time algebra (STA) [2], where the basis is a four dimensional space \( \setlr{ \gamma_\mu } \), subject to Dirac matrix like relations \( \gamma_\mu \cdot \gamma_\nu = \eta_{\mu \nu} \).

In this formalism a four vector is just the sum of the products of coordinates and basis vectors, for example, using summation convention

\begin{equation}\label{eqn:boostToParavector:160}
x = x^\mu \gamma_\mu.
\end{equation}

The invariant for a four-vector in STA is just the square of that vector

\begin{equation}\label{eqn:boostToParavector:180}
\begin{aligned}
x^2
&= (x^\mu \gamma_\mu) \cdot (x^\nu \gamma_\nu) \\
&= \sum_\mu (x^\mu)^2 (\gamma_\mu)^2 \\
&= (x^0)^2 – \sum_{k = 1}^3 (x^k)^2 \\
&= (ct)^2 – \Bx^2.
\end{aligned}
\end{equation}

Recall that a four-vector is time-like if this squared-length is positive, spacelike if negative, and light-like when zero.

Time-like projections are possible by dotting with the “lab-frame” time like basis vector \( \gamma_0 \)

\begin{equation}\label{eqn:boostToParavector:200}
ct = x \cdot \gamma_0 = x^0,
\end{equation}

and space-like projections are wedges with the same

\begin{equation}\label{eqn:boostToParavector:220}
\Bx = x \cdot \gamma_0 = x^k \sigma_k,
\end{equation}

where sums over Latin indexes \( k \in \setlr{1,2,3} \) are implied, and where the elements \( \sigma_k \)

\begin{equation}\label{eqn:boostToParavector:80}
\sigma_k = \gamma_k \gamma_0.
\end{equation}

which are bivectors in STA, can be viewed as an Euclidean vector basis \( \setlr{ \sigma_k } \).

Rotations in STA involve exponentials of space like bivectors \( \theta = a_{ij} \gamma_i \wedge \gamma_j \)

\begin{equation}\label{eqn:boostToParavector:240}
x’ = e^{ \theta/2 } x e^{ -\theta/2 }.
\end{equation}

Boosts, on the other hand, have exactly the same form, but the exponentials are with respect to space-time bivectors arguments, such as \( \theta = a \wedge \gamma_0 \), where \( a \) is any four-vector.

Observe that both boosts and rotations necessarily conserve the space-time length of a four vector (or any multivector with a scalar square).

\begin{equation}\label{eqn:boostToParavector:260}
\begin{aligned}
\lr{x’}^2
&=
\lr{ e^{ \theta/2 } x e^{ -\theta/2 } } \lr{ e^{ \theta/2 } x e^{ -\theta/2 } } \\
&=
e^{ \theta/2 } x \lr{ e^{ -\theta/2 } e^{ \theta/2 } } x e^{ -\theta/2 } \\
&=
e^{ \theta/2 } x^2 e^{ -\theta/2 } \\
&=
x^2 e^{ \theta/2 } e^{ -\theta/2 } \\
&=
x^2.
\end{aligned}
\end{equation}

Paravectors.

Paravectors, as used by Baylis [1], represent four-vectors using a Euclidean multivector basis \( \setlr{ \Be_\mu } \), where \( \Be_0 = 1 \). The conversion between STA and paravector notation requires only multiplication with the timelike basis vector for the lab frame \( \gamma_0 \)

\begin{equation}\label{eqn:boostToParavector:40}
\begin{aligned}
X
&= x \gamma_0 \\
&= \lr{ x^0 \gamma_0 + x^k \gamma_k } \gamma_0 \\
&= x^0 + x^k \gamma_k \gamma_0 \\
&= x^0 + \Bx \\
&= c t + \Bx,
\end{aligned}
\end{equation}

We need a different structure for the invariant length in paravector form. That invariant length is
\begin{equation}\label{eqn:boostToParavector:280}
\begin{aligned}
x^2
&=
\lr{ \lr{ ct + \Bx } \gamma_0 }
\lr{ \lr{ ct + \Bx } \gamma_0 } \\
&=
\lr{ \lr{ ct + \Bx } \gamma_0 }
\lr{ \gamma_0 \lr{ ct – \Bx } } \\
&=
\lr{ ct + \Bx }
\lr{ ct – \Bx }.
\end{aligned}
\end{equation}

Baylis introduces an involution operator \( \overline{{M}} \) which toggles the sign of any vector or bivector grades of a multivector. For example, if \( M = a + \Ba + I \Bb + I c \), where \( a,c \in \mathbb{R} \) and \( \Ba, \Bb \in \mathbb{R}^3 \) is a multivector with all grades \( 0,1,2,3 \), then the involution of \( M \) is

\begin{equation}\label{eqn:boostToParavector:300}
\overline{{M}} = a – \Ba – I \Bb + I c.
\end{equation}

Utilizing this operator, the invariant length for a paravector \( X \) is \( X \overline{{X}} \).

Let’s consider how boosts and rotations can be expressed in the paravector form. The half angle operator for a boost along the spacelike \( \Bv = v \vcap \) direction has the form

\begin{equation}\label{eqn:boostToParavector:120}
L = e^{ -\vcap \phi/2 },
\end{equation}

\begin{equation}\label{eqn:boostToParavector:140}
\begin{aligned}
X’
&=
c t’ + \Bx’ \\
&=
x’ \gamma_0 \\
&=
L x L^\dagger \\
&=
e^{ -\vcap \phi/2 } x^\mu \gamma_\mu
e^{ \vcap \phi/2 } \gamma_0 \\
&=
e^{ -\vcap \phi/2 } x^\mu \gamma_\mu \gamma_0
e^{ -\vcap \phi/2 } \\
&=
e^{ -\vcap \phi/2 } \lr{ x^0 + \Bx } e^{ -\vcap \phi/2 } \\
&=
L X L.
\end{aligned}
\end{equation}

Because the involution operator toggles the sign of vector grades, it is easy to see that the required invariance is maintained

\begin{equation}\label{eqn:boostToParavector:320}
\begin{aligned}
X’ \overline{{X’}}
&=
L X L
\overline{{ L X L }} \\
&=
L X L
\overline{{ L }} \overline{{ X }} \overline{{ L }} \\
&=
L X \overline{{ X }} \overline{{ L }} \\
&=
X \overline{{ X }} L \overline{{ L }} \\
&=
X \overline{{ X }}.
\end{aligned}
\end{equation}

Let’s explicitly expand the transformation of \ref{eqn:boostToParavector:140}, so we can relate the rapidity angle \( \phi \) to the magnitude of the velocity. This is most easily done by splitting the spacelike component \( \Bx \) of the four vector into its projective and rejective components

\begin{equation}\label{eqn:boostToParavector:340}
\begin{aligned}
\Bx
&= \vcap \vcap \Bx \\
&= \vcap \lr{ \vcap \cdot \Bx + \vcap \wedge \Bx } \\
&= \vcap \lr{ \vcap \cdot \Bx } + \vcap \lr{ \vcap \wedge \Bx } \\
&= \Bx_\parallel + \Bx_\perp.
\end{aligned}
\end{equation}

The exponential

\begin{equation}\label{eqn:boostToParavector:360}
e^{-\vcap \phi/2}
=
\cosh\lr{ \phi/2 }
– \vcap \sinh\lr{ \phi/2 },
\end{equation}

commutes with any scalar grades and with \( \Bx_\parallel \), but anticommutes with \( \Bx_\perp \), so

\begin{equation}\label{eqn:boostToParavector:380}
\begin{aligned}
X’
&=
\lr{ c t + \Bx_\parallel } e^{ -\vcap \phi/2 } e^{ -\vcap \phi/2 }
+
\Bx_\perp e^{ \vcap \phi/2 } e^{ -\vcap \phi/2 } \\
&=
\lr{ c t + \Bx_\parallel } e^{ -\vcap \phi }
+
\Bx_\perp \\
&=
\lr{ c t + \vcap \lr{ \vcap \cdot \Bx } } \lr{ \cosh \phi – \vcap \sinh \phi }
+
\Bx_\perp \\
&=
\Bx_\perp
+
\lr{ c t \cosh\phi – \lr{ \vcap \cdot \Bx} \sinh \phi }
+
\vcap \lr{ \lr{ \vcap \cdot \Bx } \cosh\phi – c t \sinh \phi } \\
&=
\Bx_\perp
+
\cosh\phi \lr{ c t – \lr{ \vcap \cdot \Bx} \tanh \phi }
+
\vcap \cosh\phi \lr{ \vcap \cdot \Bx – c t \tanh \phi }.
\end{aligned}
\end{equation}

Employing the argument from [3],
we want \( \phi \) defined so that this has structure of a Galilean transformation in the limit where \( \phi \rightarrow 0 \). This means we equate

\begin{equation}\label{eqn:boostToParavector:400}
\tanh \phi = \frac{v}{c},
\end{equation}

so that for small \(\phi\)

\begin{equation}\label{eqn:boostToParavector:420}
\Bx’ = \Bx – \Bv t.
\end{equation}

We can solving for \( \sinh^2 \phi \) and \( \cosh^2 \phi \) in terms of \( v/c \) using

\begin{equation}\label{eqn:boostToParavector:440}
\tanh^2 \phi
= \frac{v^2}{c^2}
=
\frac{ \sinh^2 \phi }{1 + \sinh^2 \phi}
=
\frac{ \cosh^2 \phi – 1 }{\cosh^2 \phi}.
\end{equation}

which after picking the positive root required for Galilean equivalence gives
\begin{equation}\label{eqn:boostToParavector:460}
\begin{aligned}
\cosh \phi &= \frac{1}{\sqrt{1 – (\Bv/c)^2}} \equiv \gamma \\
\sinh \phi &= \frac{v/c}{\sqrt{1 – (\Bv/c)^2}} = \gamma v/c.
\end{aligned}
\end{equation}

The Lorentz boost, written out in full is

\begin{equation}\label{eqn:boostToParavector:480}
ct’ + \Bx’
=
\Bx_\perp
+
\gamma \lr{ c t – \frac{\Bv}{c} \cdot \Bx }
+
\gamma \lr{ \vcap \lr{ \vcap \cdot \Bx } – \Bv t }
.
\end{equation}

Authors like Chappelle, et al., that also use paravectors [4], specify the form of the Lorentz transformation for the electromagnetic field, but for that transformation reversion is used instead of involution.
I plan to explore that in a later post, starting from the STA formalism that I already understand, and see if I can make sense
of the underlying rationale.

References

[1] William Baylis. Electrodynamics: a modern geometric approach, volume 17. Springer Science \& Business Media, 2004.

[2] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.

[3] L. Landau and E. Lifshitz. The Classical theory of fields. Addison-Wesley, 1951.

[4] James M Chappell, Samuel P Drake, Cameron L Seidel, Lachlan J Gunn, and Derek Abbott. Geometric algebra for electrical and electronic engineers. Proceedings of the IEEE, 102 0(9), 2014.

2D SHO xy perturbation

December 7, 2015 phy1520 , , , , ,

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

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.

Third update of aggregate notes for phy1520, Graduate Quantum Mechanics.

November 9, 2015 phy1520 , , , , , , , , , , , , , , , ,

I’ve posted a third update of my aggregate notes for PHY1520H Graduate Quantum Mechanics, taught by Prof. Arun Paramekanti. In addition to what was noted previously, this contains lecture notes up to lecture 13, my solutions for the third problem set, and some additional worked practice problems.

Most of the content was posted individually in the following locations, but those original documents will not be maintained individually any further.

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

November 5, 2015 phy1520 , , , , , , ,

[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 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 9: Dirac equation (cont.). Taught by Prof. Arun Paramekanti

October 15, 2015 phy1520 , , , , , , , ,

[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.

Where we left off

\begin{equation}\label{eqn:qmLecture9:20}
-i \Hbar \PD{t}{}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar c \PD{x}{} & m c^2 \\
m c^2 & i \Hbar c \PD{x}{} \\
\end{bmatrix}.
\end{equation}

With a potential this would be

\begin{equation}\label{eqn:qmLecture9:40}
-i \Hbar \PD{t}{}
\begin{bmatrix}
\psi_1 \\
\psi_2
\end{bmatrix}
=
\begin{bmatrix}
-i \Hbar c \PD{x}{} + V(x) & m c^2 \\
m c^2 & i \Hbar c \PD{x}{} + V(x) \\
\end{bmatrix}.
\end{equation}

This means that the potential is raising the energy eigenvalue of the system.

Free Particle

Assuming a form

\begin{equation}\label{eqn:qmLecture9:60}
\begin{bmatrix}
\psi_1(x,t) \\
\psi_2(x,t)
\end{bmatrix}
=
e^{i k x}
\begin{bmatrix}
f_1(t) \\
f_2(t) \\
\end{bmatrix},
\end{equation}

and plugging back into the Dirac equation we have

\begin{equation}\label{eqn:qmLecture9:80}
-i \Hbar \PD{t}{}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}
=
\begin{bmatrix}
k \Hbar c & m c^2 \\
m c^2 & – \Hbar k c \\
\end{bmatrix}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}.
\end{equation}

We can use a diagonalizing rotation

\begin{equation}\label{eqn:qmLecture9:100}
\begin{bmatrix}
f_1 \\
f_2
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta_k & -\sin\theta_k \\
\sin\theta_k & \cos\theta_k \\
\end{bmatrix}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}.
\end{equation}

Plugging this in reduces the system to the form

\begin{equation}\label{eqn:qmLecture9:140}
-i \Hbar \PD{t}{}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}
=
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
\begin{bmatrix}
f_{+} \\
f_{-} \\
\end{bmatrix}.
\end{equation}

Where the rotation angle is found to be given by

\begin{equation}\label{eqn:qmLecture9:160}
\begin{aligned}
\sin(2 \theta_k) &= \frac{m c^2}{\sqrt{(\Hbar k c)^2 + m^2 c^4}} \\
\cos(2 \theta_k) &= \frac{\Hbar k c}{\sqrt{(\Hbar k c)^2 + m^2 c^4}} \\
E_k &= \sqrt{(\Hbar k c)^2 + m^2 c^4}
\end{aligned}
\end{equation}

See fig. 1 for a sketch of energy vs momentum. The asymptotes are the limiting cases when \( m c^2 \rightarrow 0 \). The \( + \) branch is what we usually associate with particles. What about the other energy states. For Fermions Dirac argued that the lower energy states could be thought of as “filled up”, using the Pauli principle to leave only the positive energy states available. This was called the “Dirac Sea”. This isn’t a good solution, and won’t work for example for Bosons.

fig. 1. Dirac equation solution space

fig. 1. Dirac equation solution space

Another way to rationalize this is to employ ideas from solid state theory. For example consider a semiconductor with a valence and conduction band as sketched in fig. 2.

fig. 2. Solid state valence and conduction band transition

fig. 2. Solid state valence and conduction band transition

A photon can excite an electron from the valence band to the conduction band, leaving all the valence band states filled except for one (a hole). For an electron we can use almost the same picture, as sketched in fig. 3.

fig. 3. Pair creation

fig. 3. Pair creation

A photon with energy \( E_k – (-E_k) \) can create a positron-electron pair from the vacuum, where the energy of the electron and positron pair is \( E_k \).

At high enough energies, we can see this pair creation occur.

Zitterbewegung

If a particle is created at a non-eigenstate such as one on the asymptotes, then oscillations between the positive and negative branches are possible as sketched in fig. 4.

fig. 4. Zitterbewegung oscillation

fig. 4. Zitterbewegung oscillation

Only “vertical” oscillations between the positive and negative locations on these branches is possible since those are the points that match the particle momentum. Examining this will be the aim of one of the problem set problems.

Probability and current density

If we define a probability density

\begin{equation}\label{eqn:qmLecture9:180}
\rho(x, t) = \Abs{\psi_1}^2 + \Abs{\psi_2}^2,
\end{equation}

does this satisfy a probability conservation relation

\begin{equation}\label{eqn:qmLecture9:200}
\PD{t}{\rho} + \PD{x}{j} = 0,
\end{equation}

where \( j \) is the probability current. Plugging in the density, we have

\begin{equation}\label{eqn:qmLecture9:220}
\PD{t}{\rho}
=
\PD{t}{\psi_1^\conj} \psi_1
+
\psi_1^\conj \PD{t}{\psi_1}
+
\PD{t}{\psi_2^\conj} \psi_2
+
\psi_2^\conj \PD{t}{\psi_2}.
\end{equation}

It turns out that the probability current has the form

\begin{equation}\label{eqn:qmLecture9:240}
j(x,t) = c \lr{ \psi_1^\conj \psi_1 + \psi_2^\conj \psi_2 }.
\end{equation}

Here the speed of light \( c \) is the slope of the line in the plots above. We can think of this current density as right movers minus the left movers. Any state that is given can be thought of as a combination of right moving and left moving states, neither of which are eigenstates of the free particle Hamiltonian.

Potential step

The next logical thing to think about, as in non-relativistic quantum mechanics, is to think about what occurs when the particle hits a potential step, as in fig. 5.

fig. 5. Reflection off a potential barrier

fig. 5. Reflection off a potential barrier

The approach is the same. We write down the wave functions for the \( V = 0 \) region (I), and the higher potential region (II).

The eigenstates are found on the solid lines above the asymptotes on the right hand movers side as sketched in fig. 6. The right and left moving designations are based on the phase velocity \( \PDi{k}{E} \) (approaching \( \pm c \) on the top-right and top-left quadrants respectively).

fig. 6. Right movers and left movers

fig. 6. Right movers and left movers

For \( k > 0 \), an eigenstate for the incident wave is

\begin{equation}\label{eqn:qmLecture9:261}
\Bpsi_{\textrm{inc}}(x) =
\begin{bmatrix}
\cos\theta_k \\
\sin\theta_k
\end{bmatrix}
e^{i k x},
\end{equation}

For the reflected wave function, we pick a function on the left moving side of the positive energy branch.

\begin{equation}\label{eqn:qmLecture9:260}
\Bpsi_{\textrm{ref}}(x) =
\begin{bmatrix}
? \\
?
\end{bmatrix}
e^{-i k x},
\end{equation}

We’ll go through this in more detail next time.

Question: Calculate the right going diagonalization

Prove (7).

Answer

To determine the relations for \( \theta_k \) we have to solve

\begin{equation}\label{eqn:qmLecture9:280}
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
= R^{-1} H R.
\end{equation}

Working with \( \Hbar = c = 1 \) temporarily, and \( C = \cos\theta_k, S = \sin\theta_k \), that is

\begin{equation}\label{eqn:qmLecture9:300}
\begin{aligned}
\begin{bmatrix}
E_k & 0 \\
0 & -E_k
\end{bmatrix}
&=
\begin{bmatrix}
C & S \\
-S & C
\end{bmatrix}
\begin{bmatrix}
k & m \\
m & -k
\end{bmatrix}
\begin{bmatrix}
C & -S \\
S & C
\end{bmatrix} \\
&=
\begin{bmatrix}
C & S \\
-S & C
\end{bmatrix}
\begin{bmatrix}
k C + m S & -k S + m C \\
m C – k S & -m S – k C
\end{bmatrix} \\
&=
\begin{bmatrix}
k C^2 + m S C + m C S – k S^2 & -k S C + m C^2 -m S^2 – k C S \\
-k C S – m S^2 + m C^2 – k S C & k S^2 – m C S -m S C – k C^2
\end{bmatrix} \\
&=
\begin{bmatrix}
k \cos(2 \theta_k) + m \sin(2 \theta_k) & m \cos(2 \theta_k) – k \sin(2 \theta_k) \\
m \cos(2 \theta_k) – k \sin(2 \theta_k) & -k \cos(2 \theta_k) – m \sin(2 \theta_k) \\
\end{bmatrix}.
\end{aligned}
\end{equation}

This gives

\begin{equation}\label{eqn:qmLecture9:320}
\begin{aligned}
E_k
\begin{bmatrix}
1 \\
0
\end{bmatrix}
&=
\begin{bmatrix}
k \cos(2 \theta_k) + m \sin(2 \theta_k) \\
m \cos(2 \theta_k) – k \sin(2 \theta_k) \\
\end{bmatrix} \\
&=
\begin{bmatrix}
k & m \\
m & -k
\end{bmatrix}
\begin{bmatrix}
\cos(2 \theta_k) \\
\sin(2 \theta_k) \\
\end{bmatrix}.
\end{aligned}
\end{equation}

Adding back in the \(\Hbar\)’s and \(c\)’s this is

\begin{equation}\label{eqn:qmLecture9:340}
\begin{aligned}
\begin{bmatrix}
\cos(2 \theta_k) \\
\sin(2 \theta_k) \\
\end{bmatrix}
&=
\frac{E_k}{-(\Hbar k c)^2 -(m c^2)^2}
\begin{bmatrix}
– \Hbar k c & – m c^2 \\
– m c^2 & \Hbar k c
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\inv{E_k}
\begin{bmatrix}
\Hbar k c \\
m c^2
\end{bmatrix}.
\end{aligned}
\end{equation}

Question: Verify the Dirac current relationship.

Prove \ref{eqn:qmLecture9:240}.

Answer

The components of the Schrodinger equation are

\begin{equation}\label{eqn:qmLecture9:360}
\begin{aligned}
-i \Hbar \PD{t}{\psi_1} &= -i \Hbar c \PD{x}{\psi_1} + m c^2 \psi_2 \\
-i \Hbar \PD{t}{\psi_2} &= m c^2 \psi_1 + i \Hbar c \PD{x}{\psi_2},
\end{aligned}
\end{equation}

The conjugates of these are
\begin{equation}\label{eqn:qmLecture9:380}
\begin{aligned}
i \Hbar \PD{t}{\psi_1^\conj} &= i \Hbar c \PD{x}{\psi_1^\conj} + m c^2 \psi_2^\conj \\
i \Hbar \PD{t}{\psi_2^\conj} &= m c^2 \psi_1^\conj – i \Hbar c \PD{x}{\psi_2^\conj}.
\end{aligned}
\end{equation}

This gives
\begin{equation}\label{eqn:qmLecture9:400}
\begin{aligned}
i \Hbar \PD{t}{\rho}
&=
\lr{ i \Hbar c \PD{x}{\psi_1^\conj} + m c^2 \psi_2^\conj } \psi_1 \\
&+ \psi_1^\conj \lr{ i \Hbar c \PD{x}{\psi_1} – m c^2 \psi_2 } \\
&+ \lr{ m c^2 \psi_1^\conj – i \Hbar c \PD{x}{\psi_2^\conj} } \psi_2 \\
&+ \psi_2^\conj \lr{ -m c^2 \psi_1 – i \Hbar c \PD{x}{\psi_2} }.
\end{aligned}
\end{equation}

All the non-derivative terms cancel leaving

\begin{equation}\label{eqn:qmLecture9:420}
\inv{c} \PD{t}{\rho}
=
\PD{x}{\psi_1^\conj} \psi_1
+ \psi_1^\conj \PD{x}{\psi_1}
– \PD{x}{\psi_2^\conj} \psi_2
– \psi_2^\conj \PD{x}{\psi_2}
=
\PD{x}{}
\lr{
\psi_1^\conj \psi_1 –
\psi_2^\conj \psi_2
}.
\end{equation}