Month: September 2015

Gauge transformed probability current

September 17, 2015 math and physics play , , , ,

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Question: Gauge transformed probability current ([1] pr. 2.37 (b))

For the gauge transformed Schrodinger equation

\begin{equation}\label{eqn:gaugeTxCurrent:20}
\inv{2m} \BPi(\Bx) \cdot \BPi(\Bx) \psi(\Bx, t) + e \phi(\Bx) \psi(\Bx, t) = i \Hbar \PD{t}{}\psi(\Bx, t),
\end{equation}

where

\begin{equation}\label{eqn:gaugeTxCurrent:40}
\BPi(\Bx) = -i \Hbar \spacegrad – \frac{e}{c} \BA(\Bx),
\end{equation}

find the probability current defined by

\begin{equation}\label{eqn:gaugeTxCurrent:60}
\PD{t}{\psi} + \spacegrad \cdot \Bj.
\end{equation}

Answer

Equation \ref{eqn:gaugeTxCurrent:20} and its conjugate are

\begin{equation}\label{eqn:gaugeTxCurrent:22}
\begin{aligned}
\inv{2m} \BPi \cdot \BPi \psi + e \phi \psi &= i \Hbar \PD{t}{\psi} \\
\inv{2m} \BPi^\conj \cdot \BPi^\conj \psi^\conj + e \phi \psi^\conj &= -i \Hbar \PD{t}{\psi^\conj}
\end{aligned}
\end{equation}

which can be used immediately in a chain rule expansion of the probability time derivative

\begin{equation}\label{eqn:gaugeTxCurrent:80}
\begin{aligned}
i \Hbar \PD{t}{\rho}
&=
i \Hbar \psi^\conj \PD{t}{\psi} +
i \Hbar \psi \PD{t}{\psi^\conj} \\
&=
\psi^\conj \lr{ \inv{2m} \BPi \cdot \BPi \psi + e \phi \psi } –
\psi \lr{ \inv{2m} \BPi^\conj \cdot \BPi^\conj \psi^\conj + e \phi \psi^\conj
} \\
&=
\inv{2m} \lr{
\psi^\conj \BPi \cdot \BPi \psi
-\psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
}.
\end{aligned}
\end{equation}

We have a difference of conjugates, so can get away with expanding just the first term

\begin{equation}\label{eqn:gaugeTxCurrent:100}
\begin{aligned}
\psi^\conj \BPi \cdot \BPi \psi
&=
\psi^\conj
\psi \\
&=
\psi^\conj
\lr{ -i \Hbar \spacegrad – \frac{e}{c} \BA } \cdot \lr{ -i \Hbar \spacegrad – \frac{e}{c} \BA }
\psi \\
&=
\psi^\conj
\lr{
-\Hbar^2 \spacegrad^2 + \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad + \spacegrad \cdot \BA }
+ \frac{e^2}{c^2} \BA^2
}
\psi.
\end{aligned}
\end{equation}

Note that in the directional derivative terms, the gradient operates on everything to its right, including \( \BA \). Also note that the last term has no imaginary component, so it will not contribute to the difference of conjugates.

This gives

\begin{equation}\label{eqn:gaugeTxCurrent:120}
\begin{aligned}
\psi^\conj \BPi \cdot \BPi \psi – \psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
&=
\psi^\conj
\lr{
-\Hbar^2 \spacegrad^2 \psi + \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi + \spacegrad \cdot (\BA \psi) }
} \\
&\quad –
\psi
\lr{
-\Hbar^2 \spacegrad^2 \psi^\conj – \frac{i \Hbar e}{c} \lr{ \BA \cdot \spacegrad \psi^\conj + \spacegrad \cdot (\BA \psi^\conj) }
} \\
&=
-\Hbar^2 \lr{ \psi^\conj \spacegrad^2 \psi – \psi \spacegrad^2 \psi^\conj } \\
&\quad +
\frac{i \Hbar e}{c}
\lr{
\psi^\conj
\BA \cdot \spacegrad \psi + \psi^\conj \spacegrad \cdot (\BA \psi)
+
\psi
\BA \cdot \spacegrad \psi^\conj + \psi \spacegrad \cdot (\BA \psi^\conj)
}
\end{aligned}
\end{equation}

The first term is recognized as a divergence

\begin{equation}\label{eqn:gaugeTxCurrent:140}
\begin{aligned}
\spacegrad \cdot \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
&=
\psi^\conj \spacegrad \cdot \spacegrad \psi
+
\spacegrad \psi \cdot \spacegrad \psi^\conj
–
\psi \spacegrad \cdot \spacegrad \psi^\conj
–
\spacegrad \psi^\conj \cdot \spacegrad \psi \\
&= \psi^\conj \spacegrad^2 \psi – \psi \spacegrad^2 \psi^\conj.
\end{aligned}
\end{equation}

The second term can also be factored into a divergence operation

\begin{equation}\label{eqn:gaugeTxCurrent:160}
\begin{aligned}
\psi^\conj
\BA \cdot \spacegrad \psi &+ \psi^\conj \spacegrad \cdot (\BA \psi)
+
\psi
\BA \cdot \spacegrad \psi^\conj + \psi \spacegrad \cdot (\BA \psi^\conj) \\
&=
\lr{ \psi^\conj\BA \cdot \spacegrad \psi
+\psi \spacegrad \cdot (\BA \psi^\conj)
}
+\lr{
\psi \BA \cdot \spacegrad \psi^\conj
+\psi^\conj \spacegrad \cdot (\BA \psi)
} \\
&= 2 \spacegrad \cdot \lr{ \BA \psi \psi^\conj } \\
\end{aligned}
\end{equation}

Putting all the pieces back together we have

\begin{equation}\label{eqn:gaugeTxCurrent:180}
\begin{aligned}
\PD{t}{\rho}
&=
\inv{2m i \Hbar} \lr{
\psi^\conj \BPi \cdot \BPi \psi
-\psi \BPi^\conj \cdot \BPi^\conj \psi^\conj
} \\
&=
\spacegrad \cdot
\inv{2m i \Hbar} \lr{
-\Hbar^2
\lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
+ \frac{ i \Hbar e}{c} 2 \BA \psi \psi^\conj
} \\
&=
\spacegrad \cdot
\lr{
\frac{i \Hbar}{2 m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
+ \frac{e}{m c} \BA \psi \psi^\conj
}.
\end{aligned}
\end{equation}

From \ref{eqn:gaugeTxCurrent:60}, the probability current must be

\begin{equation}\label{eqn:gaugeTxCurrent:200}
\Bj
=
\frac{\Hbar}{2 i m} \lr{ \psi^\conj \spacegrad \psi – \psi \spacegrad \psi^\conj }
– \frac{e}{m c} \BA \psi \psi^\conj,
\end{equation}

or
\begin{equation}\label{eqn:gaugeTxCurrent:220}
\boxed{
\Bj
=
\frac{\Hbar}{m} \textrm{Im} \lr{ \psi^\conj \spacegrad \psi }
– \frac{e}{m c} \BA \psi \psi^\conj.
}
\end{equation}

References

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

Gauge transformation of free particle Hamiltonian

September 15, 2015 math and physics play , , , , , ,


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Question:

Given a gauge transformation of the free particle Hamiltonian to

\begin{equation}\label{eqn:gaugeTx:20}
H = \inv{2 m} \BPi \cdot \BPi + e \phi,
\end{equation}

where

\begin{equation}\label{eqn:gaugeTx:40}
\BPi = \Bp – \frac{e}{c} \BA,
\end{equation}

calculate \( m d\Bx/dt \), \( \antisymmetric{\Pi_i}{\Pi_j} \), and \( m d^2\Bx/dt^2 \), where \( \Bx \) is the Heisenberg picture position operator, and the fields are functions only of position \( \phi = \phi(\Bx), \BA = \BA(\Bx) \).

Answer

The final results for these calculations are found in [1], but seem worth deriving to exercise our commutator muscles.

Heisenberg picture velocity operator

The first order of business is the Heisenberg picture velocity operator, but first note

\begin{equation}\label{eqn:gaugeTx:60}
\begin{aligned}
\BPi \cdot \BPi
&= \lr{ \Bp – \frac{e}{c} \BA} \cdot \lr{ \Bp – \frac{e}{c} \BA} \\
&= \Bp^2 – \frac{e}{c} \lr{ \BA \cdot \Bp + \Bp \cdot \BA } + \frac{e^2}{c^2} \BA^2.
\end{aligned}
\end{equation}

The time evolution of the Heisenberg picture position operator is therefore

\begin{equation}\label{eqn:gaugeTx:80}
\begin{aligned}
\ddt{\Bx}
&= \inv{i\Hbar} \antisymmetric{\Bx}{H} \\
&= \inv{i\Hbar 2 m} \antisymmetric{\Bx}{\BPi^2} \\
&= \inv{i\Hbar 2 m} \antisymmetric{\Bx}{\Bp^2 – \frac{e}{c} \lr{ \BA \cdot \Bp
+ \Bp \cdot \BA } + \frac{e^2}{c^2} \BA^2 } \\
&= \inv{i\Hbar 2 m}
\lr{
\antisymmetric{\Bx}{\Bp^2}
– \frac{e}{c} \antisymmetric{\Bx}{ \BA \cdot \Bp + \Bp \cdot \BA }
}
.
\end{aligned}
\end{equation}

For the \( \Bp^2 \) commutator we have

\begin{equation}\label{eqn:gaugeTx:100}
\antisymmetric{x_r}{\Bp^2}
=
i \Hbar \PD{p_r}{\Bp^2}
=
2 i \Hbar p_r,
\end{equation}

or
\begin{equation}\label{eqn:gaugeTx:120}
\antisymmetric{\Bx}{\Bp^2}
=
2 i \Hbar \Bp.
\end{equation}

Computing the remaining commutator, we’ve got

\begin{equation}\label{eqn:gaugeTx:140}
\begin{aligned}
\antisymmetric{x_r}{\Bp \cdot \BA + \BA \cdot \Bp}
&= x_r p_s A_s – p_s A_s x_r \\
&\quad+ x_r A_s p_s – A_s p_s x_r \\
&= \lr{ \antisymmetric{x_r}{p_s} + p_s x_r } A_s – p_s A_s x_r \\
&\quad+ x_r A_s p_s – A_s \lr{ \antisymmetric{p_s}{x_r} + x_r p_s } \\
&= \antisymmetric{x_r}{p_s} A_s + {p_s A_s x_r – p_s A_s x_r} \\
&\quad+ {x_r A_s p_s – x_r A_s p_s} + A_s \antisymmetric{x_r}{p_s} \\
&= 2 i \Hbar \delta_{r s} A_s \\
&= 2 i \Hbar A_r,
\end{aligned}
\end{equation}

so

\begin{equation}\label{eqn:gaugeTx:160}
\antisymmetric{\Bx}{\Bp \cdot \BA + \BA \cdot \Bp} = 2 i \Hbar \BA.
\end{equation}

Assembling these results gives

\begin{equation}\label{eqn:gaugeTx:180}
\boxed{
\ddt{\Bx} = \inv{m} \lr{ \Bp – \frac{e}{c} \BA } = \inv{m} \BPi,
}
\end{equation}

as asserted in the text.

Kinetic Momentum commutators

\begin{equation}\label{eqn:gaugeTx:200}
\begin{aligned}
\antisymmetric{\Pi_r}{\Pi_s}
&=
\antisymmetric{p_r – e A_r/c}{p_s – e A_s/c} \\
&=
{\antisymmetric{p_r}{p_s}}
– \frac{e}{c} \lr{ \antisymmetric{p_r}{A_s} + \antisymmetric{A_r}{p_s}}
+ \frac{e^2}{c^2} {\antisymmetric{A_r}{A_s}} \\
&=
– \frac{e}{c} \lr{ (-i\Hbar) \PD{x_r}{A_s} + (i\Hbar) \PD{x_s}{A_r} } \\
&=
– \frac{i e \Hbar}{c} \lr{ -\PD{x_r}{A_s} + \PD{x_s}{A_r} } \\
&=
– \frac{i e \Hbar}{c} \epsilon_{t s r} B_t,
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:gaugeTx:220}
\boxed{
\antisymmetric{\Pi_r}{\Pi_s}
=
\frac{i e \Hbar}{c} \epsilon_{r s t} B_t.
}
\end{equation}

Quantum Lorentz force

For the force equation we have

\begin{equation}\label{eqn:gaugeTx:240}
\begin{aligned}
m \frac{d^2 \Bx}{dt^2}
&= \ddt{\BPi} \\
&= \inv{i \Hbar} \antisymmetric{\BPi}{H} \\
&= \inv{i \Hbar 2 m } \antisymmetric{\BPi}{\BPi^2}
+ \inv{i \Hbar } \antisymmetric{\BPi}{e \phi}.
\end{aligned}
\end{equation}

For the \( \phi \) commutator consider one component

\begin{equation}\label{eqn:gaugeTx:260}
\begin{aligned}
\antisymmetric{\Pi_r}{e \phi}
&=
e \antisymmetric{p_r – \frac{e}{c} A_r}{\phi} \\
&=
e \antisymmetric{p_r}{\phi} \\
&=
e (-i\Hbar) \PD{x_r}{\phi},
\end{aligned}
\end{equation}

or
\begin{equation}\label{eqn:gaugeTx:280}
\inv{i \Hbar} \antisymmetric{\BPi}{e \phi}
=
– e \spacegrad \phi
=
e \BE.
\end{equation}

For the \( \BPi^2 \) commutator I initially did this the hard way (it took four notebook pages, plus two for a false start.) Realizing that I didn’t use \ref{eqn:gaugeTx:220} for that expansion was the clue to doing this more expediently.

Considering a single component

\begin{equation}\label{eqn:gaugeTx:300}
\begin{aligned}
\antisymmetric{\Pi_r}{\BPi^2}
&=
\antisymmetric{\Pi_r}{\Pi_s \Pi_s} \\
&=
\Pi_r \Pi_s \Pi_s – \Pi_s \Pi_s \Pi_r \\
&=
\lr{ \antisymmetric{\Pi_r}{\Pi_s} + {\Pi_s \Pi_r} }
\Pi_s
– \Pi_s
\lr{ \antisymmetric{\Pi_s}{\Pi_r} + {\Pi_r \Pi_s} } \\
&= i \Hbar \frac{e}{c} \epsilon_{r s t}
\lr{ B_t \Pi_s + \Pi_s B_t },
\end{aligned}
\end{equation}

or

\begin{equation}\label{eqn:gaugeTx:320}
\begin{aligned}
\inv{ i \Hbar 2 m} \antisymmetric{\BPi}{\BPi^2}
&= \frac{e}{2 m c } \epsilon_{r s t} \Be_r
\lr{ B_t \Pi_s + \Pi_s B_t } \\
&= \frac{e}{ 2 m c }
\lr{
\BPi \cross \BB
– \BB \cross \BPi
}.
\end{aligned}
\end{equation}

Putting all the pieces together we’ve got the quantum equivalent of the Lorentz force equation

\begin{equation}\label{eqn:gaugeTx:340}
\boxed{
m \frac{d^2 \Bx}{dt^2} = e \BE + \frac{e}{2 c} \lr{
\frac{d\Bx}{dt} \cross \BB
– \BB \cross \frac{d\Bx}{dt}
}.
}
\end{equation}

While this looks equivalent to the classical result, all the vectors here are Heisenberg picture operators dependent on position.

References

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

Notes from reading Sakurai’s QM, prep for graduate Quantum Mechanics (PHY1520H)

September 14, 2015 phy1520 , ,

I start the lectures for grad Quantum Mechanics (PHY1520H) tomorrow, taught by Prof. Arun Paramekanti.

The curriculum has scary range of material, although some of it will be review. As prep for that, I’ve been working some problems, and have also explored a few interesting details and digressions. My notes for that preparatory work, collecting a number of individual posts from over the summer, is now posted.

questions to my ward 3 council rep: city council finances question

September 10, 2015 Incoherent ramblings , ,

Hi Mr Hamilton,

I’ve lived in Markham since 2001 in various locations. It appears you are my current ward 3 council representative. I have a couple questions:

1) Where can I find the total property taxes that Markham collected in a given year?
2) Over what time period was the money given away for the attempt to establish a Markham NHL arena?
3) What was the exact amount that was spent on that study?

If council will not even disclose how they spent the money, then I think it is fair to send them a bill for reimbursement of the portion of property taxes that I spent in that time interval. I’d like this information to prepare such a bill.

Sincerely, Peeter Joot.
—
Note. It is my personal policy to make a copy of all political correspondence publicly available on the internet.

Generalized Gaussian integrals

September 10, 2015 phy1520 , , , ,

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Both [3] and [4] use Gaussian integrals with both (negative) real, and imaginary arguments, which give the impression that the following is true:

\begin{equation}\label{eqn:generalizedGaussian:20}
\int_{-\infty}^\infty \exp\lr{ a x^2 } dx = \sqrt{\frac{-\pi}{a}},
\end{equation}

even when \( a \) is not a real negative constant, and in particular, with values \( a = \pm i \). Clearly this doesn’t follow by just making a substition \( x \rightarrow x/\sqrt{a} \), since that moves the integration range onto a rotated path in the complex plane when \( a \) is \( \pm i \). However, with some care, it can be shown that \ref{eqn:generalizedGaussian:20} holds provided \( \textrm{Re} \, a \le 0 \).

The first special case is \( \int_{-\infty}^\infty \exp\lr{ – x^2 } dx = \sqrt{\pi} \) which is easy to derive using the usual square it and integrate in circular coordinates trick.

Purely imaginary cases.

Let’s handle the \( a = \pm i \) cases next. These can be evaluated by considering integrals over the contours of fig. 1, where the upper plane contour is used for \( a = i \) and the lower plane contour for \( a = -i \).

twoContours45DegreesFig1

fig. 1. Contours for a = i,-i

Since there are no poles, the integral over either such contour is zero. Credit for figuring out how to tackle that integral and what contour to use goes to Dr MV, on stackexchange [2].

For the upper plane contour we have

\begin{equation}\label{eqn:generalizedGaussian:40}
\begin{aligned}
0
&= \oint \exp\lr{ i z^2 } dz \\
&= \int_0^R \exp\lr{ i x^2 } dx
+ \int_0^{\pi/4} \exp\lr{ i R^2 e^{2 i \theta} } R i e^{i\theta} d\theta
+ \int_R^0 \exp\lr{ i^2 t^2 } e^{i\pi/4} dt.
\end{aligned}
\end{equation}

Observe that \( i e^{2 i \theta} = i\cos(2 \theta) – \sin(2\theta) \) which has a negative real part for all values of \( \theta \ne 0 \). Provided the contour is slightly deformed from the axis, that second integral has a term of the form \( \sim R e^{-R^2} \) which tends to zero as \( R \rightarrow \infty \). So in the limit, this is

\begin{equation}\label{eqn:generalizedGaussian:60}
\int_0^\infty \exp\lr{ i x^2 } dx
= \sqrt{\pi} e^{i\pi/4}/2,
\end{equation}

or
\begin{equation}\label{eqn:generalizedGaussian:80}
\int_{-\infty}^\infty \exp\lr{ i x^2 } dx
= \sqrt{i \pi},
\end{equation}

a special case of \ref{eqn:generalizedGaussian:20} as desired. For \( a = -i \) integrating around the lower plane contour, we have

\begin{equation}\label{eqn:generalizedGaussian:100}
\begin{aligned}
0
&= \oint \exp\lr{ -i z^2 } dz \\
&= \int_0^R \exp\lr{ i x^2 } dx
+ \int_0^{-\pi/4} \exp\lr{ -i R^2 e^{2 i \theta} } R i e^{i\theta} d\theta
+ \int_R^0 \exp\lr{ -i (-i) t^2 } e^{-i\pi/4} dt.
\end{aligned}
\end{equation}

This time, in the second integral we also have \( -i R^2 e^{2 i \theta} = i R^2 \cos(2 \theta) + \sin(2 \theta) \), which also has a negative real part for \( \theta \in (0, \pi/4] \). Again the contour needs to be infinitesimally deformed, placed just lower than the axis.

This time we find

\begin{equation}\label{eqn:generalizedGaussian:120}
\int_{-\infty}^\infty \exp\lr{ -i x^2 } dx
= \sqrt{-i \pi},
\end{equation}

another special case of \ref{eqn:generalizedGaussian:20}.

Note.

Distorting the contour in this fashion seems somewhat like handwaving. A better approach would probably follow [1] where Jordan’s lemma is covered. It doesn’t look like Jordan’s lemma applies as is to this case, but the arguments look like they could be adapted appropriately.

Completely complex case.

A similar trick can be used to evaluate the more general cases, but a bit of thought is required to figure out the contours required. More precisely, while these contours will still have a wedge of pie shape, as sketched in fig. 2, we have to figure out the angle subtended by the edge of this piece of pie.

twoContoursThetaFig2

fig. 2. Contours for complex a.

To evaluate an integral consider

\begin{equation}\label{eqn:generalizedGaussian:140}
\begin{aligned}
0
&= \oint \exp\lr{ e^{i\phi} z^2 } dz \\
&= \int_0^R \exp\lr{ e^{i\phi} x^2 } dx
+ \int_0^{\theta} \exp\lr{ e^{i\phi} R^2 e^{2 i \mu} } R i e^{i\mu} d\mu
+ \int_R^0 \exp\lr{ e^{i\phi} e^{2 i \theta} t^2 } e^{i\theta} dt,
\end{aligned}
\end{equation}

where \( \phi \in (\pi/2, \pi) \cup (\pi,3\pi/2) \). We have a hope of evaluating this last integral if \( \phi + 2 \theta = \pi \), or

\begin{equation}\label{eqn:generalizedGaussian:160}
\theta = (\pi -\phi)/2,
\end{equation}

giving

\begin{equation}\label{eqn:generalizedGaussian:180}
\int_0^R \exp\lr{ e^{i\phi} x^2 } dx
=
e^{i\lr{\pi – \phi}/2} \int_0^R \exp\lr{ -t^2 } dt
– \int_0^{\theta} \exp\lr{ R^2 \lr{ \cos\lr{\phi + 2 \mu} + i \sin\lr{\phi + 2 \mu}} } R i e^{i\mu} d\mu.
\end{equation}

If the cosine is always negative on the chosen contours, then that integral will vanish in the \( R \rightarrow \infty \) limit. This turns out to be the case, which can be confirmed by considering each of the contours in sequence. If the upper plane contour is used to evaluate the integral for the \( \phi \in (\pi/2,\pi) \) case, we have

\begin{equation}\label{eqn:generalizedGaussian:200}
\theta \in (0, \pi/4).
\end{equation}

Since \( \phi + 2\theta = \pi \), we have

\begin{equation}\label{eqn:generalizedGaussian:220}
\phi + 2 \mu \in (\pi/2, \pi),
\end{equation}

and find that the cosine is strictly negative on that contour for that range of \( \phi \). Picking the lower plane contour for the \( \phi \in (\pi, 3\pi/2) \) range, we have

\begin{equation}\label{eqn:generalizedGaussian:240}
\theta \in (-\pi/4, 0),
\end{equation}

and so

\begin{equation}\label{eqn:generalizedGaussian:260}
\phi + 2 \mu \in (\pi/2, 3\pi/2).
\end{equation}

For this range of \( \phi \) the cosine on the lower plane contour is again negative as desired, so in the infinite \( R \) limit we have

\begin{equation}\label{eqn:generalizedGaussian:280}
\int_0^\infty \exp\lr{ e^{i\phi} x^2 } dx
=
\inv{2} \sqrt{ -\pi e^{-i\phi} }.
\end{equation}

The points at \( \phi = \pi/2, \pi, 3\pi/2 \) were omitted, but we’ve found the same result at those points, completing the verification of \ref{eqn:generalizedGaussian:20} for all \( \textrm{Re} a \le 0 \).

References

[1] W.R. Le Page and W.R. LePage. Complex Variables and the Laplace Transform for Engineers, chapter 8-10. A Theorem for Trigonometric Integrals. Courier Dover Publications, 1980.

[2] Dr. MV. Evaluating definite integral of exp(i t^2), 2015. URL https://math.stackexchange.com/a/1411084/359. [Online; accessed 10-September-2015].

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

[4] A. Zee. Quantum field theory in a nutshell. Universities Press, 2005.