Expand knowledge base

1 files changed, 97 insertions, 14 deletions

diff --git a/content/know/concept/time-dependent-perturbation-theory/index.pdc b/content/know/concept/time-dependent-perturbation-theory/index.pdc
index fbb71b2..b16e3ee 100644
--- a/content/know/concept/time-dependent-perturbation-theory/index.pdc
+++ b/content/know/concept/time-dependent-perturbation-theory/index.pdc
@@ -17,7 +17,7 @@ markup: pandoc
 In quantum mechanics, **time-dependent perturbation theory** exists to deal
 with time-varying perturbations to the Schrödinger equation.
 This is in contrast to [time-independent perturbation theory](/know/concept/time-independent-perturbation-theory/),
-where the perturbation is is stationary.
+where the perturbation is stationary.
 
 Let $\hat{H}_0$ be the base time-independent
 Hamiltonian, and $\hat{H}_1$ be a time-varying perturbation, with
@@ -32,14 +32,14 @@ $\hat{H}_0 \ket{n} = E_n \ket{n}$ has already been solved, such that the
 full solution is:
 
 $$\begin{aligned}
-    \ket{\Psi_0(t)} = \sum_{n} c_n \ket{n} \exp(- i E_n t / \hbar)
+    \ket{\Psi_0(t)} = \sum_{n} c_n \ket{n} \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 Since these $\ket{n}$ form a complete basis, the perturbed wave function
 can be written in the same form, but with time-dependent coefficients $c_n(t)$:
 
 $$\begin{aligned}
-    \ket{\Psi(t)} = \sum_{n} c_n(t) \ket{n} \exp(- i E_n t / \hbar)
+    \ket{\Psi(t)} = \sum_{n} c_n(t) \ket{n} \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 We insert this ansatz in the time-dependent Schrödinger equation, and
@@ -50,32 +50,32 @@ $$\begin{aligned}
     &= \hat{H}_0 \ket{\Psi(t)} + \lambda \hat{H}_1 \ket{\Psi(t)} - i \hbar \dv{t} \ket{\Psi(t)}
     \\
     &= \sum_{n}
-    \Big( c_n \hat{H}_0 \ket{n} + \lambda c_n \hat{H}_1 \ket{n} - c_n E_n \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar)
+    \Big( c_n \hat{H}_0 \ket{n} + \lambda c_n \hat{H}_1 \ket{n} - c_n E_n \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp\!(- i E_n t / \hbar)
     \\
-    &= \sum_{n} \Big( \lambda c_n \hat{H}_1 \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp(- i E_n t / \hbar)
+    &= \sum_{n} \Big( \lambda c_n \hat{H}_1 \ket{n} - i \hbar \dv{c_n}{t} \ket{n} \Big) \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 We then take the inner product with an arbitrary stationary basis state $\ket{m}$:
 
 $$\begin{aligned}
     0
-    &= \sum_{n} \Big( \lambda c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \frac{d c_n}{dt} \braket{m}{n} \Big) \exp(- i E_n t / \hbar)
+    &= \sum_{n} \Big( \lambda c_n \matrixel{m}{\hat{H}_1}{n} - i \hbar \frac{d c_n}{dt} \braket{m}{n} \Big) \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 Thanks to orthonormality, this removes the latter term from the summation:
 
 $$\begin{aligned}
-    i \hbar \frac{d c_m}{dt} \exp(- i E_m t / \hbar)
-    &= \lambda \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} \exp(- i E_n t / \hbar)
+    i \hbar \frac{d c_m}{dt} \exp\!(- i E_m t / \hbar)
+    &= \lambda \sum_{n} c_n \matrixel{m}{\hat{H}_1}{n} \exp\!(- i E_n t / \hbar)
 \end{aligned}$$
 
 We divide by the left-hand exponential and define
-$\omega_{mn} = (E_m - E_n) / \hbar$ to get:
+$\omega_{mn} \equiv (E_m - E_n) / \hbar$ to get:
 
 $$\begin{aligned}
     \boxed{
         i \hbar \frac{d c_m}{dt}
-        = \lambda \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} \exp(i \omega_{mn} t)
+        = \lambda \sum_{n} c_n(t) \matrixel{m}{\hat{H}_1(t)}{n} \exp\!(i \omega_{mn} t)
     }
 \end{aligned}$$
 
@@ -85,7 +85,7 @@ Furthermore, it is useful to write this equation in integral form instead:
 
 $$\begin{aligned}
     c_m(t)
-    = c_m(0) - \lambda \frac{i}{\hbar} \sum_{n} \int_0^t c_n(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    = c_m(0) - \lambda \frac{i}{\hbar} \sum_{n} \int_0^t c_n(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp\!(i \omega_{mn} \tau) \dd{\tau}
 \end{aligned}$$
 
 If this cannot be solved exactly, we must approximate it. We expand
@@ -100,15 +100,15 @@ We then insert this into the integral and collect the non-zero orders of $\lambd
 
 $$\begin{aligned}
     c_m^{(1)}(t)
-    &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(0)} \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    &= - \frac{i}{\hbar} \sum_{n} \int_0^t c_n^{(0)} \matrixel{m}{\hat{H}_1(\tau)}{n} \exp\!(i \omega_{mn} \tau) \dd{\tau}
     \\
     c_m^{(2)}(t)
     &= - \frac{i}{\hbar} \sum_{n}
-    \int_0^t c_n^{(1)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    \int_0^t c_n^{(1)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp\!(i \omega_{mn} \tau) \dd{\tau}
     \\
     c_m^{(3)}(t)
     &= - \frac{i}{\hbar} \sum_{n}
-    \int_0^t c_n^{(2)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp(i \omega_{mn} \tau) \dd{\tau}
+    \int_0^t c_n^{(2)}(\tau) \matrixel{m}{\hat{H}_1(\tau)}{n} \exp\!(i \omega_{mn} \tau) \dd{\tau}
 \end{aligned}$$
 
 And so forth. The pattern here is clear: we can calculate the $(j\!+\!1)$th
@@ -116,7 +116,90 @@ correction using only our previous result for the $j$th correction.
 We cannot go any further than this without considering a specific perturbation $\hat{H}_1(t)$.
 
 
+## Sinusoidal perturbation
+
+Arguably the most important perturbation
+is a sinusoidally-varying potential, which represents
+e.g. incoming electromagnetic waves,
+or an AC voltage being applied to the system.
+In this case, $\hat{H}_1$ has the following form:
+
+$$\begin{aligned}
+    \hat{H}_1(\vec{r}, t)
+    \equiv V(\vec{r}) \sin\!(\omega t)
+    = \frac{1}{2 i} V(\vec{r}) \: \big( \exp\!(i \omega t) - \exp\!(-i \omega t) \big)
+\end{aligned}$$
+
+We abbreviate $V_{mn} = \matrixel{m}{V}{n}$,
+and take the first-order correction formula:
+
+$$\begin{aligned}
+    c_m^{(1)}(t)
+    &= - \frac{1}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)}
+    \int_0^t \exp\!\big(i \tau (\omega_{mn} \!+\! \omega)\big) - \exp\big(i \tau (\omega_{mn} \!-\! \omega)\big) \dd{\tau}
+    \\
+    &= \frac{i}{2 \hbar} \sum_{n} V_{mn} c_n^{(0)}
+    \bigg( \frac{\exp\!\big(i t (\omega_{mn} \!+\! \omega) \big) - 1}{\omega_{mn} + \omega}
+    + \frac{\exp\!\big(i t (\omega_{mn} \!-\! \omega) \big) - 1}{\omega_{mn} - \omega} \bigg)
+\end{aligned}$$
+
+For simplicity, we let the system start in a known state $\ket{a}$,
+such that $c_n^{(0)} = \delta_{na}$,
+and we assume that the driving frequency is close to resonance $\omega \approx \omega_{ma}$,
+such that the second term dominates the first, which can then be neglected.
+We thus get:
+
+$$\begin{aligned}
+    c_m^{(1)}(t)
+    &= i \frac{V_{ma}}{2 \hbar} \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) \big) - 1}{\omega_{ma} - \omega}
+    \\
+    &= i \frac{V_{ma}}{2 \hbar}
+    \frac{\exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big) - \exp\!\big(\!-\! i t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega}
+    \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big)
+    \\
+    &= - \frac{V_{ma}}{\hbar}
+    \frac{\sin\!\big( t (\omega_{ma} \!-\! \omega) / 2 \big)}{\omega_{ma} - \omega}
+    \: \exp\!\big(i t (\omega_{ma} \!-\! \omega) / 2 \big)
+\end{aligned}$$
+
+Taking the norm squared yields the **transition probability**:
+the probability that a particle that started in state $\ket{a}$
+will be found in $\ket{m}$ at time $t$:
+
+$$\begin{aligned}
+    \boxed{
+        P_{a \to m}
+        = |c_m^{(1)}(t)|^2
+        = \frac{|V_{ma}|^2}{\hbar^2} \frac{\sin^2\!\big( (\omega_{ma} - \omega) t / 2 \big)}{(\omega_{ma} - \omega)^2}
+    }
+\end{aligned}$$
+
+The result would be the same if $\hat{H}_1 \equiv V \cos\!(\omega t)$.
+However, if instead $\hat{H}_1 \equiv V \exp\!(- i \omega t)$,
+the result is larger by a factor of $4$,
+which can cause confusion when comparing literature.
+
+In any case, the probability oscillates as a function of $t$
+with period $T = 2 \pi / (\omega_{ma} \!-\! \omega)$,
+so after one period the particle is certain to be back in $\ket{a}$.
+
+However, when regarded as a function of $\omega$,
+the probability takes the form of
+a sinc-function centred around $(\omega_{ma} \!-\! \omega)$,
+so it is highest for transitions with energy $\hbar \omega = E_m \!-\! E_a$.
+
+Also note that the sinc-distribution becomes narrower over time,
+which roughly means that it takes some time
+for the system to "notice" that
+it is being driven periodically.
+In other words, there is some "inertia" to it.
+
+
+
 ## References
 1.  D.J. Griffiths, D.F. Schroeter,
     *Introduction to quantum mechanics*, 3rd edition,
     Cambridge.
+2.  R. Shankar,
+    *Principles of quantum mechanics*, 2nd edition,
+    Springer.