-
+
+{% include proof/start.html id="proof-mean" -%}
The trick is to treat $$p$$ and $$q$$ as independent until the last moment:
$$\begin{aligned}
@@ -62,8 +59,8 @@ $$\begin{aligned}
\end{aligned}$$
Inserting $$q = 1 - p$$ then gives the desired result.
-
-
+{% include proof/end.html id="proof-mean" %}
+
Meanwhile, we find the following variance $$\sigma^2$$,
with $$\sigma$$ being the standard deviation:
@@ -74,12 +71,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
-We use the same trick to calculate $$\overline{n^2}$$
+
+{% include proof/start.html id="proof-var" -%}
(the mean squared number of successes):
$$\begin{aligned}
@@ -106,8 +99,8 @@ $$\begin{aligned}
\end{aligned}$$
By inserting $$q = 1 - p$$, we arrive at the desired expression.
-
-
+{% include proof/end.html id="proof-var" %}
+
As $$N \to \infty$$, the binomial distribution
turns into the continuous normal distribution,
@@ -119,11 +112,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-normal" -%}
We take the Taylor expansion of $$\ln\!\big(P_N(n)\big)$$
around the mean $$\mu = Np$$:
@@ -211,8 +201,7 @@ $$\begin{aligned}
\end{aligned}$$
Taking $$\exp$$ of this expression then yields a normalized Gaussian distribution.
-
-
+
+{% include proof/start.html id="proof-moment0" -%}
We insert $$Q = m$$ into our prototype,
and since $$m$$ is constant, the rest is trivial:
@@ -159,9 +156,8 @@ $$\begin{aligned}
\\
&= \pdv{\rho}{t} + \nabla \cdot \big(\rho \Expval{\vb{v}}\big) - 0
\end{aligned}$$
+{% include proof/end.html id="proof-moment0" %}
-
-
If we instead choose the momentum $$Q = m \vb{v}$$,
we find that the **first moment** of the BTE describes conservation of momentum,
@@ -174,11 +170,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-moment1" -%}
We insert $$Q = m \vb{v}$$ into our prototype and recognize $$\rho$$ wherever possible:
$$\begin{aligned}
@@ -220,9 +213,8 @@ $$\begin{aligned}
0
&= \pdv{}{t}\big(\rho \vb{V}\big) + \nabla \cdot \big(\rho \vb{V} \vb{V} + \hat{P}\big) - n \vb{F}
\end{aligned}$$
+{% include proof/end.html id="proof-moment1" %}
-
-
Finally, if we choose the kinetic energy $$Q = m |\vb{v}|^2 / 2$$,
we find that the **second moment** gives conservation of energy,
@@ -237,11 +229,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-moment2" -%}
We insert $$Q = m |\vb{v}|^2 / 2$$ into our prototype and recognize $$\rho$$ wherever possible:
$$\begin{aligned}
@@ -349,9 +338,7 @@ $$\begin{aligned}
\end{bmatrix}
= \sum_{i=1}^{3} \sum_{j=1}^{3} \pdv{P_{ij}}{x_j} V_i
\end{aligned}$$
-
-
-
+{% include proof/end.html id="proof-moment2" %}
diff --git a/source/know/concept/convolution-theorem/index.md b/source/know/concept/convolution-theorem/index.md
index 742c8ff..510417a 100644
--- a/source/know/concept/convolution-theorem/index.md
+++ b/source/know/concept/convolution-theorem/index.md
@@ -12,6 +12,8 @@ is equal to a product in the frequency domain. This is especially useful
for computation, replacing an $$\mathcal{O}(n^2)$$ convolution with an
$$\mathcal{O}(n \log(n))$$ transform and product.
+
+
## Fourier transform
The convolution theorem is usually expressed as follows, where
@@ -27,11 +29,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-fourier" -%}
We expand the right-hand side of the theorem and
rearrange the integrals:
@@ -57,8 +56,8 @@ $$\begin{aligned}
&= B \int_{-\infty}^\infty \tilde{g}(k') \: \tilde{f}(k - k') \dd{k'}
= B \cdot (\tilde{f} * \tilde{g})(k)
\end{aligned}$$
-
-
+
+{% include proof/start.html id="proof-laplace" -%}
We expand the left-hand side.
Note that the lower integration limit is 0 instead of $$-\infty$$,
because we set both $$f(t)$$ and $$g(t)$$ to zero for $$t < 0$$:
@@ -106,8 +102,7 @@ $$\begin{aligned}
&= \int_0^\infty \tilde{f}(s) \: g(t') \exp(- s t') \dd{t'}
= \tilde{f}(s) \: \tilde{g}(s)
\end{aligned}$$
-
-
+{% include proof/end.html id="proof-laplace" %}
diff --git a/source/know/concept/curvilinear-coordinates/index.md b/source/know/concept/curvilinear-coordinates/index.md
index cb22e43..48a5a72 100644
--- a/source/know/concept/curvilinear-coordinates/index.md
+++ b/source/know/concept/curvilinear-coordinates/index.md
@@ -48,6 +48,7 @@ we derive general formulae to convert expressions
from Cartesian coordinates to the new orthogonal system $$(x_1, x_2, x_3)$$.
+
## Basis vectors
Consider the the vector form of the line element $$\dd{\ell}$$,
@@ -86,6 +87,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## Gradient
In an orthogonal coordinate system,
@@ -102,11 +104,8 @@ $$\begin{gathered}
}
\end{gathered}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-grad" -%}
For a direction $$\dd{\ell}$$, we know that
$$\idv{f}{\ell}$$ is the component of $$\nabla f$$ in that direction:
@@ -127,9 +126,8 @@ $$\begin{gathered}
+ \vu{e}_2 \dv{x_2}{\ell} \pdv{f}{x_2}
+ \vu{e}_3 \dv{x_3}{\ell} \pdv{f}{x_3}
\end{gathered}$$
+{% include proof/end.html id="proof-grad" %}
-
-
+
+{% include proof/start.html id="proof-div" -%}
As preparation, we rewrite $$\vb{V}$$ as follows
to introduce the scale factors:
@@ -222,8 +217,8 @@ $$\begin{aligned}
After repeating this procedure for the other components of $$\vb{V}$$,
we get the desired general expression for the divergence.
-
-
+{% include proof/end.html id="proof-div" %}
+
## Laplacian
@@ -246,6 +241,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## Curl
The curl of a vector $$\vb{V}$$ is as follows
@@ -264,11 +260,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-curl" -%}
The curl is found in a similar way as the divergence.
We rewrite $$\vb{V}$$ like so:
@@ -317,8 +310,8 @@ $$\begin{aligned}
If we go through the same process for the other components of $$\vb{V}$$
and add up the results, we get the desired expression for the curl.
-
-
+{% include proof/end.html id="proof-curl" %}
+
## Differential elements
diff --git a/source/know/concept/detailed-balance/index.md b/source/know/concept/detailed-balance/index.md
index b89d5da..98f9bd3 100644
--- a/source/know/concept/detailed-balance/index.md
+++ b/source/know/concept/detailed-balance/index.md
@@ -103,11 +103,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-reversibility" -%}
Consider the following weighted inner product,
whose weight function is a stationary distribution $$\pi$$
satisfying detailed balance,
@@ -222,8 +219,7 @@ $$\begin{aligned}
Where the integral gave the expectation value at $$X_0$$,
since $$\pi$$ does not change in time.
-
-
+
+{% include proof/start.html id="proof-scale" -%}
Because it is symmetric, $$\delta(s x) = \delta(|s| x)$$.
Then by substituting $$\sigma = |s| x$$:
@@ -77,9 +74,8 @@ $$\begin{aligned}
\int \delta(|s| x) \dd{x}
&= \frac{1}{|s|} \int \delta(\sigma) \dd{\sigma} = \frac{1}{|s|}
\end{aligned}$$
+{% include proof/end.html id="proof-scale" %}
-
-
An even more impressive property is the behaviour of the derivative of $$\delta(x)$$:
@@ -89,11 +85,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-dv1" -%}
Note which variable is used for the
differentiation, and that $$\delta'(x - \xi) = - \delta'(\xi - x)$$:
@@ -102,9 +95,8 @@ $$\begin{aligned}
&= \dv{}{x}\int f(\xi) \: \delta(x - \xi) \dd{x}
= f'(x)
\end{aligned}$$
+{% include proof/end.html id="proof-dv1" %}
-
-
This property also generalizes nicely for the higher-order derivatives:
diff --git a/source/know/concept/dynkins-formula/index.md b/source/know/concept/dynkins-formula/index.md
index c0d20c5..307f098 100644
--- a/source/know/concept/dynkins-formula/index.md
+++ b/source/know/concept/dynkins-formula/index.md
@@ -39,11 +39,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-kolmogorov" -%}
We define a new process $$Y_t \equiv h(X_t)$$, and then apply Itō's lemma, leading to:
$$\begin{aligned}
@@ -84,9 +81,8 @@ $$\begin{aligned}
\hat{L}\{h(X_0)\}
\approx \frac{1}{t} \mathbf{E}[Y_t - Y_0| X_0]
\end{aligned}$$
+{% include proof/end.html id="proof-kolmogorov" %}
-
-
The general definition of resembles that of a classical derivative,
and indeed, the generator $$\hat{A}$$ can be thought of as a differential operator.
@@ -104,11 +100,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-dynkin" -%}
The proof is similar to the one above.
Define $$Y_t = h(X_t)$$ and use Itō’s lemma:
@@ -136,9 +129,9 @@ $$\begin{aligned}
= \mathbf{E}\bigg[ Y_\tau - Y_0 - \int_0^\tau \hat{L}\{h(X_t)\} \dd{t} \bigg| X_0 \bigg]
\end{aligned}$$
-Isolating this equation for $$\mathbf{E}[Y_\tau | X_0]$$ then gives Dynkin's formula.
-
-
+Isolating this equation for $$\mathbf{E}[Y_\tau \!\mid\! X_0]$$ then gives Dynkin's formula.
+{% include proof/end.html id="proof-dynkin" %}
+
A common application of Dynkin's formula is predicting
when the stopping time $$\tau$$ occurs, and in what state $$X_\tau$$ this happens.
diff --git a/source/know/concept/equation-of-motion-theory/index.md b/source/know/concept/equation-of-motion-theory/index.md
index 02ed856..c1ed8da 100644
--- a/source/know/concept/equation-of-motion-theory/index.md
+++ b/source/know/concept/equation-of-motion-theory/index.md
@@ -63,11 +63,8 @@ $$\begin{aligned}
= - \sum_{\nu''} u_{\nu \nu''} \hat{c}_{\nu''}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-commutator" -%}
Using the commutator identity for $$\comm{A B}{C}$$,
we decompose it like so:
@@ -105,9 +102,8 @@ $$\begin{aligned}
- 2 \acomm{\hat{f}_{\!\nu'}^\dagger}{\hat{f}_{\!\nu}} \hat{f}_{\!\nu''} \Big)
= - \sum_{\nu''} u_{\nu \nu''} \hat{f}_{\!\nu''}
\end{aligned}$$
+{% include proof/end.html id="proof-commutator" %}
-
-
Substituting this into $$G_{\nu \nu'}^R$$'s equation of motion,
we recognize another Green's function $$G_{\nu'' \nu'}^R$$:
diff --git a/source/know/concept/euler-bernoulli-law/index.md b/source/know/concept/euler-bernoulli-law/index.md
index dad67ca..5a6c38d 100644
--- a/source/know/concept/euler-bernoulli-law/index.md
+++ b/source/know/concept/euler-bernoulli-law/index.md
@@ -81,11 +81,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-field" -%}
By integrating the above strains $$u_{ii} = \ipdv{u_i}{i}$$,
we get the components of $$\va{u}$$:
@@ -171,8 +168,8 @@ $$\begin{aligned}
Inserting this into the components $$u_x$$, $$u_y$$ and $$u_z$$
then yields the full displacement field.
-
-
+{% include proof/end.html id="proof-field" %}
+
In any case, the beam experiences a bending torque with an $$x$$-component $$T_x$$ given by:
diff --git a/source/know/concept/fourier-transform/index.md b/source/know/concept/fourier-transform/index.md
index 0bc849b..c86d997 100644
--- a/source/know/concept/fourier-transform/index.md
+++ b/source/know/concept/fourier-transform/index.md
@@ -67,6 +67,7 @@ on whether the analysis is for forward ($$s > 0$$) or backward-propagating
($$s < 0$$) waves.
+
## Derivatives
The FT of a derivative has a very useful property.
@@ -113,6 +114,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## Multiple dimensions
The Fourier transform is straightforward to generalize to $$N$$ dimensions.
@@ -150,11 +152,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-constants-ndim" -%}
The inverse FT of the forward FT of $$f(\vb{x})$$ must be equal to $$f(\vb{x})$$ again, so:
$$\begin{aligned}
@@ -180,9 +179,8 @@ $$\begin{aligned}
&= \frac{(2 \pi)^N A B}{|s|^N} \int f(\vb{x}') \: \delta(\vb{x}' - \vb{x}) \ddn{N}{\vb{x}'}
= \frac{(2 \pi)^N A B}{|s|^N} f(\vb{x})
\end{aligned}$$
+{% include proof/end.html id="proof-constants-ndim" %}
-
-
Differentiation is more complicated for $$N > 1$$,
but the FT is still useful,
@@ -197,11 +195,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-laplacian" -%}
We insert $$\nabla^2 f$$ into the FT,
decompose the exponential and the Laplacian,
and then integrate by parts (limits $$\pm \infty$$ omitted):
@@ -236,9 +231,7 @@ $$\begin{aligned}
&= - A s^2 \sum_{n = 1}^N k_n^2 \int f \exp(i s \vb{k} \cdot \vb{x}) \ddn{N}{\vb{x}}
= - s^2 \sum_{n = 1}^N k_n^2 \tilde{f}
\end{aligned}$$
-
-
-
+
+{% include proof/start.html id="proof-solution" -%}
$$\hat{L}$$ only acts on $$x$$, so $$x' \in ]a, b[$$ is simply a parameter,
meaning we are free to multiply the definition of $$G$$
by the constant $$f(x')$$ on both sides,
@@ -72,8 +69,8 @@ $$\begin{aligned}
By definition, $$\hat{L}$$'s response $$u(x)$$ to $$f(x)$$
satisfies $$\hat{L}\{ u(x) \} = f(x)$$, recognizable here.
-
-
+{% include proof/end.html id="proof-solution" %}
+
While the impulse response is typically used for initial value problems,
the fundamental solution $$G$$ is used for boundary value problems.
@@ -117,11 +114,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-reciprocity" -%}
Consider two parameters $$x_1'$$ and $$x_2'$$.
The self-adjointness of $$\hat{L}$$ means that:
@@ -135,9 +129,7 @@ $$\begin{aligned}
G^*(x_2', x_1')
&= G(x_1', x_2')
\end{aligned}$$
-
-
-
+{% include proof/end.html id="proof-reciprocity" %}
diff --git a/source/know/concept/greens-functions/index.md b/source/know/concept/greens-functions/index.md
index ddba2cd..eda5671 100644
--- a/source/know/concept/greens-functions/index.md
+++ b/source/know/concept/greens-functions/index.md
@@ -21,6 +21,7 @@ but in general they are not the same,
except in a special case, see below.
+
## Single-particle functions
If the two operators are single-particle creation/annihilation operators,
@@ -146,11 +147,8 @@ $$\begin{gathered}
G_{\nu \nu'}^<(t, t') = G_{\nu \nu'}^<(t - t')
\end{gathered}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-time-delta" -%}
We will prove that the thermal expectation value
$$\expval{\hat{A}(t) \hat{B}(t')}$$ only depends on $$t - t'$$
for arbitrary $$\hat{A}$$ and $$\hat{B}$$,
@@ -189,8 +187,7 @@ because $$\hat{H}$$ is time-independent by assumption.
Note that thermodynamic equilibrium is crucial:
intuitively, if the system is not in equilibrium,
then it evolves in some transient time-dependent way.
-
-
+{% include proof/end.html id="proof-time-delta" %}
If the Hamiltonian is both time-independent and non-interacting,
then the time-dependence of $$\hat{c}_\nu$$
@@ -214,6 +211,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## As fundamental solutions
In the absence of interactions,
@@ -237,11 +235,8 @@ $$\begin{aligned}
= \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-commutator" -%}
In the second quantization,
the Hamiltonian $$\hat{H}_0$$ is written like so:
@@ -307,9 +302,8 @@ $$\begin{aligned}
&= \frac{\hbar^2}{2 m} \sum_{\nu'} \hat{c}_{\nu'} \nabla^2 \psi_{\nu'}(\vb{r})
= \frac{\hbar^2}{2 m} \nabla^2 \hat{\Psi}(\vb{r})
\end{aligned}$$
+{% include proof/end.html id="proof-commutator" %}
-
-
After substituting this into the equation of motion,
we recognize $$G^R(\vb{r}, t; \vb{r}', t')$$ itself:
diff --git a/source/know/concept/gronwall-bellman-inequality/index.md b/source/know/concept/gronwall-bellman-inequality/index.md
index 8096aaf..da1bcad 100644
--- a/source/know/concept/gronwall-bellman-inequality/index.md
+++ b/source/know/concept/gronwall-bellman-inequality/index.md
@@ -26,11 +26,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-original" -%}
We define $$w(t)$$ to equal the upper bounds above
on both $$w'(t)$$ and $$w(t)$$ itself:
@@ -63,8 +60,8 @@ $$\begin{aligned}
Since $$u' \le \beta u$$ as a condition,
the above derivative is always negative.
-
-
+{% include proof/end.html id="proof-original" %}
+
Grönwall's inequality can be generalized to non-differentiable functions.
Suppose we know:
@@ -84,11 +81,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-integral" -%}
We start by defining $$w(t)$$ as follows,
which will act as shorthand:
@@ -138,8 +132,8 @@ $$\begin{aligned}
\end{aligned}$$
Insert this into the condition under which the Grönwall-Bellman inequality holds.
-
-
+{% include proof/end.html id="proof-integral" %}
+
In the special case where $$\alpha(t)$$ is non-decreasing with $$t$$,
the inequality reduces to:
@@ -151,11 +145,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-special" -%}
Starting from the "ordinary" Grönwall-Bellman inequality,
the fact that $$\alpha(t)$$ is non-decreasing tells us that
$$\alpha(s) \le \alpha(t)$$ for all $$s \le t$$, so:
@@ -194,9 +185,7 @@ $$\begin{aligned}
\\
&\le \alpha(t) - \alpha(t) + \alpha(t) \exp\!\bigg( \int_0^t \beta(r) \dd{r} \bigg)
\end{aligned}$$
-
-
-
+{% include proof/end.html id="proof-special" %}
diff --git a/source/know/concept/guiding-center-theory/index.md b/source/know/concept/guiding-center-theory/index.md
index 5368966..412c88b 100644
--- a/source/know/concept/guiding-center-theory/index.md
+++ b/source/know/concept/guiding-center-theory/index.md
@@ -72,6 +72,7 @@ we can use this average to approximately remove the finer dynamics,
and focus only on the guiding center.
+
## Uniform electric and magnetic field
Consider the case where $$\vb{E}$$ and $$\vb{B}$$ are both uniform,
@@ -149,6 +150,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## Non-uniform magnetic field
Next, consider a more general case, where $$\vb{B}$$ is non-uniform,
@@ -193,11 +195,8 @@ $$\begin{aligned}
\approx - \frac{u_L^2}{2 \omega_c} \nabla B
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-averages" -%}
We know what $$\vb{x}_L$$ is,
so we can write out $$(\vb{x}_L \cdot \nabla) \vb{B}$$
for $$\vb{B} = (B_x, B_y, B_z)$$:
@@ -290,9 +289,8 @@ $$\begin{aligned}
\end{pmatrix}
= - \frac{u_L^2}{2 \omega_c} \nabla B
\end{aligned}$$
+{% include proof/end.html id="proof-averages" %}
-
-
With this, the guiding center's equation of motion
is reduced to the following:
@@ -332,11 +330,8 @@ $$\begin{aligned}
\approx - u_{gc\parallel} \frac{\vb{R}_c}{R_c^2}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-curvature" -%}
Assuming that $$\vu{b}$$ does not explicitly depend on time,
i.e. $$\ipdv{\vu{b}}{t} = 0$$,
we can rewrite the derivative using the chain rule:
@@ -381,9 +376,8 @@ $$\begin{aligned}
= - \frac{\vu{R}_c}{R_c}
= - \frac{\vb{R}_c}{R_c^2}
\end{aligned}$$
+{% include proof/end.html id="proof-curvature" %}
-
-
With this, we arrive at the following equation of motion
for the guiding center:
diff --git a/source/know/concept/hamiltonian-mechanics/index.md b/source/know/concept/hamiltonian-mechanics/index.md
index 19e55b0..03ff2dd 100644
--- a/source/know/concept/hamiltonian-mechanics/index.md
+++ b/source/know/concept/hamiltonian-mechanics/index.md
@@ -15,6 +15,7 @@ It is built on the shoulders of [Lagrangian mechanics](/know/concept/lagrangian-
which is in turn built on [variational calculus](/know/concept/calculus-of-variations/).
+
## Definitions
In Lagrangian mechanics, use a Lagrangian $$L$$,
@@ -90,6 +91,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## Canonical equations
Lagrangian mechanics has a single Euler-Lagrange equation per object,
@@ -105,11 +107,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-canonical" -%}
For the first equation,
we differentiate $$H$$ with respect to $$q_n$$,
and use the chain rule:
@@ -148,9 +147,8 @@ $$\begin{aligned}
- 0 \pdv{L}{q_j} - p_j \pdv{\dot{q}_j}{p_n} \Big)
= \dot{q}_n
\end{aligned}$$
+{% include proof/end.html id="proof-canonical" %}
-
-
Just like in Lagrangian mechanics, if $$H$$ does not explicitly contain $$q_n$$,
then $$q_n$$ is called a **cyclic coordinate**, and leads to the conservation of $$p_n$$:
@@ -175,11 +173,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-dv-t" -%}
We differentiate via the multivariate chain rule,
insert the canonical equations,
and eventually recognize the PB definition:
@@ -192,9 +187,8 @@ $$\begin{aligned}
\\
&= \sum_{n} \Big( \pdv{A}{q_n} \pdv{H}{p_n} - \pdv{A}{p_n} \pdv{H}{q_n} \Big) + \pdv{A}{t}
\end{aligned}$$
+{% include proof/end.html id="proof-dv-t" %}
-
-
Assuming that $$H$$ does not explicitly depend on $$t$$,
the above property naturally leads us to an alternative
@@ -247,11 +241,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-transformation" -%}
Assuming that $$Q_n$$, $$P_n$$ and $$H$$ do not explicitly depend on $$t$$,
we use our expression for the $$t$$-derivative of an arbitrary quantity,
and apply the multivariate chain rule to it:
@@ -296,8 +287,8 @@ if and only if $$\{P_n, P_j\} = 0$$,
and $$\{Q_n, P_j\} = - \delta_{nj}$$.
The PB is anticommutative,
i.e. $$\{A, B\} = - \{B, A\}$$.
-
-
+{% include proof/end.html id="proof-transformation" %}
+
If you have experience with quantum mechanics,
the latter equation should look suspiciously similar
diff --git a/source/know/concept/heaviside-step-function/index.md b/source/know/concept/heaviside-step-function/index.md
index 15d1729..9f5d4ec 100644
--- a/source/know/concept/heaviside-step-function/index.md
+++ b/source/know/concept/heaviside-step-function/index.md
@@ -57,11 +57,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-fourier" -%}
In this case, it is easiest to use $$\Theta(0) = 1/2$$,
such that the Heaviside step function can be expressed
using the signum function $$\mathrm{sgn}(t)$$:
@@ -88,9 +85,8 @@ $$\begin{aligned}
&= A \pi \delta(s \omega) + \frac{A}{2} \pv{\int_{-\infty}^\infty \mathrm{sgn}(t) \exp(i s \omega t) \dd{t}}
= \frac{A}{|s|} \pi \delta(\omega) + i \frac{A}{s} \pv{\frac{1}{\omega}}
\end{aligned}$$
+{% include proof/end.html id="proof-fourier" %}
-
-
The use of $$\pv{}$$ without an integral is an abuse of notation,
and means that this result only makes sense when wrapped in an integral.
diff --git a/source/know/concept/holomorphic-function/index.md b/source/know/concept/holomorphic-function/index.md
index 5dde240..cf252c0 100644
--- a/source/know/concept/holomorphic-function/index.md
+++ b/source/know/concept/holomorphic-function/index.md
@@ -61,6 +61,7 @@ and imaginary parts satisfy these equations. This gives an idea of how
strict the criteria are to qualify as holomorphic.
+
## Integration formulas
Holomorphic functions satisfy **Cauchy's integral theorem**, which states
@@ -73,11 +74,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-int-theorem" -%}
Just like before, we decompose $$f(z)$$ into its real and imaginary parts:
$$\begin{aligned}
@@ -97,8 +95,8 @@ $$\begin{aligned}
Since $$f(z)$$ is holomorphic, $$u$$ and $$v$$ satisfy the Cauchy-Riemann
equations, such that the integrands disappear and the final result is zero.
-
-
+{% include proof/end.html id="proof-int-theorem" %}
+
An interesting consequence is **Cauchy's integral formula**, which
states that the value of $$f(z)$$ at an arbitrary point $$z_0$$ is
@@ -110,11 +108,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-int-formula" -%}
Thanks to the integral theorem, we know that the shape and size
of $$C$$ is irrelevant. Therefore we choose it to be a circle with radius $$r$$,
such that the integration variable becomes $$z = z_0 + r e^{i \theta}$$. Then
@@ -133,9 +128,8 @@ $$\begin{aligned}
&= \frac{f(z_0)}{2 \pi} \int_0^{2 \pi} \dd{\theta}
= f(z_0)
\end{aligned}$$
+{% include proof/end.html id="proof-int-formula" %}
-
-
Similarly, **Cauchy's differentiation formula**,
or **Cauchy's integral formula for derivatives**
@@ -149,11 +143,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-dv-formula" -%}
By definition, the first derivative $$f'(z)$$ of a
holomorphic function exists and is:
@@ -186,6 +177,5 @@ $$\begin{aligned}
Since the second-order derivative $$f''(z)$$ is simply the derivative of $$f'(z)$$,
this proof works inductively for all higher orders $$n$$.
-
-
+
+{% include proof/start.html id="proof-theorem" -%}
Starting from the definition of $$u_p(t)$$,
we shift the argument by some constant $$\tau$$,
and multiply both sides by the constant $$f(\tau)$$:
@@ -60,9 +57,8 @@ $$\begin{aligned}
\hat{L} \int_0^\infty f(\tau) \: u_p(t - \tau) \dd{\tau}
&= (f * u_p)(t) = \hat{L}\{ u(t) \} = f(t)
\end{aligned}$$
+{% include proof/end.html id="proof-theorem" %}
-
-
This is useful for solving initial value problems,
because any initial condition can be satisfied
diff --git a/source/know/concept/ito-integral/index.md b/source/know/concept/ito-integral/index.md
index f087f97..4a725e1 100644
--- a/source/know/concept/ito-integral/index.md
+++ b/source/know/concept/ito-integral/index.md
@@ -29,6 +29,7 @@ and $$\mathbf{E}[G_t^2]$$ is integrable for $$t \in [a, b]$$.
If $$I_t$$ exists, $$G_t$$ is said to be **Itō-integrable** with respect to $$B_t$$.
+
## Motivation
Consider the following simple first-order differential equation for $$X_t$$,
@@ -99,7 +100,8 @@ $$\begin{aligned}
\end{aligned}$$
For more information about applying the Itō integral in this way,
-see the [Itō calculus](/know/concept/ito-calculus/).
+see the [Itō calculus](/know/concept/ito-process/).
+
## Properties
@@ -138,11 +140,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-isometry" -%}
We write out the left-hand side of the Itō isometry,
where eventually $$h \to 0$$:
@@ -208,20 +207,16 @@ $$\begin{aligned}
\longrightarrow
\int_a^b \mathbf{E} \big[ G_t^2 \big] \dd{t}
\end{aligned}$$
+{% include proof/end.html id="proof-isometry" %}
-
-
Furthermore, Itō integrals are [martingales](/know/concept/martingale/),
meaning that the average noise contribution is zero,
which makes intuitive sense,
since true white noise cannot be biased.
-
-
-
-
-
+
+{% include proof/start.html id="proof-martingale" -%}
We will prove that an arbitrary Itō integral $$I_t$$ is a martingale.
Using additivity, we know that the increment $$I_t \!-\! I_s$$
is as follows, given information $$\mathcal{F}_s$$:
@@ -259,8 +254,7 @@ so from the Itō isometry we have $$\mathbf{E}[I]^2 < \infty$$,
and therefore $$\mathbf{E}[I] < \infty$$,
so $$I_t$$ has all the properties of a Martingale,
since it is trivially $$\mathcal{F}_t$$-adapted.
-
-
+{% include proof/end.html id="proof-martingale" %}
diff --git a/source/know/concept/ito-process/index.md b/source/know/concept/ito-process/index.md
index f192e28..2756e33 100644
--- a/source/know/concept/ito-process/index.md
+++ b/source/know/concept/ito-process/index.md
@@ -61,6 +61,7 @@ since only the current value of $$X_t$$ determines the future,
and $$B_t$$ is also a Markov process.
+
## Itō's lemma
Classically, given $$y \equiv h(x(t), t)$$,
@@ -83,11 +84,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-lemma" -%}
We start by applying the classical chain rule,
but we go to second order in $$x$$.
This is also valid classically,
@@ -133,14 +131,15 @@ $$\begin{aligned}
Where $$\chi_1^2(\dd{t})$$ is the generalized chi-squared distribution
with one term of variance $$\dd{t}$$.
-
-
+{% include proof/end.html id="proof-lemma" %}
+
The most important application of Itō's lemma
is to perform coordinate transformations,
to make the solution of a given Itō SDE easier.
+
## Coordinate transformations
The simplest coordinate transformation is a scaling of the time axis.
@@ -208,6 +207,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## Existence and uniqueness
It is worth knowing under what condition a solution to a given SDE exists,
@@ -232,11 +232,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-existence" -%}
If we define $$Y_t \equiv X_t^2$$,
then Itō's lemma tells us that the following holds:
@@ -275,9 +272,8 @@ $$\begin{aligned}
\\
&\le (Y_0 + 3 K t) \exp\!\big(3 K t\big)
\end{aligned}$$
+{% include proof/end.html id="proof-existence" %}
-
-
If a solution exists, it is also worth knowing whether it is unique.
Suppose that $$f$$ and $$g$$ satisfy the following inequalities,
@@ -301,11 +297,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-uniqueness" -%}
We define $$D_t \equiv X_t \!-\! Y_t$$ and $$Z_t \equiv D_t^2 \ge 0$$,
together with $$F_t \equiv f(X_t) \!-\! f(Y_t)$$ and $$G_t \equiv g(X_t) \!-\! g(Y_t)$$,
such that Itō's lemma states:
@@ -347,9 +340,8 @@ $$\begin{aligned}
\\
&\le Z_0 \exp\!\Big( \big( 2 K \!+\! K^2 \big) t \Big)
\end{aligned}$$
+{% include proof/end.html id="proof-uniqueness" %}
-
-
Using these properties, it can then be shown
that if all of the above conditions are satisfied,
diff --git a/source/know/concept/laplace-transform/index.md b/source/know/concept/laplace-transform/index.md
index c7f352a..94c3742 100644
--- a/source/know/concept/laplace-transform/index.md
+++ b/source/know/concept/laplace-transform/index.md
@@ -35,6 +35,7 @@ using [partial fraction decomposition](/know/concept/partial-fraction-decomposit
and then looking up the individual terms.
+
## Derivatives
The derivative of a transformed function is the transform
@@ -55,11 +56,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-dv-s" -%}
The exponential $$\exp(- s t)$$ is the only thing that depends on $$s$$ here:
$$\begin{aligned}
@@ -69,9 +67,8 @@ $$\begin{aligned}
&= \int_0^\infty (-t)^n f(t) \exp(- s t) \dd{t}
= (-1)^n \hat{\mathcal{L}}\{t^n f(t)\}
\end{aligned}$$
+{% include proof/end.html id="proof-dv-s" %}
-
-
The Laplace transform of a derivative introduces the initial conditions into the result.
Notice that $$f(0)$$ is the initial value in the original $$t$$-domain:
@@ -98,11 +95,8 @@ and $$f^{(0)}(t) = f(t)$$.
As an example, $$\hat{\mathcal{L}}\{f'''(t)\}$$ becomes
$$- f''(0) - s f'(0) - s^2 f(0) + s^3 \tilde{f}(s)$$.
-
-
-
-
-
+
+{% include proof/start.html id="proof-dv-t" -%}
We integrate by parts and use the fact that $$\lim_{x \to \infty} \exp(-x) = 0$$:
$$\begin{aligned}
@@ -116,8 +110,7 @@ $$\begin{aligned}
And so on.
By partially integrating $$n$$ times in total we arrive at the conclusion.
-
-
+
+{% include proof/start.html id="proof-density" -%}
Starting from the general definition of $$\hat{n}$$,
we write out the field operators $$\hat{\Psi}(\vb{r})$$,
and insert the known non-interacting single-electron orbitals
@@ -210,8 +207,8 @@ $$\begin{aligned}
The summation variable $$\vb{k}$$ has an associated spin $$\sigma$$,
and $$\hat{n}$$ does not carry any spin.
-
-
+{% include proof/end.html id="proof-density" %}
+
When neglecting interactions, it is tradition to rename $$\chi$$ to $$\chi_0$$.
We insert $$\hat{n}$$, suppressing spin:
@@ -290,12 +287,10 @@ $$\begin{aligned}
= \hat{c}_{\vb{k}}^\dagger \hat{c}_{\vb{k}} - \hat{c}_{\vb{k} + \vb{q}}^\dagger \hat{c}_{\vb{k} + \vb{q}}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-commutator" -%}
In general, for any single-particle states labeled by $$m$$, $$n$$, $$o$$ and $$p$$, we have:
+
$$\begin{aligned}
\comm{\hat{c}_m^\dagger \hat{c}_n}{\hat{c}_o^\dagger \hat{c}_p}
&= \hat{c}_m^\dagger \hat{c}_n \hat{c}_o^\dagger \hat{c}_p - \hat{c}_o^\dagger \hat{c}_p \hat{c}_m^\dagger \hat{c}_n
@@ -319,8 +314,8 @@ $$\begin{aligned}
In this case, $$m = p = \vb{k}$$ and $$n = o = \vb{k} \!+\! \vb{q}$$,
so the Kronecker deltas are unnecessary.
-
-
+{% include proof/end.html id="proof-commutator" %}
+
We substitute this result into $$\chi_0$$,
and reintroduce the spin index $$\sigma$$ associated with $$\vb{k}$$:
diff --git a/source/know/concept/matsubara-greens-function/index.md b/source/know/concept/matsubara-greens-function/index.md
index fdcadb3..fd46abf 100644
--- a/source/know/concept/matsubara-greens-function/index.md
+++ b/source/know/concept/matsubara-greens-function/index.md
@@ -83,11 +83,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-period" -%}
First $$\tau \!-\! \tau' < 0$$.
We insert the argument $$\tau \!-\! \tau' \!+\! \hbar \beta$$,
and use the cyclic property:
@@ -133,9 +130,8 @@ $$\begin{aligned}
\\
&= \pm C_{AB}(\tau \!-\! \tau')
\end{aligned}$$
+{% include proof/end.html id="proof-period" %}
-
-
Due to this limited domain $$\tau \in [-\hbar \beta, \hbar \beta]$$,
the [Fourier transform](/know/concept/fourier-transform/)
@@ -157,11 +153,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-fourier-def" -%}
We will prove that one is indeed the inverse of the other.
We demand that the inverse FT of the forward FT of $$C_{AB}(\tau)$$
is simply $$C_{AB}(\tau)$$ again:
@@ -198,9 +191,8 @@ $$\begin{aligned}
\\
&= C_{AB}(\tau)
\end{aligned}$$
+{% include proof/end.html id="proof-fourier-def" %}
-
-
Let us now define the **Matsubara frequencies** $$\omega_n$$
as a species-dependent subset of $$k_n$$:
@@ -228,11 +220,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-fourier-alt" -%}
We split the integral, shift its limits,
and use the (anti)periodicity of $$C_{AB}$$:
@@ -265,9 +254,8 @@ $$\begin{aligned}
\\
&= \frac{1}{2} \big( 1 \pm e^{-i k_n \hbar \beta} \big) \int_{-\hbar \beta}^0 C_{AB}(\tau) \: e^{i k_n \tau} \dd{\tau}
\end{aligned}$$
+{% include proof/end.html id="proof-fourier-alt" %}
-
-
If we actually evaluate this,
we obtain the following form of $$C_{AB}$$,
@@ -283,11 +271,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-lehmann" -%}
For $$\tau \!-\! \tau' > 0$$, we start by expanding
in the many-particle eigenstates $$\Ket{n}$$:
@@ -363,8 +348,8 @@ $$\begin{aligned}
\end{aligned}$$
Where swapping $$n$$ and $$n'$$ gives the desired result.
-
-
+{% include proof/end.html id="proof-lehmann" %}
+
This gives us the primary use of the Matsubara Green's function $$C_{AB}$$:
calculating the retarded $$C_{AB}^R$$ and advanced $$C_{AB}^A$$.
diff --git a/source/know/concept/maxwell-bloch-equations/index.md b/source/know/concept/maxwell-bloch-equations/index.md
index b306c7d..ba8a677 100644
--- a/source/know/concept/maxwell-bloch-equations/index.md
+++ b/source/know/concept/maxwell-bloch-equations/index.md
@@ -296,11 +296,8 @@ $$\begin{aligned}
\equiv \frac{\gamma_g - \gamma_e}{\gamma_g + \gamma_e}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-inversion-decay" -%}
We introduce some new terms, and reorganize the expression:
$$\begin{aligned}
@@ -324,9 +321,8 @@ $$\begin{aligned}
\\
&= \gamma_\parallel ( d_0 - d )
\end{aligned}$$
+{% include proof/end.html id="proof-inversion-decay" %}
-
-
With this, the equation for the population inversion $$d$$
takes the following final form:
diff --git a/source/know/concept/multi-photon-absorption/index.md b/source/know/concept/multi-photon-absorption/index.md
index 5dd9887..80dbc9b 100644
--- a/source/know/concept/multi-photon-absorption/index.md
+++ b/source/know/concept/multi-photon-absorption/index.md
@@ -73,11 +73,8 @@ $$\begin{aligned}
= 2 \pi \: \delta(x) \: t
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-relation" -%}
First, observe that we can rewrite the fraction using an integral:
$$\begin{aligned}
@@ -119,9 +116,8 @@ $$\begin{aligned}
= 4 \pi^2 \delta^2(x)
= 2 \pi \: \delta(x) \: t
\end{aligned}$$
+{% include proof/end.html id="proof-relation" %}
-
-
## One-photon absorption
@@ -187,6 +183,7 @@ Note that this transition is only possible when $$\matrixel{u}{\vu{p}}{0} \neq 0
i.e. for any odd-numbered final state $$\Ket{u}$$.
+
## Two-photon absorption
Next, we go to second-order perturbation theory.
@@ -255,6 +252,7 @@ Notice that the rate is proportional to $$|\vb{E}|^4$$,
so this effect is only noticeable at high light intensities.
+
## Three-photon absorption
For third-order perturbation theory,
@@ -327,6 +325,7 @@ The rate is proportional to $$|\vb{E}|^6$$,
so this effect only appears at extremely high light intensities.
+
## N-photon absorption
A pattern has appeared in these calculations:
diff --git a/source/know/concept/parsevals-theorem/index.md b/source/know/concept/parsevals-theorem/index.md
index df90244..377f3a1 100644
--- a/source/know/concept/parsevals-theorem/index.md
+++ b/source/know/concept/parsevals-theorem/index.md
@@ -24,11 +24,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-fourier" -%}
We insert the inverse FT into the defintion of the inner product:
$$\begin{aligned}
@@ -68,9 +65,8 @@ $$\begin{aligned}
&= \frac{2 \pi A^2}{|s|} \int_{-\infty}^\infty f^*(x) \: g(x) \dd{x}
= \frac{2 \pi A^2}{|s|} \Inprod{f}{g}
\end{aligned}$$
+{% include proof/end.html id="proof-fourier" %}
-
-
For this reason, physicists like to define the Fourier transform
with $$A\!=\!B\!=\!1 / \sqrt{2\pi}$$ and $$|s|\!=\!1$$, because then it nicely
diff --git a/source/know/concept/repetition-code/index.md b/source/know/concept/repetition-code/index.md
index 678211e..89e6f4d 100644
--- a/source/know/concept/repetition-code/index.md
+++ b/source/know/concept/repetition-code/index.md
@@ -164,63 +164,12 @@ while $$ZZI$$ cannot protect the 3rd qubit.
But by using both, we know exactly which qubit was flipped
thanks to the eigenvalues:
-
-
-
- Error
-
-
- $$ZZI$$
-
-
- $$IZZ$$
-
-
-
-
- $$I$$
-
-
- $$+1$$
-
-
- $$+1$$
-
-
-
-
- $$X_1$$
-
-
- $$-1$$
-
-
- $$+1$$
-
-
-
-
- $$X_2$$
-
-
- $$-1$$
-
-
- $$-1$$
-
-
-
-
- $$X_1$$
-
-
- $$+1$$
-
-
- $$-1$$
-
-
-
+| **Error** | $$ZZI$$ | $$IZZ$$ |
+| :-: | :-: | :-: |
+| $$I$$ | $$+1$$ | $$+1$$ |
+| $$X_1$$ | $$-1$$ | $$+1$$ |
+| $$X_2$$ | $$-1$$ | $$-1$$ |
+| $$X_1$$ | $$+1$$ | $$-1$$ |
Where e.g. $$X_3$$ denotes that the 3rd qubit was flipped.
The measurement outcomes on the last three rows are called **error syndromes**,
diff --git a/source/know/concept/residue-theorem/index.md b/source/know/concept/residue-theorem/index.md
index b58e3c2..a0f515e 100644
--- a/source/know/concept/residue-theorem/index.md
+++ b/source/know/concept/residue-theorem/index.md
@@ -41,11 +41,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-theorem" -%}
From the definition of a meromorphic function,
we know that we can decompose $$f(z)$$ like so,
where $$h(z)$$ is holomorphic and $$z_p$$ are all its poles:
@@ -62,9 +59,8 @@ $$\begin{aligned}
&= \oint_C h(z) \dd{z} + \sum_{p} R_p \oint_C \frac{1}{z - z_p} \dd{z}
= \sum_{p} R_p \: 2 \pi i
\end{aligned}$$
+{% include proof/end.html id="proof-theorem" %}
-
-
This theorem might not seem very useful,
but in fact, by cleverly choosing the contour $$C$$,
diff --git a/source/know/concept/selection-rules/index.md b/source/know/concept/selection-rules/index.md
index 373486e..620e345 100644
--- a/source/know/concept/selection-rules/index.md
+++ b/source/know/concept/selection-rules/index.md
@@ -25,6 +25,7 @@ between $$\ell_i$$, $$\ell_f$$, $$m_i$$ and $$m_f$$, which, if not met,
guarantee that the above matrix element is zero.
+
## Parity rules
Let $$\hat{O}$$ denote any operator which is odd under spatial inversion
@@ -73,6 +74,7 @@ $$\begin{aligned}
\end{aligned}$$
+
## Dipole rules
Arguably the most common operator found in such matrix elements
@@ -87,11 +89,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-dipole-m" -%}
We know that the angular momentum $$z$$-component operator $$\hat{L}_z$$ satisfies:
$$\begin{aligned}
@@ -166,8 +165,8 @@ whenever $$\matrixel{f}{\hat{z}}{i} \neq 0$$.
Only if $$\matrixel{f}{\hat{z}}{i} = 0$$
does the previous rule $$\Delta m = \pm 1$$ hold,
in which case the inner products of $$\hat{x}$$ and $$\hat{y}$$ are nonzero.
-
-
+{% include proof/end.html id="proof-dipole-m" %}
+
Meanwhile, for the total angular momentum $$\ell$$ we have the following:
@@ -177,11 +176,8 @@ $$\begin{aligned}
}
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-dipole-l" -%}
We start from the following relation
(which is already quite a chore to prove):
@@ -190,11 +186,8 @@ $$\begin{aligned}
= 2 \hbar^2 (\vu{r} \hat{L}^2 + \hat{L}^2 \vu{r})
\end{aligned}$$
-
-
-
-
-
+
+{% include proof/start.html id="proof-dipole-l-commutator" -%}
To begin with, we want to find the commutator of $$\hat{L}^2$$ and $$\hat{x}$$:
$$\begin{aligned}
@@ -364,8 +357,8 @@ $$\begin{aligned}
At last, this brings us to the desired equation for $$\comm{\hat{L}^2}{\comm{\hat{L}^2}{\vu{r}}}$$,
with $$\vu{r} = (\hat{x}, \hat{y}, \hat{z})$$.
-
-
+{% include proof/end.html id="proof-dipole-l-commutator" %}
+
We then multiply this relation by $$\Bra{f} = \Bra{\ell_f m_f}$$ on the left
and $$\Ket{i} = \Ket{\ell_i m_i}$$ on the right,
@@ -458,9 +451,8 @@ $$\begin{aligned}
(\ell_f - \ell_i)^2
= 1
\end{aligned}$$
+{% include proof/end.html id="proof-dipole-l" %}
-
-
+
+{% include proof/start.html id="proof-rotation-scalar" -%}
Firstly, we look at the commutator of $$\hat{s}$$ with the $$z$$-component $$\hat{L}_z$$:
$$\begin{aligned}
0
@@ -578,8 +567,8 @@ $$\begin{aligned}
Which means that the value of the matrix element
does not depend on $$m_i$$ (or $$m_f$$) at all.
-
-
+{% include proof/end.html id="proof-rotation-scalar" %}
+
Similarly, given a general (pseudo)vector operator $$\vu{V}$$,
which, by nature, must satisfy the following commutation relations,
@@ -631,6 +620,7 @@ $$\begin{gathered}
\end{gathered}$$
+
## Superselection rule
Selection rules are not always about atomic electron transitions, or angular momenta even.
diff --git a/source/know/concept/superdense-coding/index.md b/source/know/concept/superdense-coding/index.md
index ba6e898..4338205 100644
--- a/source/know/concept/superdense-coding/index.md
+++ b/source/know/concept/superdense-coding/index.md
@@ -27,63 +27,12 @@ Based on the values of the two classical bits $$(a_1, a_2)$$,
Alice performs the following operations on her side $$A$$
of the Bell state:
-