quantum mechanics - Deriving the optical Bloch equations from the von Neumann equations

Is it possible to derive the optical Bloch equations for a 2-level-system driven by an oscillating EM-Field from the von Neumann equation for the density operator?

I'm assuming a system consisting of the states $|g \rangle$ and $ e \rangle$. Those are eigenstates of the Hamiltonian $\hat{H}_0$ with energies $E_e$ and $E_g$. The whole hamiltonian should be a sum of $\hat{H}_0$ and $\hat{H}_E = \vec{r_z} E_0 \cos{\omega t}$. Because of the dipole operator, the diagonal matrix elements of $\hat{H}_E$ will disappear:

$$ \langle g | \hat{H}_E | g \rangle = \langle e | \hat{H}_E | e \rangle = 0 \\ $$ $$ \langle g | \hat{H}_E | e \rangle = \langle e | \hat{H}_E | g \rangle^* = \Omega_{Rabi} \hbar \cos{\omega t} \\ $$

Let's say i'm interested in the differential equation for the first density matrix element $\rho_gg$, and I know it's supposed to look like this (According to my professor): $$ \frac{d \rho_{gg}}{dt} = \frac i2 \Omega^*_{Rabi}e^{-i(\omega-\omega_0)t} \rho_{ge} - \frac i2 \Omega_{Rabi}e^{i(\omega-\omega_0)t} \rho_{eg} $$

However, if I try to derive this:

$$ \frac{d}{dt}\hat{\rho} = \frac{i}{\hbar} (\hat{\rho} (\hat{H}_0 + \hat{H}_E) - (\hat{H}_0 + \hat{H}_E)\hat{\rho}) \\ \frac{d}{dt} \rho_{gg} = \frac{i}{\hbar} \langle g |(\hat{\rho} (\hat{H}_0 + \hat{H}_E) - (\hat{H}_0 + \hat{H}_E)\hat{\rho})| g\rangle \\ = \frac{i}{\hbar} ( \langle g |\hat{\rho} \hat{H}_0| g\rangle + \langle g |\hat{\rho} \hat{H}_E| g\rangle - \langle g | \hat{H}_0 \hat{\rho}| g\rangle + \langle g | \hat{H}_E \hat{\rho}| g\rangle) \\ = \frac{i}{\hbar} (E_g \langle g |\hat{\rho}| g\rangle + \langle g |\hat{\rho} \hat{H}_E| g\rangle - E_g \langle g | \hat{\rho}| g\rangle + \langle g | \hat{H}_E \hat{\rho}| g\rangle) \\ = \frac{i}{\hbar} ( \langle g |\hat{\rho} \hat{H}_E| g\rangle - \langle g | \hat{H}_E \hat{\rho}| g\rangle) \\ = \frac{i}{\hbar} ( \langle g |\hat{\rho} |g \rangle \langle g | \hat{H}_E| g\rangle + \langle g |\hat{\rho} | e \rangle \langle e| \hat{H}_E| g\rangle - \langle g | \hat{H}_E |g \rangle \langle g | \hat{\rho}| g\rangle - \langle g | \hat{H}_E |e \rangle \langle e | \hat{\rho}| g\rangle) \\ = \frac{i}{\hbar} ( \rho_{ge} \Omega_{Rabi}^* \cos{\omega t} - \rho_{eg} \Omega_{Rabi} \cos{\omega t}) $$

So now here I stand and don't know wether I made a mistake, or wether it's not possible without additional assumptions. I don't know how possibly something like $e^{i\omega_0 t}$ should appear in this equation.

Answer

To arrive at the equations that my professor gave, I have to assume different states for $| e \rangle$ and $| g \rangle$. While I used the time independent, you can also use the same states multiplied by a phase-factor $| e \rangle =e^{-i\omega_g t}| e\rangle$ and $| g \rangle =g^{-i\omega_g t}| e\rangle$. Using them, the matrix elements of $\hat{H}_E$ are:

$$ \langle \tilde{g} | \hat{H}_E | \tilde{e} \rangle = \langle \tilde{e} | \hat{H}_E | \tilde{g} \rangle^* = \Omega_{Rabi} \hbar \cos{\omega t}e^{i \omega_0 t} \\ $$

With $\omega_0 = \omega_e - \omega_g$. The rotating wave approximation yields the desired result in the question:

$$ \frac{d \rho_{gg}}{dt} = \frac i2 \Omega^*_{Rabi}e^{-i(\omega-\omega_0)t} \rho_{eg} - \frac i2 \Omega_{Rabi}e^{i(\omega-\omega_0)t} \rho_{ge} $$

If conservation of energy was wrong, how would we know about it?

Suppose you just started learning physics and you've been introduced to conservation of energy and kinetic energy. Apart from those concepts you know next to nothing. Then you observe an inelastic collision. You measure the speeds of the objects before and after the collision and you are puzzled because kinetic energy is the only form of energy you know and you see it's clearly not conserved. You conclude that either:

a) Conservation of energy is wrong.

b) The formula $E_k = mv^2/2$ is wrong.

c) There is some other form of energy you didn't account for.

HOW do you know which one of those scenarios is true? Can you measure the total amount of energy contained in those two objects before and after the collision and reassure yourself that everything is okay, energy hasn't gone anywhere, it just changed its form? If you observe an object that seems to gain energy from nothing, how will you know whether conservation of energy fails or there is some undiscovered form of energy that you don't know how to measure yet?

homework and exercises - Synchronizing Pendulums

Assume we have a frictionless pendulum of length $l$ with mass $m$. This pendulum hangs from some weightless contraption, which is itself bolted to a platform. This platform can move horizontally in the direction of the swing of the pendulum. There are no other forces than gravity.

If the pendulum is set in motion, as it swings on one direction, the whole platform and the pendulum move to the direction of swing and when it swings to the other direction, again the whole system moves to that direction.

Question 1

At time $t$, what is the angle $\theta(t)$ of swing of the pendulum away from a perpendicular line through the point the pendulum swings?

Assume we have another pendulum, identical to the other one, and it too is then bolted to the same platform as the other one. When set in motion, the pendulums have same frequency. Assume that pendulum 1 starts at angle $\theta_1(0)$ and the other at $\theta_2(0).$

Question 2

What are $\theta_1(t)$ and $\theta_2(t)$?

Now, if $\theta_1(0) = \theta_2(0)$, then it would seem intuitive that $\theta_1(t) = \theta_2(t).$

Also, if $\theta_1(0) = -\theta_2(0)$ it would seem that $\theta_1(t) = -\theta_2(t).$ Now the movement of the whole system, platform included, is $0$, as the movements cancel each other out.

The motivation for this question is the video showing a large number of metronomes, out of sync, on a moving platform, synchronizing over time. This was demonstrated on recent episode of Mythbusters. They used metronomes, I should think pendulums are identical when it comes to this property.

Answer

This is actually an interesting problem in classical mechanics, dating back to Huygens. We'll work with the three variables you define in the question, namely $(x,\theta_1, \theta_2)$. Also we'll set your $m = l = g = 1$ for simplicity.

Kinetic Energy

The position vector of the first pendulum bob is

$$\mathbb{r}_1 = (x+\sin\theta_1, \cos\theta_1)$$

whence we deduce its kinetic energy to be

$$2T_1 = \dot{x}^2 + 2\dot{x}\dot{\theta_1}\cos\theta_1 + \dot{\theta_1}^2$$

We can similarly find the kinetic energy of the second bob.

Finally we must take into account the kinetic energy of the support with mass $M$.

$$2T_3 = M\dot{x}^2$$

It's interesting to keep $M\neq 0$ since different values of $M$ give different behaviour.

Potential Energy

We assume gravity acts on the bobs as usual, producing potential energy terms of form $\cos\theta$. We also assume an elastic potential $kx^2$ pulling the table back to equilibrium. Overall we have

$$V = kx^2 - \cos\theta_1-\cos\theta_2$$

Equations of Motion

One can easily write down the Lagrangian $L=T-V$ and from it deduce the equations of motion

$$\ddot{\theta}+\ddot{x}\cos\theta+\sin\theta-\dot{x}\dot{\theta}\sin{\theta} = 0$$

$$(M+2)\ddot{x}+\ddot{\theta_1}\cos\theta_1+\ddot{\theta_2}\cos\theta_2 - \dot{\theta_1}^2\sin\theta_1-\dot{\theta_2}^2\sin{\theta_2} + kx = 0$$

Interestingly when I put these into Mathematica, there was no synchronization! It turns out the missing ingredient is damping.

Damping

Intuitively the phase difference between the pendulums must drift in a periodic way in the absence of any dissipative effects. Indeed that's what you see when numerically solving the above equations with Mathematica.

Recall that damping is usually modelled as an additive term proportional to the velocity. Adding in such terms for $\theta_1$, $\theta_2$ and $x$ now does produce the desired synchronization behaviour. For my initial conditions and choice of constants we get antiphase locking.

Summary of the Physics

Momentum transfer through a connecting medium coupled with dissipative effects leads to synchronization.

Better Models

To fully model the video mentioned you'd need a forcing term from the escapement mechanism of the metronomes. You can read about such an approach here. See also this Wolfram demonstration and the papers it references.

Towards Chaos

Evidently this setup is nonlinear and so generically displays chaotic behaviour. The study of such systems is particularly important in chemistry and biology. Here is a good introduction.

If you want to play around with this behaviour yourself, here's my rudimentary Mathematica code. Try playing with the constants and initial conditions.

sol = NDSolve[{30 x''[t] + y''[t] Cos[y[t]] + z''[t] Cos[z[t]] - 
     y'[t]^2 Sin[y[t]] - z'[t]^2 Sin[z[t]] + 30 x[t] + 2 x'[t] == 0, 
   y''[t] + x''[t] Cos[y[t]] + Sin[y[t]] - x'[t] y'[t] Sin[y[t]] + 
     0.02 y'[t] == 0, 
   z''[t] + x''[t] Cos[z[t]] + Sin[z[t]] - x'[t] z'[t] Sin[z[t]] + 
     0.02 z'[t] == 0, x[0] == 0, x'[0] == 0,  y[0] == Pi/10, 
   y'[0] == 0, z[0.5] == 1, z'[2] == 0}, {x, y, z}, {t, 0, 1000}]
Plot[{Evaluate[y[t] /. sol], Evaluate[z[t] /. sol]}, {t, 0, 250}, 
 PlotRange -> All]

quantum field theory - CPT transformation for bilinears

In the page 5 of the document 'CPT Symmetry and Its Violation' by Ralf Lehnert (https://core.ac.uk/download/pdf/80103866.pdf), appears a discussion about how the spin-statistics theorem applies to the CPT theorem proof. It is said that for 2 spinors $\chi, \psi$, CPT transformations looks like:

$$ \bar{\chi}\psi \rightarrow -\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T} = \dots = (\bar{\chi} \psi)^\dagger $$

Nevertheless, from the left hand side of the first equal symbol I derive,

$$ -\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T} = (-\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T})^{\dagger\ *} $$

Since a bilinear and its transpose is the same thing. Now I'm going to use introduce inside bracket the conjugation operation represented by $*$. Then,

$$ (-\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T})^{\dagger\ *} = -(\chi^\dagger \gamma^0 \psi)^\dagger = -(\bar{\chi}\psi)^\dagger $$

So, my result has different sign from the one in the document. It is no conflict with the usual CPT result that says $\bar{\psi}\psi \rightarrow \bar{\psi}\psi$ since you can choose $\chi = \psi$ and due to anti-commutation of the 'bar' fields with fields you get precisely that result. Otherwise, it would be, $\bar{\psi}\psi \rightarrow -\bar{\psi}\psi$

Am I right or I'm loosing something?

Answer

In the text, you can see that the CPT transformation can be written as

$$ \bar{\chi}\psi \rightarrow -\chi^{\dagger\ T \ \dagger} \gamma^0 \psi^{\dagger\ T} = -(\psi^T \gamma^0 \chi^{T\ \dagger})^\dagger $$

If you go on with that expression,

$$-(\psi^T \gamma^0 \chi^{T\ \dagger}) = -(\psi^T \gamma^0 \chi^*) = -(\psi^T \gamma^0 \chi^{T\ \dagger}) = -(\chi^\dagger \gamma^0 \psi)^T $$

And,

$$ -(\chi^\dagger \gamma^0 \psi)^T = -(\psi^T \gamma^0 \chi^*) = -\psi_i(\gamma^0)_{ij}\chi^*_j = +\chi^*_j(\gamma^0)_{ji}\psi_i = -\chi^\dagger\gamma^0\psi = \bar{\chi}\psi $$

$\gamma^0_{ij} = \gamma^0_{ji}$ and since $\gamma^0_{ii} = 0$ you can use without Dirac deltas the anti-commutation between $\chi$ and $\psi$ even if $\chi = \psi$

So under CPT,

$$ \bar{\chi}\psi \rightarrow (\bar{\chi}\psi)^\dagger $$

The key is not to consider that transpose or adjoint introduces sign. It's just as simple as if $A, B$ are fermion fields, then

$$ (AB)^T = B^T A^T,\quad (AB)^\dagger = B^\dagger A^\dagger \tag{A}$$

The second one comes from the definion of adjoint operator, i.e., if ${\cal O}$ is an operator, its adjoint ${\cal O}^\dagger$ is given by

$$ \langle f|{\cal O}g \rangle = \langle {\cal O}^\dagger f|g \rangle $$

So, if ${\cal O} = AB$ you have that,

$$ \langle f|ABg \rangle = \langle {A}^\dagger f|Bg \rangle = \langle B^\dagger A^\dagger f|g \rangle $$

The first one of Eq. (A) it's now a corollary that comes from the definition of adjoint as transpose plus complex conjugation.

A last remark is that it's NOT true that $(\bar{\chi}\psi)^T = \bar{\chi}\psi$, so in general

$$ (\bar{\chi}\psi)^T \neq \bar{\chi}\psi $$

This is due to $\bar{\chi}\psi$ is not a number, it's an operator and it's not true, in general, that an operator and its transpose is the same thing. I write this because I've seen it in other post related to similar questions about transposition and adjoit of bilinears, and I think that I have already proved it to be wrong in this answer. I recommend to visit Transposition of spinors

visible light - How are photons made?

I mean in manufacturing a bicycle we know how to "ensemble" a bicycle, what actions and "assembly of parts". So what steps are needed for make a photon?

Also is there a limit on how many photons for an emisor can make?

quantum mechanics - How can one see that the Hydrogen atom has $SO(4)$ symmetry?

1. 
   

   For solving hydrogen atom energy level by $SO(4)$ symmetry, where does the symmetry come from?

   
   


2. 
   

   How can one see it directly from the Hamiltonian?

   
   

Answer

The Hamiltonian for the hydrogen atom $$ H = \frac{\mathbf{p}^2}{2m} - \frac{k}{r} $$ describes an electron in a central $1/r$ potential. This has the same form as the Kepler problem, and the symmetries are similar. There is an obvious $SO(3)$ generated by the angular momentum $\mathbf{L} = \mathbf{r} \times \mathbf{p}$. In other words, the components of $\mathbf{L}$ satisfy $$ [L_i,L_j] = i \hbar \epsilon_{ijk}L_k . $$ A more subtle symmetry is given by the Laplace-Runge-Lenz vector $$ \mathbf{A} = \frac{1}{2m} ( \mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p}) - k \frac{\mathbf{r}}{r}. $$ The commutation relations involving $\mathbf{L}$ and $\mathbf{A}$ are $$ [L_i,A_j] = i\hbar \epsilon_{ijk} A_k \\ [A_i,A_j] = -i\hbar\epsilon_{ijk} \frac{2H}{m} L_k . $$ Up to the normalization of $\mathbf{L}$ this is the commutation relations of $SO(4)$. (Here I assume that we are considering a bound state whose energy $E$ is negative. If $E>0$ the above relation generate a non-compact $SO(3,1)$ symmetry.)

Furthermore, both $\mathbf{L}$ and $\mathbf{A}$ commute with the Hamiltonian, $$ [H,L_i] = 0, \qquad [H,A_i] = 0 $$ showing that they indeed generate symmetries of the hydrogen atom.

quantum field theory - A certain regularization and renormalization scheme

In a certain lecture of Witten's about some QFT in $1+1$ dimensions, I came across these two statements of regularization and renormalization, which I could not prove,

(1) $\int ^\Lambda \frac{d^2 k}{(2\pi)^2}\frac{1}{k^2 + q_i ^2 \vert \sigma \vert ^2} = - \frac{1}{2\pi} ln \vert q _ i \vert - \frac{1}{2\pi}ln \frac{\vert \sigma\vert}{\mu}$

(..there was an overall $\sum _i q_i$ in the above but I don't think that is germane to the point..)

(2) $\int ^\Lambda \frac{d^2 k}{(2\pi)^2}\frac{1}{k^2 + \vert \sigma \vert ^2} = \frac{1}{2\pi} (ln \frac{\Lambda}{\mu} - ln \frac{\vert \sigma \vert }{\mu} )$

I tried doing dimensional regularization and Pauli-Villar's (motivated by seeing that $\mu$ which looks like an IR cut-off) but nothing helped me reproduce the above equations.

I would glad if someone can help prove these above two equations.

Answer

Let's just look at the integral $$\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2}.$$ The other integrals should follow from this one. Introduce the Pauli-Villars regulator, $$\begin{eqnarray*} \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} &\rightarrow& \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} - \int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\Lambda^2} \\ &=& (\Lambda^2-\alpha^2)\int \frac{d^2k}{(2\pi)^2} \frac{1}{(k^2+\alpha^2)(k^2+\Lambda^2)} \\ &=& (\Lambda^2-\alpha^2)\int_0^1 dx\, \int\frac{d^2k}{(2\pi)^2} \frac{1}{(k^2 + \beta^2)^2} \\ &=& (\Lambda^2-\alpha^2)\int_0^1 dx\, \frac{1}{2} \frac{2\pi}{(2\pi)^2} \int_0^\infty dk^2\,\frac{1}{(k^2 + \beta^2)^2} \\ &=& (\Lambda^2-\alpha^2) \frac{1}{4\pi} \int_0^1 dx\, \frac{1}{\beta^2} \\ &=& (\Lambda^2-\alpha^2) \frac{1}{4\pi} \int_0^1 dx\, \frac{1}{\Lambda^2 - x(\Lambda^2-\alpha^2)} \\ &=& -\frac{1}{2\pi} \ln \frac{|\alpha|}{\Lambda} \end{eqnarray*}$$ Where we have combined denominators with the Feynman parameter $x$, with the intermediate variable $\beta^2 = \Lambda^2 - x(\Lambda^2-\alpha^2)$. Of course, this could also be approached with dimensional regularization with the same result.

Addendum: After regularization we must renormalize. Using the minimal subtraction prescription we find $$\int \frac{d^2k}{(2\pi)^2} \frac{1}{k^2+\alpha^2} \rightarrow -\frac{1}{2\pi} \ln \frac{|\alpha|}{\mu},$$ as required.