perturbation

PHY2403H Quantum Field Theory. Lecture 14: Time evolution, Hamiltonian perturbation, ground state. Taught by Prof. Erich Poppitz

October 29, 2018 phy2403 No comments , , , , , ,

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DISCLAIMER: Very rough notes from class, with some additional side notes.

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall 2018.

Review

Given a field \( \phi(t_0, \Bx) \), satisfying the commutation relations
\begin{equation}\label{eqn:qftLecture14:20}
\antisymmetric{\pi(t_0, \Bx)}{\phi(t_0, \By)} = -i \delta(\Bx – \By)
\end{equation}
we introduced an interaction picture field given by
\begin{equation}\label{eqn:qftLecture14:40}
\phi_I(t, x) = e^{i H_0(t- t_0)} \phi(t_0, \Bx) e^{-iH_0(t – t_0)}
\end{equation}
related to the Heisenberg picture representation by
\begin{equation}\label{eqn:qftLecture14:60}
\phi_H(t, x)
= e^{i H(t- t_0)} \phi(t_0, \Bx) e^{-iH(t – t_0)}
= U^\dagger(t, t_0) \phi_I(t, \Bx) U(t, t_0),
\end{equation}
where \( U(t, t_0) \) is the time evolution operator.
\begin{equation}\label{eqn:qftLecture14:80}
U(t, t_0) =
e^{i H_0(t – t_0)}
e^{-i H(t – t_0)}
\end{equation}
We argued that
\begin{equation}\label{eqn:qftLecture14:100}
i \PD{t}{} U(t, t_0) = H_{\text{I,int}}(t) U(t, t_0)
\end{equation}
We found the glorious expression
\begin{equation}\label{eqn:qftLecture14:120}
\boxed{
\begin{aligned}
U(t, t_0)
&= T \exp{\lr{ -i \int_{t_0}^t H_{\text{I,int}}(t’) dt’}} \\
&=
\sum_{n = 0}^\infty \frac{(-i)^n}{n!} \int_{t_0}^t dt_1 dt_2 \cdots dt_n T\lr{ H_{\text{I,int}}(t_1) H_{\text{I,int}}(t_2) \cdots H_{\text{I,int}}(t_n) }
\end{aligned}
}
\end{equation}

However, what we are really after is
\begin{equation}\label{eqn:qftLecture14:140}
\bra{\Omega} T(\phi(x_1) \cdots \phi(x_n)) \ket{\Omega}
\end{equation}
Such a product has many labels and names, and we’ll describe it as “vacuum expectation values of time-ordered products of arbitrary #’s of local Heisenberg operators”.

Perturbation

Following section 4.2, [1].

\begin{equation}\label{eqn:qftLecture14:160}
\begin{aligned}
H &= \text{exact Hamiltonian} = H_0 + H_{\text{int}}
\\
H_0 &= \text{free Hamiltonian.
}
\end{aligned}
\end{equation}
We know all about \( H_0 \) and assume that it has a lowest (ground state) \( \ket{0} \), the “vacuum” state of \( H_0 \).

\( H \) has eigenstates, in particular \( H \) is assumed to have a unique ground state \( \ket{\Omega} \) satisfying
\begin{equation}\label{eqn:qftLecture14:180}
H \ket{\Omega} = \ket{\Omega} E_0,
\end{equation}
and has states \( \ket{n} \), representing excited (non-vacuum states with energies > \( E_0 \)).
These states are assumed to be a complete basis
\begin{equation}\label{eqn:qftLecture14:200}
\mathbf{1} = \ket{\Omega}\bra{\Omega} + \sum_n \ket{n}\bra{n} + \int dn \ket{n}\bra{n}.
\end{equation}
The latter terms may be written with a superimposed sum-integral notation as
\begin{equation}\label{eqn:qftLecture14:440}
\sum_n + \int dn
=
{\int\kern-1em\sum}_n,
\end{equation}
so the identity operator takes the more compact form
\begin{equation}\label{eqn:qftLecture14:460}
\mathbf{1} = \ket{\Omega}\bra{\Omega} + {\int\kern-1em\sum}_n \ket{n}\bra{n}.
\end{equation}

For some time \( T \) we have
\begin{equation}\label{eqn:qftLecture14:220}
e^{-i H T} \ket{0} = e^{-i H T}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n \ket{n}\braket{n}{0}
}.
\end{equation}

We now wish to argue that the \( {\int\kern-1em\sum}_n \) term can be ignored.

Argument 1:

This is something of a fast one, but one can consider a formal transformation \( T \rightarrow T(1 – i \epsilon) \), where \( \epsilon \rightarrow 0^+ \), and consider very large \( T \). This gives
\begin{equation}\label{eqn:qftLecture14:240}
\begin{aligned}
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i H T(1 – i \epsilon)} \ket{0}
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i H T(1 – i \epsilon)}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n \ket{n}\braket{n}{0}
} \\
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i E_0 T – E_0 \epsilon T}
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n e^{-i E_n T – \epsilon E_n T} \ket{n}\braket{n}{0} \\
&=
\lim_{T \rightarrow \infty, \epsilon \rightarrow 0^+}
e^{-i E_0 T – E_0 \epsilon T}
\lr{
\ket{\Omega}\braket{\Omega}{0} + {\int\kern-1em\sum}_n e^{-i (E_n -E_0) T – \epsilon T (E_n – E_0)} \ket{n}\braket{n}{0}
}
\end{aligned}
\end{equation}
The limits are evaluated by first taking \( T \) to infinity, then only after that take \( \epsilon \rightarrow 0^+ \). Doing this, the sum is dominated by the ground state contribution, since each excited state also has a \( e^{-\epsilon T(E_n – E_0)} \) suppression factor (in addition to the leading suppression factor).

Argument 2:

With the hand waving required for the argument above, it’s worth pointing other (less formal) ways to arrive at the same result. We can write
\begin{equation}\label{eqn:qftLecture14:260}
sectionumInt \ket{n}\bra{n} \rightarrow
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \ket{\Bp, k}\bra{\Bp, k}
\end{equation}
where \( k \) is some unknown quantity that we are summing over.
If we have
\begin{equation}\label{eqn:qftLecture14:280}
H \ket{\Bp, k} = E_{\Bp, k} \ket{\Bp, k},
\end{equation}
then
\begin{equation}\label{eqn:qftLecture14:300}
e^{-i H T} sectionumInt \ket{n}\bra{n}
=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \ket{\Bp, k} e^{-i E_{\Bp, k}} \bra{\Bp, k}.
\end{equation}
If we take matrix elements
\begin{equation}\label{eqn:qftLecture14:320}
\begin{aligned}
\bra{A}
e^{-i H T} sectionumInt \ket{n}\bra{n} \ket{B}
&=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} \braket{A}{\Bp, k} e^{-i E_{\Bp, k}} \braket{\Bp, k}{B} \\
&=
\sum_k \int \frac{d^3 p}{(2 \pi)^3} e^{-i E_{\Bp, k}} f(\Bp).
\end{aligned}
\end{equation}
If we assume that \( f(\Bp) \) is a well behaved smooth function, we have “infinite” frequency oscillation within the envelope provided by the amplitude of that function, as depicted in fig. 1.
The Riemann-Lebesgue lemma [2] describes such integrals, the result of which is that such an integral goes to zero. This is a different sort of hand waving argument, but either way, we can argue that only the ground state contributes to the sum \ref{eqn:qftLecture14:220} above.

fig. 1. High frequency oscillations within envelope of well behaved function.

 

Ground state of the perturbed Hamiltonian.

With the excited states ignored, we are left with
\begin{equation}\label{eqn:qftLecture14:340}
e^{-i H T} \ket{0} = e^{-i E_0 T} \ket{\Omega}\braket{\Omega}{0}
\end{equation}
in the \( T \rightarrow \infty(1 – i \epsilon) \) limit. We can now write the ground state as

\begin{equation}\label{eqn:qftLecture14:360}
\begin{aligned}
\ket{\Omega}
&=
\evalbar{
\frac{ e^{i E_0 T – i H T } \ket{0} }{
\braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) } \\
&=
\evalbar{
\frac{ e^{- i H T } \ket{0} }{
e^{-i E_0 T} \braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) }.
\end{aligned}
\end{equation}
Shifting the very large \( T \rightarrow T + t_0 \) shouldn’t change things, so
\begin{equation}\label{eqn:qftLecture14:480}
\ket{\Omega}
=
\evalbar{
\frac{ e^{- i H (T + t_0) } \ket{0} }{
e^{-i E_0 (T + t_0) } \braket{\Omega}{0}
}
}{ T \rightarrow \infty(1 – i \epsilon) }.
\end{equation}

A bit of manipulation shows that the operator in the numerator has the structure of a time evolution operator.

Claim: (DIY):

\Cref{eqn:qftLecture14:80}, \ref{eqn:qftLecture14:120} may be generalized to
\begin{equation}\label{eqn:qftLecture14:400}
U(t, t’) = e^{i H_0(t – t_0)} e^{-i H(t – t’)} e^{-i H_0(t’ – t_0)} =
T \exp{\lr{ -i \int_{t’}^t H_{\text{I,int}}(t”) dt”}}.
\end{equation}
Observe that we recover \ref{eqn:qftLecture14:120} when \( t’ = t_0 \).  Using \ref{eqn:qftLecture14:400} we find
\begin{equation}\label{eqn:qftLecture14:520}
\begin{aligned}
U(t_0, -T) \ket{0}
&= e^{i H_0(t_0 – t_0)} e^{-i H(t_0 + T)} e^{-i H_0(-T – t_0)} \ket{0} \\
&= e^{-i H(t_0 + T)} e^{-i H_0(-T – t_0)} \ket{0} \\
&= e^{-i H(t_0 + T)} \ket{0},
\end{aligned}
\end{equation}
where we use the fact that \( e^{i H_0 \tau} \ket{0} = \lr{ 1 + i H_0 \tau + \cdots } \ket{0} = 1 \ket{0}, \) since \( H_0 \ket{0} = 0 \).

We are left with
\begin{equation}\label{eqn:qftLecture14:420}
\boxed{
\ket{\Omega}
= \frac{U(t_0, -T) \ket{0} }{e^{-i E_0(t_0 – (-T))} \braket{\Omega}{0}}.
}
\end{equation}

We are close to where we want to be. Wednesday we finish off, and then start scattering and Feynman diagrams.

References

[1] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

[2] Wikipedia contributors. Riemann-lebesgue lemma — Wikipedia, the free encyclopedia, 2018. URL https://en.wikipedia.org/w/index.php?title=Riemann%E2%80%93Lebesgue_lemma&oldid=856778941. [Online; accessed 29-October-2018].

2D SHO xy perturbation

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

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Q: [1] pr. 5.4

Given a 2D SHO with Hamiltonian

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

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

  • (b)
    With perturbation

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

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

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

A: part (a)

Recall that we have

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

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

A: part (b)

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

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

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

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

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

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

This diagonalizes with

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

The unperturbed Hamiltonian in the original basis is

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

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

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

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

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

The new energies are

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

A: part (c)

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

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

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

For the commutators of the new operators we have

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

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

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

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

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

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

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

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

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

The perturbed Hamiltonian is

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

In this coordinate system, the corresponding eigensystem is

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

For small \( \delta \)

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

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

The lowest order perturbed energy levels are

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

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

References

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

Simplest perturbation two by two Hamiltonian

December 7, 2015 phy1520 No comments , , ,

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Q: two state Hamiltonian.

Given a two-state system

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:20}
H = H_0 + \lambda V
=
\begin{bmatrix}
E_1 & \lambda \Delta \\
\lambda \Delta & E_2
\end{bmatrix}
\end{equation}

  • (a) Solve the system exactly.
  • (b) Find the first order perturbed states and second order energy shifts, and compare to the exact solution.
  • (c) Solve the degenerate case for \( E_1 = E_2 \), and compare to the exact solution.

A: part (a)

The energy eigenvalues \( \epsilon \) are given by

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:40}
0
=
\lr{ E_1 – \epsilon }
\lr{ E_2 – \epsilon }
– (\lambda \Delta)^2,
\end{equation}

or

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:60}
\epsilon^2 – \epsilon\lr{ E_1 + E_2 } + E_1 E_2 = (\lambda \Delta)^2.
\end{equation}

After rearranging this is
\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:80}
\epsilon = \frac{ E_1 + E_2 }{2} \pm \sqrt{ \lr{ \frac{ E_1 – E_2 }{2} }^2 + (\lambda \Delta)^2 }.
\end{equation}

Notice that for \( E_2 = E_1 \) we have

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:100}
\epsilon = E_1 \pm \lambda \Delta.
\end{equation}

Since a change of basis can always put the problem in a form so that \( E_1 > E_2 \), let’s assume that to make an approximation of the energy eigenvalues for \( \Abs{\lambda \Delta} \ll \ifrac{ (E_1 – E_2) }{2} \)

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:120}
\begin{aligned}
\epsilon
&=
\frac{ E_1 + E_2 }{2} \pm \frac{ E_1 – E_2 }{2} \sqrt{ 1 + \frac{(2 \lambda \Delta)^2}{(E_1 – E_2)^2} } \\
&\approx
\frac{ E_1 + E_2 }{2} \pm \frac{ E_1 – E_2 }{2} \lr{ 1 + 2 \frac{(\lambda
\Delta)^2}{(E_1 – E_2)^2} } \\
&=
\frac{ E_1 + E_2 }{2} \pm \frac{ E_1 – E_2 }{2}
\pm
\frac{(\lambda \Delta)^2}{E_1 – E_2} \\
&=
E_1 + \frac{(\lambda \Delta)^2}{E_1 – E_2}, E_2 + \frac{(\lambda \Delta)^2}{E_2 – E_1}.
\end{aligned}
\end{equation}

For the perturbed states, starting with the plus case, if

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:140}
\ket{+} \propto
\begin{bmatrix}
a \\
b
\end{bmatrix},
\end{equation}

we must have
\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:160}
\begin{aligned}
0
&=
\biglr{ E_1 – \lr{ E_1 + \frac{(\lambda \Delta)^2}{E_1 – E_2} } } a + \lambda
\Delta b \\
&=
\biglr{ – \frac{(\lambda \Delta)^2}{E_1 – E_2} } a + \lambda \Delta b,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:180}
\ket{+} \rightarrow
\begin{bmatrix}
1 \\
\frac{\lambda \Delta}{E_1 – E_2}
\end{bmatrix}
= \ket{+} + \frac{\lambda \Delta}{E_1 – E_2} \ket{-}.
\end{equation}

Similarly for the minus case we must have

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:200}
\begin{aligned}
0
&=
\lambda \Delta a + \biglr{ E_2 – \lr{ E_2 + \frac{(\lambda \Delta)^2}{E_2 – E_1} } } b \\
&=
\lambda \Delta b + \biglr{ – \frac{(\lambda \Delta)^2}{E_2 – E_1} } b,
\end{aligned}
\end{equation}

for
\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:220}
\ket{-} \rightarrow
\ket{-} + \frac{\lambda \Delta}{E_2 – E_1} \ket{+}.
\end{equation}

A: part (b)

For the perturbation the first energy shift for perturbation of the \( \ket{+} \) state is

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:240}
\begin{aligned}
E_{+}^{(1)}
&= \ket{+} V \ket{+} \\
&=
\lambda \Delta
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 & 1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\lambda \Delta
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 \\
1
\end{bmatrix} \\
&=
0.
\end{aligned}
\end{equation}

The first order energy shift for the perturbation of the \( \ket{-} \) state is also zero. The perturbed \( \ket{+} \) state is

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:260}
\begin{aligned}
\ket{+}^{(1)}
&= \frac{\overline{{P}}_{+}}{E_1 – H_0} V \ket{+} \\
&= \frac{\ket{-}\bra{-}}{E_1 – E_2} V \ket{+}
\end{aligned}
\end{equation}

The numerator matrix element is

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:280}
\begin{aligned}
\bra{-} V \ket{+}
&=
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
0 & \Delta \\
\Delta & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
0
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
0 \\
\Delta
\end{bmatrix} \\
&=
\Delta,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:300}
\ket{+} \rightarrow \ket{+} + \ket{-} \frac{\Delta}{E_1 – E_2}.
\end{equation}

Observe that this matches the first order series expansion of the exact value above.

For the perturbation of \( \ket{-} \) we need the matrix element

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:320}
\begin{aligned}
\bra{+} V \ket{-}
&=
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 & \Delta \\
\Delta & 0
\end{bmatrix}
\begin{bmatrix}
0 \\
1
\end{bmatrix} \\
&=
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
\Delta \\
0 \\
\end{bmatrix} \\
&=
\Delta,
\end{aligned}
\end{equation}

so it’s clear that the perturbed ket is

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:340}
\ket{-} \rightarrow \ket{-} + \ket{+} \frac{\Delta}{E_2 – E_1},
\end{equation}

also matching the approximation found from the exact computation. The second order energy shifts can now be calculated

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:360}
\begin{aligned}
\bra{+} V \ket{+}’
&=
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
0 & \Delta \\
\Delta & 0
\end{bmatrix}
\begin{bmatrix}
1 \\
\frac{\Delta}{E_1 – E_2}
\end{bmatrix} \\
&=
\begin{bmatrix}
1 & 0
\end{bmatrix}
\begin{bmatrix}
\frac{\Delta^2}{E_1 – E_2} \\
\Delta
\end{bmatrix} \\
&=
\frac{\Delta^2}{E_1 – E_2},
\end{aligned}
\end{equation}

and

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:380}
\begin{aligned}
\bra{-} V \ket{-}’
&=
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
0 & \Delta \\
\Delta & 0
\end{bmatrix}
\begin{bmatrix}
\frac{\Delta}{E_2 – E_1} \\
1 \\
\end{bmatrix} \\
&=
\begin{bmatrix}
0 & 1
\end{bmatrix}
\begin{bmatrix}
\Delta \\
\frac{\Delta^2}{E_2 – E_1} \\
\end{bmatrix} \\
&=
\frac{\Delta^2}{E_2 – E_1},
\end{aligned}
\end{equation}

The energy perturbations are therefore
\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:400}
\begin{aligned}
E_1 &\rightarrow E_1 + \frac{(\lambda \Delta)^2}{E_1 – E_2} \\
E_2 &\rightarrow E_2 + \frac{(\lambda \Delta)^2}{E_2 – E_1}.
\end{aligned}
\end{equation}

This is what we had by approximating the exact case.

A: part (c)

For the \( E_2 = E_1 \) case, we’ll have to diagonalize the perturbation potential. That is

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:420}
\begin{aligned}
V &= U \bigwedge U^\dagger \\
\bigwedge &=
\begin{bmatrix}
\Delta & 0 \\
0 & -\Delta
\end{bmatrix} \\
U &= U^\dagger = \inv{\sqrt{2}}
\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}.
\end{aligned}
\end{equation}

A change of basis for the Hamiltonian is

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:440}
\begin{aligned}
H’
&=
U^\dagger H U \\
&=
U^\dagger H_0 U + \lambda U^\dagger V U \\
&=
E_1 U^\dagger + \lambda U^\dagger V U \\
&=
H_0 + \lambda \bigwedge.
\end{aligned}
\end{equation}

We can now calculate the perturbation energy with respect to the new basis, say \( \setlr{ \ket{1}, \ket{2} } \). Those energy shifts are

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:460}
\begin{aligned}
\Delta^{(1)} &= \bra{1} V \ket{1} = \Delta \\
\Delta^{(2)} &= \bra{2} V \ket{2} = -\Delta.
\end{aligned}
\end{equation}

The perturbed energies are therefore

\begin{equation}\label{eqn:simplestTwoByTwoPerturbation:480}
\begin{aligned}
E_1 &\rightarrow E_1 + \lambda \Delta \\
E_2 &\rightarrow E_2 – \lambda \Delta,
\end{aligned}
\end{equation}

which matches \ref{eqn:simplestTwoByTwoPerturbation:100}, the exact result.

References

Harmonic oscillator with energy shift

December 5, 2015 phy1520 No comments , ,

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

Q: [1] pr 5.1

Given a perturbed 1D SHO Hamiltonian

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

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

A:

The first order energy shift is seen to be zero

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

The first order perturbation to the ground state is

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

The second order ground state energy perturbation is

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

so the total energy perturbation up to second order is

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

To compare to the exact result, rewrite the Hamiltonian as

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

The Hamiltonian is subject to a constant energy shift

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

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

References

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

PHY1520H Graduate Quantum Mechanics. Lecture 21: Non-degenerate perturbation. Taught by Prof. Arun Paramekanti

December 4, 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 [2] chap. 5 content.

Non-degenerate perturbation theory. Recap.

\begin{equation}\label{eqn:qmLecture21:20}
\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:qmLecture21:40}
\Delta_{n} = \Delta_{n_0}
+ \lambda \Delta_{n_1}
+ \lambda^2 \Delta_{n_2}
+ \lambda^3 \Delta_{n_3} + \cdots
\end{equation}

\begin{equation}\label{eqn:qmLecture21:60}
\begin{aligned}
\Delta_{n_1} &= \bra{n^{(0)}} V \ket{n^{(0)}} \\
\ket{n_0} &= \ket{n^{(0)}}
\end{aligned}
\end{equation}

\begin{equation}\label{eqn:qmLecture21:80}
\begin{aligned}
\Delta_{n_2} &= \sum_{m \ne n} \frac{\Abs{\bra{n^{(0)}} V \ket{m^{(0)}}}^2}{E_n^{(0)} – E_m^{(0)}} \\
\ket{n_1} &= \sum_{m \ne n} \frac{ \ket{m^{(0)}} V_{mn} }{E_n^{(0)} – E_m^{(0)}}
\end{aligned}
\end{equation}

Example: Stark effect

\begin{equation}\label{eqn:qmLecture21:100}
H = H_{\textrm{atom}} + e \mathcal{E} z,
\end{equation}

where \( H_{\textrm{atom}} \) is assumed to be Hydrogen-like with Hamiltonian

\begin{equation}\label{eqn:qmLecture21:120}
H_{\textrm{atom}} = \frac{\BP^2}{2m} – \frac{e^2}{4 \pi \epsilon_0 r},
\end{equation}

and wave functions

\begin{equation}\label{eqn:qmLecture21:140}
\braket{\Br}{\psi_{n l m}} = R_{n l}(r) Y_{lm}( \theta, \phi )
\end{equation}

For the first level correction to the energy

\begin{equation}\label{eqn:qmLecture21:160}
\begin{aligned}
\Delta_1
&= \bra{\psi_{100}} e \mathcal{E} z \ket{ \psi_{100}} \\
&= e \mathcal{E} \int \frac{d\Omega}{4 \pi} \cos \theta \int dr r^2 R_{100}^2(r)
\end{aligned}
\end{equation}

The cosine integral is obliterated, so we have \( \Delta_1 = 0 \).

How about the second order energy correction? That is

\begin{equation}\label{eqn:qmLecture21:180}
\Delta_2 = \sum_{n l m \ne 100} \frac{
\Abs{ \bra{\psi_{100}} e \mathcal{E} z \ket{ n l m }}^2
}{
E_{100}^{(0)} – E_{n l m}
}
\end{equation}

The matrix element in the numerator is the absolute square of

\begin{equation}\label{eqn:qmLecture21:200}
V_{100,nlm}
=
e \mathcal{E} \int d\Omega \inv{\sqrt{ 4 \pi } }
\cos\theta Y_{l m}(\theta, \phi)
\int dr r^3 R_{100}(r) R_{n l}(r).
\end{equation}

For all \( m \ne 0 \), \( Y_{lm} \) includes a \( e^{i m \phi} \) factor, so this cosine integral is zero. For \( m = 0 \), each of the \( Y_{lm} \) functions appears to contain either even or odd powers of cosines. For example:

\begin{equation}\label{eqn:qmLecture21:760}
\begin{aligned}
Y_{00} &= \frac{1}{2 \sqrt{\pi}} \\
Y_{10} &= \frac{1}{2} \sqrt{\frac{3}{\pi }} \cos(t) \\
Y_{20} &= \frac{1}{4} \sqrt{\frac{5}{\pi }} \lr{(3 \cos^2(t)-1} \\
Y_{30} &= \frac{1}{4} \sqrt{\frac{7}{\pi }} \lr{(5 \cos^3(t)-3 \cos(t)} \\
Y_{40} &= \frac{3 \lr{(35 \cos^4(t)-30 \cos^2(t)+3}}{16 \sqrt{\pi }} \\
Y_{50} &= \frac{1}{16} \sqrt{\frac{11}{\pi }} \lr{(63 \cos^5(t)-70 \cos^3(t)+15 \cos(t)} \\
Y_{60} &= \frac{1}{32} \sqrt{\frac{13}{\pi }} \lr{(231 \cos^6(t)-315 \cos^4(t)+105 \cos^2(t)-5} \\
Y_{70} &= \frac{1}{32} \sqrt{\frac{15}{\pi }} \lr{(429 \cos^7(t)-693 \cos^5(t)+315 \cos^3(t)-35 \cos(t)} \\
Y_{80} &= \frac{1}{256} \sqrt{\frac{17}{\pi }} \lr{(6435 \cos^8(t)-12012 \cos^6(t)+6930 \cos^4(t)-1260 \cos^2(t)+35 } \\
\end{aligned}
\end{equation}

This shows that for even \( 2k = l \), the cosine integral is zero

\begin{equation}\label{eqn:qmLecture21:780}
\int_0^\pi \sin\theta \cos\theta \sum_k a_k \cos^{2k}\theta d\theta
=
0,
\end{equation}

since \( \cos^{2k}(\theta) \) is even and \( \sin\theta \cos\theta \) is odd over the same interval. We find zero for \( \int_0^\pi \sin\theta \cos\theta Y_{30}(\theta, \phi) d\theta \), and Mathematica appears to show that the rest of these integrals for \( l > 1 \) are also zero.

FIXME: find the property of the spherical harmonics that can be used to prove that this is true in general for \( l > 1 \).

This leaves

\begin{equation}\label{eqn:qmLecture21:220}
\begin{aligned}
\Delta_2
&= \sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} e \mathcal{E} z \ket{ n 1 0 }}^2
}{
E_{100}^{(0)} – E_{n 1 0}
} \\
&=
-e^2 \mathcal{E}^2
\sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} z \ket{ n 1 0 }}^2
}{
E_{n 1 0}
-E_{100}^{(0)}
}.
\end{aligned}
\end{equation}

This is sometimes written in terms of a polarizability \( \alpha \)

\begin{equation}\label{eqn:qmLecture21:260}
\Delta_2 = -\frac{\mathcal{E}^2}{2} \alpha,
\end{equation}

where

\begin{equation}\label{eqn:qmLecture21:280}
\alpha =
2 e^2
\sum_{n \ne 1} \frac{
\Abs{ \bra{\psi_{100}} z \ket{ n 1 0 }}^2
}{
E_{n 1 0}
-E_{100}^{(0)}
}.
\end{equation}

With
\begin{equation}\label{eqn:qmLecture21:840}
\BP = \alpha \boldsymbol{\mathcal{E}},
\end{equation}

the energy change upon turning on the electric field from \( 0 \rightarrow \mathcal{E} \) is simply \( – \BP \cdot d\boldsymbol{\mathcal{E}} \) integrated from \( 0 \rightarrow \mathcal{E} \). Putting \( \BP = \alpha \mathcal{E} \zcap \), we have

\begin{equation}\label{eqn:qmLecture21:400}
\begin{aligned}
– \int_0^\mathcal{E} p_z d\mathcal{E}
&=
– \int_0^\mathcal{E} \alpha \mathcal{E} d\mathcal{E} \\
&=
– \inv{2} \alpha \mathcal{E}^2
\end{aligned}
\end{equation}

leading to an energy change \( – \alpha \mathcal{E}^2/2 \), so we can directly compute \( \expectation{\BP} \) or we can compute change in energy, and both contain information about the polarization factor \( \alpha \).

There is an exact answer to the sum \ref{eqn:qmLecture21:280}, but we aren’t going to try to get it here. Instead let’s look for bounds

\begin{equation}\label{eqn:qmLecture21:240}
\Delta_2^{\mathrm{min}} < \Delta_2 < \Delta_2^{\mathrm{max}} \end{equation} \begin{equation}\label{eqn:qmLecture21:320} \alpha^{\mathrm{min}} = 2 e^2 \frac{ \Abs{ \bra{\psi_{100}} z \ket{\psi_{210}} }^2 }{E_{210}^{(0)} - E_{100}^{(0)}} \end{equation} For the hydrogen atom we have \begin{equation}\label{eqn:qmLecture21:820} E_n = -\frac{ e^2}{ 2 n^2 a_0 }, \end{equation} allowing any difference of energy levels to be expressed as a fraction of the ground state energy, such as \begin{equation}\label{eqn:qmLecture21:340} E_{210}^{(0)} = \inv{4} E_{100}^{(0)} = \inv{4} \frac{ -\Hbar^2 }{ 2 m a_0^2 } \end{equation} So \begin{equation}\label{eqn:qmLecture21:360} E_{210}^{(0)} - E_{100}^{(0)} = \frac{3}{4} \frac{ \Hbar^2 }{ 2 m a_0^2 } \end{equation} In the numerator we have \begin{equation}\label{eqn:qmLecture21:380} \begin{aligned} \bra{\psi_{100}} z \ket{\psi_{210}} &= \int r^2 d\Omega \lr{ \inv{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0} } r \cos\theta \lr{ \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \cos\theta } \\ &= (2 \pi) \inv{\sqrt{\pi}} \inv{4 \sqrt{2 \pi} } a_0 \int_0^\pi d\theta \sin\theta \cos^2\theta \int_0^\infty \frac{dr}{a_0} \frac{r^4}{a_0^4} e^{-r/a_0 - r/2 a_0} \\ &= (2 \pi) \inv{\sqrt{\pi}} \inv{4 \sqrt{2 \pi} } a_0 \lr{ \evalrange{-\frac{u^3}{3}}{1}{-1} } \int_0^\infty s^4 ds e^{- 3 s/2 } \\ &= 2 \inv{4 \sqrt{2} } a_0 \lr{ \evalrange{-\frac{u^3}{3}}{1}{-1} } \int_0^\infty s^4 ds e^{- 3 s/2 } \\ &= \inv{2 \sqrt{2}} \frac{2}{3} a_0 \frac{256}{81} \\ &= \frac{1}{3 \sqrt{2} } \frac{ 256}{81} a_0 \approx 0.75 a_0. \end{aligned} \end{equation} This gives \begin{equation}\label{eqn:qmLecture21:420} \begin{aligned} \alpha^{\mathrm{min}} &= \frac{ 2 e^2 (0.75)^2 a_0^2 }{ \frac{3}{4} \frac{\Hbar^2}{2 m a_0^2} } \\ &= \frac{6}{4} \frac{2 m e^2 a_0^4}{ \Hbar^2 } \\ &= 3 \frac{m e^2 a_0^4}{ \Hbar^2 } \\ &= 3 \frac{ 4 \pi \epsilon_0 }{a_0} a_0^4 \\ &\approx 4 \pi \epsilon_0 a_0^3 \times 3. \end{aligned} \end{equation} The factor \( 4 \pi \epsilon_0 a_0^3 \) are the natural units for the polarizability. There is a neat trick that generalizes to many problems to find the upper bound. Recall that the general polarizability was \begin{equation}\label{eqn:qmLecture21:440} \alpha = 2 e^2 \sum_{nlm \ne 100} \frac{ \Abs{ \bra{100} z \ket{ n l m }}^2 }{ E_{n l m} -E_{100}^{(0)} }. \end{equation} If we are looking for the upper bound, and replace the denominator by the smallest energy difference that will be encountered, it can be brought out of the sum, for \begin{equation}\label{eqn:qmLecture21:460} \alpha^{\mathrm{max}} = 2 e^2 \inv{E_{2 1 0} -E_{100}^{(0)} } \sum_{nlm \ne 100} \bra{100} z \ket{ n l m } \bra{nlm} z \ket{ 100 } \end{equation} Because \( \bra{nlm} z \ket{100} = 0 \), the constraint in the sum can be removed, and the identity summation evaluated \begin{equation}\label{eqn:qmLecture21:480} \begin{aligned} \alpha^{\mathrm{max}} &= 2 e^2 \inv{E_{2 1 0} -E_{100}^{(0)} } \sum_{nlm} \bra{100} z \ket{ n l m } \bra{nlm} z \ket{ 100 } \\ &= \frac{2 e^2 }{ \frac{3}{4} \frac{\Hbar^2}{ 2 m a_0^2} } \bra{100} z^2 \ket{ 100 } \\ &= \frac{16 e^2 m a_0^2 }{ 3 \Hbar^2 } \times a_0^2 \\ &= 4 \pi \epsilon_0 a_0^3 \times \frac{16}{3}. \end{aligned} \end{equation} The bounds are \begin{equation}\label{eqn:qmLecture21:520} \boxed{ 3 \ge \frac{\alpha}{\alpha^{\mathrm{at}}} < \frac{16}{3}, } \end{equation} where \begin{equation}\label{eqn:qmLecture21:560} \alpha^{\mathrm{at}} = 4 \pi \epsilon_0 a_0^3. \end{equation} The actual value is \begin{equation}\label{eqn:qmLecture21:580} \frac{\alpha}{\alpha^{\mathrm{at}}} = \frac{9}{2}. \end{equation}

Example: Computing the dipole moment

\begin{equation}\label{eqn:qmLecture21:600}
\expectation{P_z}
= \alpha \mathcal{E}
= \bra{\psi_{100}} e z \ket{\psi_{100}}.
\end{equation}

Without any perturbation this is zero. After perturbation, retaining only the terms that are first order in \( \delta \psi_{100} \) we have

\begin{equation}\label{eqn:qmLecture21:620}
\bra{\psi_{100} + \delta \psi_{100}} e z \ket{\psi_{100} + \delta \psi_{100}}
\approx
\bra{\psi_{100}} e z \ket{\delta \psi_{100}}
+
\bra{\delta \psi_{100}} e z \ket{\psi_{100}}.
\end{equation}

Next time: Van der Walls

We will look at two hyrdogenic atomic systems interacting where the pair of nuclei are supposed to be infinitely heavy and stationary. The wave functions each set of atoms are individually known, but we can consider the problem of the interactions of atom 1’s electrons with atom 2’s nucleus and atom 2’s electrons, and also the opposite interactions of atom 2’s electrons with atom 1’s nucleus and its electrons. This leads to a result that is linear in the electric field (unlike the above result, which is called the quadratic Stark effect).

Appendix. Hydrogen wavefunctions

From [3], with the \( a_0 \) factors added in.

\begin{equation}\label{eqn:qmLecture21:660}
\psi_{1 s} = \psi_{100} = \inv{\sqrt{\pi} a_0^{3/2}} e^{-r/a_0}
\end{equation}
\begin{equation}\label{eqn:qmLecture21:680}
\psi_{2 s} = \psi_{200} = \inv{4 \sqrt{2 \pi} a_0^{3/2}} \lr{ 2 – \frac{r}{a_0} } e^{-r/2a_0}
\end{equation}
\begin{equation}\label{eqn:qmLecture21:700}
\psi_{2 p_x} = \inv{\sqrt{2}} \lr{ \psi_{2,1,1} – \psi_{2,1,-1} }
= \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \sin\theta\cos\phi
\end{equation}
\begin{equation}\label{eqn:qmLecture21:720}
\psi_{2 p_y} = \frac{i}{\sqrt{2}} \lr{ \psi_{2,1,1} + \psi_{2,1,-1} }
= \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \sin\theta\sin\phi
\end{equation}
\begin{equation}\label{eqn:qmLecture21:740}
\psi_{2 p_z} = \psi_{210} = \inv{4 \sqrt{2 \pi} a_0^{3/2}} \frac{r}{a_0} e^{-r/2a_0} \cos\theta
\end{equation}

I looked to [1] to see where to add in the \( a_0 \) factors.

References

[1] Carl R. Nave. Hydrogen Wavefunctions, 2015. URL http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html. [Online; accessed 03-Dec-2015].

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

[3] Robert Field Troy Van Voorhis. Hydrogen Atom, 2013. URL http://ocw.mit.edu/courses/chemistry/5-61-physical-chemistry-fall-2013/lecture-notes/MIT5_61F13_Lecture19-20.pdf. [Online; accessed 03-Dec-2015].

PHY1520H Graduate Quantum Mechanics. Lecture 19: Variational method. Taught by Prof. Arun Paramekanti

November 27, 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{{5}} [1] content.

Variational method

Today we want to use the variational degree of freedom to try to solve some problems that we don’t have analytic solutions for.

Anharmonic oscillator

\begin{equation}\label{eqn:qmLecture19:20}
V(x) = \inv{2} m \omega^2 x^2 + \lambda x^4, \qquad \lambda \ge 0.
\end{equation}

With the potential growing faster than the harmonic oscillator, which had a ground state solution

\begin{equation}\label{eqn:qmLecture19:40}
\psi(x) = \inv{\pi^{1/4}} \inv{a_0^{1/2} } e^{- x^2/2 a_0^2},
\end{equation}

where
\begin{equation}\label{eqn:qmLecture19:60}
a_0 = \sqrt{\frac{\Hbar}{m \omega}}.
\end{equation}

Let’s try allowing \( a_0 \rightarrow a \), to be a variational degree of freedom

\begin{equation}\label{eqn:qmLecture19:80}
\psi_a(x) = \inv{\pi^{1/4}} \inv{a^{1/2} } e^{- x^2/2 a^2},
\end{equation}

\begin{equation}\label{eqn:qmLecture19:100}
\bra{\psi_a} H \ket{\psi_a}
=
\bra{\psi_a} \frac{p^2}{2m} + \inv{2} m \omega^2 x^2 + \lambda x^4 \ket{\psi_a}
\end{equation}

We can find
\begin{equation}\label{eqn:qmLecture19:120}
\expectation{x^2} = \inv{2} a^2
\end{equation}
\begin{equation}\label{eqn:qmLecture19:140}
\expectation{x^4} = \frac{3}{4} a^4
\end{equation}

Define

\begin{equation}\label{eqn:qmLecture19:160}
\tilde{\omega} = \frac{\Hbar}{m a^2},
\end{equation}

so that

\begin{equation}\label{eqn:qmLecture19:180}
\overline{{E}}_a
=
\bra{\psi_a} \lr{ \frac{p^2}{2m} + \inv{2} m \tilde{\omega}^2 x^2 }
+ \lr{
\inv{2} m \lr{ \omega^2 – \tilde{\omega}^2 } x^2
+
\lambda x^4 }
\ket{\psi_a}
=
\inv{2} \Hbar \tilde{\omega} + \inv{2} m \lr{ \omega^2 – \tilde{\omega}^2 } \inv{2} a^2 + \frac{3}{4} \lambda a^4.
\end{equation}

Write this as
\begin{equation}\label{eqn:qmLecture19:200}
\overline{{E}}_{\tilde{\omega}}
=
\inv{2} \Hbar \tilde{\omega} + \inv{4} \frac{\Hbar}{\tilde{\omega}} \lr{ \omega^2 – \tilde{\omega}^2 } + \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \tilde{\omega}^2 }.
\end{equation}

This might look something like fig. 1.

fig. 1: Energy after perturbation.

fig. 1: Energy after perturbation.

Demand that

\begin{equation}\label{eqn:qmLecture19:220}
0
= \PD{\tilde{\omega}}{ \overline{{E}}_{\tilde{\omega}}}
=
\frac{\Hbar}{2} – \frac{\Hbar}{4} \frac{\omega^2}{\tilde{\omega}^2}
– \frac{\Hbar}{4}
+ \frac{3}{4} (-2) \frac{\lambda \Hbar^2}{m^2 \tilde{\omega}^3}
=
\frac{\Hbar}{4}
\lr{
1 – \frac{\omega^2}{\tilde{\omega}^2}
– 6 \frac{\lambda \Hbar}{m^2 \tilde{\omega}^3}
}
\end{equation}

or
\begin{equation}\label{eqn:qmLecture19:260}
\tilde{\omega}^3 – \omega^2 \tilde{\omega} – \frac{6 \lambda \Hbar}{m^2} = 0.
\end{equation}

for \( \lambda a_0^4 \ll \Hbar \omega \), we have something like \( \tilde{\omega} = \omega + \epsilon \). Expanding \ref{eqn:qmLecture19:260} to first order in \( \epsilon \), this gives

\begin{equation}\label{eqn:qmLecture19:280}
\omega^3 + 3 \omega^2 \epsilon – \omega^2 \lr{ \omega + \epsilon } – \frac{6 \lambda \Hbar}{m^2} = 0,
\end{equation}

so that

\begin{equation}\label{eqn:qmLecture19:300}
2 \omega^2 \epsilon = \frac{6 \lambda \Hbar}{m^2},
\end{equation}

and

\begin{equation}\label{eqn:qmLecture19:320}
\Hbar \epsilon = \frac{ 3 \lambda \Hbar^2}{m^2 \omega^2 } = 3 \lambda a_0^4.
\end{equation}

Plugging into

\begin{equation}\label{eqn:qmLecture19:340}
\overline{{E}}_{\omega + \epsilon}
=
\inv{2} \Hbar \lr{ \omega + \epsilon }
+ \inv{4} \frac{\Hbar}{\omega} \lr{ -2 \omega \epsilon + \epsilon^2 } + \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \omega^2 }
\approx
\inv{2} \Hbar \lr{ \omega + \epsilon }
– \inv{2} \Hbar \epsilon
+ \frac{3}{4} \lambda \frac{\Hbar^2}{m^2 \omega^2 }
=
\inv{2} \Hbar \omega + \frac{3}{4} \lambda a_0^4.
\end{equation}

With \ref{eqn:qmLecture19:320}, that is

\begin{equation}\label{eqn:qmLecture19:540}
\overline{{E}}_{\tilde{\omega} = \omega + \epsilon} \approx \inv{2} \Hbar \lr{ \omega + \frac{\epsilon}{2} }.
\end{equation}

The energy levels are shifted slightly for each shift in the Hamiltonian frequency.

What do we have in the extreme anharmonic limit, where \( \lambda a_0^4 \gg \Hbar \omega \). Now we get

\begin{equation}\label{eqn:qmLecture19:360}
\tilde{\omega}^\conj = \lr{ \frac{ 6 \Hbar \lambda }{m^2} }^{1/3},
\end{equation}

and
\begin{equation}\label{eqn:qmLecture19:380}
\overline{{E}}_{\tilde{\omega}^\conj} = \frac{\Hbar^{4/3} \lambda^{1/3}}{m^{2/3}} \frac{3}{8} 6^{1/3}.
\end{equation}

(this last result is pulled from a web treatment somewhere of the anharmonic oscillator). Note that the first factor in this energy, with \( \Hbar^4 \lambda/m^2 \) traveling together could have been worked out on dimensional grounds.

This variational method tends to work quite well in these limits. For a system where \( m = \omega = \Hbar = 1 \), for this problem, we have

Capture

tab. 1: Comparing numeric and variational solutions.

Example: (sketch) double well potential

lecture19Fig2

fig. 2: Double well potential.

\begin{equation}\label{eqn:qmLecture19:400}
V(x) = \frac{m \omega^2}{8 a^2} \lr{ x – a }^2\lr{ x + a}^2.
\end{equation}

Note that this potential, and the Hamiltonian, both commute with parity.

We are interested in the regime where \( a_0^2 = \frac{\Hbar}{m \omega} \ll a^2 \).

Near \( x = \pm a \), this will be approximately

\begin{equation}\label{eqn:qmLecture19:420}
V(x) = \inv{2} m \omega^2 \lr{ x \pm a }^2.
\end{equation}

Guessing a wave function that is an eigenstate of parity

\begin{equation}\label{eqn:qmLecture19:440}
\Psi_{\pm} = g_{\pm} \lr{ \phi_{\textrm{R}}(x) \pm \phi_{\textrm{L}}(x) }.
\end{equation}

perhaps looking like the even and odd functions sketched in fig. 3, and fig. 4.

fig. 3. Even double well function

fig. 3. Even double well function

fig. 4. Odd double well function

fig. 4. Odd double well function

Using harmonic oscillator functions

\begin{equation}\label{eqn:qmLecture19:460}
\begin{aligned}
\phi_{\textrm{L}}(x) &= \Psi_{{\textrm{H}}.{\textrm{O}}.}(x + a) \\
\phi_{\textrm{R}}(x) &= \Psi_{{\textrm{H}}.{\textrm{O}}.}(x – a)
\end{aligned}
\end{equation}

After doing a lot of integral (i.e. in the problem set), we will see a splitting of the variational energy levels as sketched in fig. 5.

fig. 5. Splitting for double well potential.

fig. 5. Splitting for double well potential.

This sort of level splitting was what was used in the very first mazers.

Perturbation theory (outline)

Given

\begin{equation}\label{eqn:qmLecture19:480}
H = H_0 + \lambda V,
\end{equation}

where \( \lambda V \) is “small”. We want to figure out the eigenvalues and eigenstates of this Hamiltonian

\begin{equation}\label{eqn:qmLecture19:500}
H \ket{n} = E_n \ket{n}.
\end{equation}

We don’t know what these are, but do know that

\begin{equation}\label{eqn:qmLecture19:520}
H_0 \ket{n^{(0)}} = E_n^{(0)} \ket{n^{(0)}}.
\end{equation}

We are hoping that the level transitions have adiabatic transitions between the original and perturbed levels as sketched in fig. 6.

fig. 6. Adiabatic transitions.

fig. 6. Adiabatic transitions.

and not crossed level transitions as sketched in fig. 7.

fig. 7. Crossed level transitions.

fig. 7. Crossed level transitions.

If we have level crossings (which can in general occur), as opposed to adiabatic transitions, then we have no hope of using perturbation theory.

References

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

Final notes for ECE1254, Modelling of Multiphysics Systems

December 27, 2014 ece1254 No comments , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , ,

Capture

I’ve now finished my first grad course, Modelling of Multiphysics Systems, taught by Prof Piero Triverio.

I’ve posted notes for lectures and other material as I was taking the course, but now have an aggregated set of notes for the whole course posted.
This is now updated with all my notes from the lectures, solved problems, additional notes on auxillary topics I wanted to explore (like SVD), plus the notes from the Harmonic Balance report that Mike and I will be presenting in January.

This version of my notes also includes all the matlab figures regenerating using http://www.mathworks.com/matlabcentral/fileexchange/23629-export-fig, which allows a save-as pdf, which rescales much better than Matlab saveas() png’s when embedded in latex.  I’m not sure if that’s the best way to include Matlab figures in latex, but they are at least not fuzzy looking now.

All in all, I’m pretty pleased with my notes for this course.  They are a lot more readable than any of the ones I’ve done for the physics undergrad courses I was taking (http://peeterjoot.com/writing/).  While there was quite a lot covered in this course, the material really only requires an introductory circuits course and some basic math (linear algebra and intro calculus), so is pretty accessible.

This was a fun course.  I recall, back in ancient times when I was a first year student, being unsatisfied with all the ad-hoc strategies we used to solve circuits problems.  This finally answers the questions of how to tackle things more systematically.

Here’s the contents outline for these notes:

Preface
Lecture notes
1 nodal analysis
1.1 In slides
1.2 Mechanical structures example
1.3 Assembling system equations automatically. Node/branch method
1.4 Nodal Analysis
1.5 Modified nodal analysis (MNA)
2 solving large systems
2.1 Gaussian elimination
2.2 LU decomposition
2.3 Problems
3 numerical errors and conditioning
3.1 Strict diagonal dominance
3.2 Exploring uniqueness and existence
3.3 Perturbation and norms
3.4 Matrix norm
4 singular value decomposition, and conditioning number
4.1 Singular value decomposition
4.2 Conditioning number
5 sparse factorization
5.1 Fill ins
5.2 Markowitz product
5.3 Markowitz reordering
5.4 Graph representation
6 gradient methods
6.1 Summary of factorization costs
6.2 Iterative methods
6.3 Gradient method
6.4 Recap: Summary of Gradient method
6.5 Conjugate gradient method
6.6 Full Algorithm
6.7 Order analysis
6.8 Conjugate gradient convergence
6.9 Gershgorin circle theorem
6.10 Preconditioning
6.11 Symmetric preconditioning
6.12 Preconditioned conjugate gradient
6.13 Problems
7 solution of nonlinear systems
7.1 Nonlinear systems
7.2 Richardson and Linear Convergence
7.3 Newton’s method
7.4 Solution of N nonlinear equations in N unknowns
7.5 Multivariable Newton’s iteration
7.6 Automatic assembly of equations for nonlinear system
7.7 Damped Newton’s method
7.8 Continuation parameters
7.9 Singular Jacobians
7.10 Struts and Joints, Node branch formulation
7.11 Problems
8 time dependent systems
8.1 Assembling equations automatically for dynamical systems
8.2 Numerical solution of differential equations
8.3 Forward Euler method
8.4 Backward Euler method
8.5 Trapezoidal rule (TR)
8.6 Nonlinear differential equations
8.7 Analysis, accuracy and stability (Dt ! 0)
8.8 Residual for LMS methods
8.9 Global error estimate
8.10 Stability
8.11 Stability (continued)
8.12 Problems
9 model order reduction
9.1 Model order reduction
9.2 Moment matching
9.3 Model order reduction (cont).
9.4 Moment matching
9.5 Truncated Balanced Realization (1000 ft overview)
9.6 Problems
Final report
10 harmonic balance
10.1 Abstract
10.2 Introduction
10.2.1 Modifications to the netlist syntax
10.3 Background
10.3.1 Discrete Fourier Transform
10.3.2 Harmonic Balance equations
10.3.3 Frequency domain representation of MNA equations
10.3.4 Example. RC circuit with a diode.
10.3.5 Jacobian
10.3.6 Newton’s method solution
10.3.7 Alternative handling of the non-linear currents and Jacobians
10.4 Results
10.4.1 Low pass filter
10.4.2 Half wave rectifier
10.4.3 AC to DC conversion
10.4.4 Bridge rectifier
10.4.5 Cpu time and error vs N
10.4.6 Taylor series non-linearities
10.4.7 Stiff systems
10.5 Conclusion
10.6 Appendices
10.6.1 Discrete Fourier Transform inversion
Appendices
a singular value decomposition
b basic theorems and definitions
c norton equivalents
d stability of discretized linear differential equations
e laplace transform refresher
f discrete fourier transform
g harmonic balance, rough notes
g.1 Block matrix form, with physical parameter ordering
g.2 Block matrix form, with frequency ordering
g.3 Representing the linear sources
g.4 Representing non-linear sources
g.5 Newton’s method
g.6 A matrix formulation of Harmonic Balance non-linear currents
h matlab notebooks
i mathematica notebooks
Index
Bibliography