symmetry breaking - Coupling constant in electroweak theory

Electroweak theory has two coupling constants before and after Spontaneous Symmetry Breaking (SSB) each one for $SU(2)_L$ and $U(1)_Y$, though they are connected by Weinberg angle after SSB. My question is, how is unification complete with two independent couplings before SSB. The motive for unification is a single unified force with certain range and (coupling)strength.

Answer

The two parts of the electroweak gauge group do not separately describe the weak and the electromagnetic force.

The unified group here is not $$ SU(2)_\mathrm{weak} \times U(1)_\mathrm{em}$$ but rather $$ SU(2)_L \times U(1)_Y $$ and the electric charge arises as a linear combination of hypercharge and weak isospin.

Therefore, although the group is not simple, the weak and electromagnetic forces have been unified, giving rise to two other forces.

Edit: To adress the matter of coupling constants:

Indeed, before SSB there are two independent coupling constants $g'$ (for the $U(1)$) and $g$ (for $SU(2)$). One way to relate them to parameters after SSB is to think of the couplings constant $g$ vanishing, but a new parameter arising: the Weinberg angle. The Weinberg angle $\theta_W$ determines, what linear combination of of the neutral vector bosons $W_3$ from $SU(2)$ and $B$ from $U(1)$ turn into the massive $Z$ boson and what combination turns into the massless $\gamma$.

The Weinberg angle is determined through the gauge couplings as $$ \cos\theta_w = \frac{g}{\sqrt{g^2 + g'^2}}.$$ In other words, in the breaking $SU(2)_L \times U(1)_Y$ both groups get broken, but there exists a linear combination of generators that remains unbroken. The $U(1)$ spanned by this generator does not relate 1:1 to either gauge group before SSB, though!

The coupling constant for the photon now relates to the couplings before SSB through the Weinberg angle $$ e = g \sin\theta_w = g' \cos\theta_w.$$

lateral thinking - A double-agent with a conundrum

You are a double-agent in the most important war of the 20th century, The Falklands war. For years you have been playing the United Kingdom and Argentinian governments against each other, taking bribes from both sides and it has caught up with you.

One day travelling you are snatched off the street, kidnapped, drugged and placed on a private plane. You wake up in a cell, you know you have been taken by one of the governments to their capital for interrogation and treason charges but you cannot tell which has abducted you. You have perfected both the UK and Argentinian languages and accents and you know you can convince the guards to let you go if you were to speak in their native tongue asking for your government contact. However speaking the wrong language first will certainly turn out very poorly for you indeed.

You look around your cell, it is a large room with high ceilings, a musty smell, and thick dust covering everything. Based on being unable to hear anything and the constant temperature you conclude you are underground. Nothing about the architecture is distinctive. The only things in the cell are a single metal chair bolted to the floor in the center of the room with a spotlight hanging from a cord right above the chair illuminating the seat brightly leaving the rest of the room in shadow. You know it is a matter of time before you are tied into the chair and interrogated by shadowy agents of one of the governments.

You search desperately for a clue as to where you are. You kick the chair and stub your toe; it is solid. You claw at the light, looking for a marking or a way to open it and see the voltage rating but cannot find either. You yank and pull on it to test how secure it is and cannot dislodge it from the ceiling to get at the cord. You try to swing it far enough to smash against a wall but it will not reach. You search the chair for a makers mark but can find none. You know much more action will invite suspicion even if you know the right language so you stand back and think.

The guard watching on CCTV notices, but decides it is normal behaviour for someone waking up in a cell regardless of guilt so is not immediately suspicious. As far as he can tell you then pace around the room randomly for a few hours deep in thought.

You suddenly walk towards the door and knock, with 100% confidence you correctly ask in perfect Spanish for your government handler who promptly releases you.

How did you know what government to ask for?

• You can assume you are correct in all your assumptions mentioned in the puzzle.


• You have no items other than the clothes and shoes you were wearing when abducted. no cell phone, no watch, no pocket lint etc.

Answer

I don't know how scientifically accurate this is, but it might be possible to use the light as a foucault pendulum to determine which hemisphere you are in. Of course, if there is a toilet (which is not mentioned), a flush would do the trick.

What does conservation of probability mean in Classical Mechanics and why is it true?

In the context of the Liouville equation, regularly the conservation of probability is invoked. (Of course, the overall probability is always conserved but this is a truism and not what is meant here. See also the discussion here.)

But what concretely does this means in the context of Classical Mechanics and why is it true?

Let's say that for some reason only three initial states are possible $A$, $B$, and $C$ and we can describe our system using the probability distribution: \begin{align} \rho(t=0,A) &= 0.7 \notag \\ \rho(t=0,B) &= 0.2 \notag \\ \rho(t=0,C) &= 0.1 \end{align} The total probability to find our system in a region $R$ which contains $A$ and $B$ but not $C$ is therefore $P(R,t=0)=90$%.

Now, as time passes on ($t\to t=t_1$) $R$ gets dragged by the Hamiltonian flow and becomes $\tilde R$. Moreover, our phase space points $A$, $B$, $C$ also get dragged and become $\tilde A$, $\tilde B$, $\tilde C$. The statement regularly used in the derivation of the Liouville equation is $$ P(\tilde R, t= t_1) = P(R,t=0)=90 \% \, . $$

Is this correct because the time-evolution of the phase space points $A$, $B$, $C$ and the time-evolution of the region $R$ are both described by Hamilton's equation? (Formulated simpler: because we move them around equally as time passes on?)

And secondly, what does it imply for our concrete probability distribution? Liouville's theorem tells us the phase space volume is constant. And this in combination with the conservation of probability tells us that $\frac{d \rho}{dt}=0$. But does this tells us that while the probabilities at our original locations can be wildly different ($\frac{\partial \rho }{\partial t}\neq 0$), e.g.

\begin{align} \rho(t=t_1,A) &= 0.3 \notag \\ \rho(t=t_1,B) &= 0.5 \notag \\ \rho(t=t_1,C) &= 0.2 \end{align}

we certainly have

\begin{align} \rho(t=t_1,\tilde A) &= 0.7 \notag \\ \rho(t=t_1,\tilde B) &= 0.2 \notag \\ \rho(t=t_1,\tilde C) &= 0.1 \quad ? \end{align}

quantum mechanics - How exactly is "normal-ordering an operator" defined?

(In this question, I'm only talking about the second-quantization version of normal ordering, not the CFT version.)

Most sources (e.g. Wikipedia) very quickly define normal-ordering as "reordering all the ladder operators so that all of the creation operators are to the left of all of the annihilation operators." This definition is extremely vague, and I want to make sure I understand the actual definition.

If I understand correctly, people use the phrase "normal-order an operator" to mean two inequivalent things. Sometimes they mean "use the (anti)commutation relations to rewrite the operator so that it is normal-ordered (without changing the operator itself)." Under this (unambiguous) definition, we have that the normal-ordered form of the operator $a a^\dagger$ is $a^\dagger a + 1$. We can use this definition to put any operator into canonical form (up to a sign, in the fermionic case. We can fix this sign ambiguity by specifying a canonical ordering of the single-site Hilbert spaces.)

But sometimes the verb "normal-order" is used in a different way, which can actually change the operator. I believe that this definition is the one usually represented by surrounding the operator with colons. If I understand correctly, this procedure is defined as "use the (anti)commutation relations $\left[ a_i, a_j \right]_\pm = \left[ a_i^\dagger, a_j^\dagger \right]_\pm = 0$ to move all the creation operators to the left of all the annihilation operators, while ignoring the $\left[ a_i, a_j^\dagger \right]_\pm = \delta_{ij}$ (anti)commutation relation and pretending that its RHS were zero."

This procedure obviously seems a bit arbitrary and unmotivated. Moreover, it doesn't seem entirely well-defined. It's fine for products of ladder operators, but the problem is that under this definition, normal-ordering does not distribute over addition:

$$ {:} a^\dagger a{:} \ =\ a^\dagger a\ =\ a a^\dagger - 1\ $$ but $${:} a a^\dagger{:} -1\ = a^\dagger a - 1.$$

It's therefore not clear how to define normal-ordering for a general operator, i.e. a general linear combination of products of ladder operators. And of course, whether or not an operator is a nontrivial sum of products of ladder operators depends on how you write it; we can equivalently write the same operator as $a^\dagger a$ (only one summand) or as $a a^\dagger - 1$ (multiple summands).

From this, I conclude that (under the second definition) "normal-ordering an operator" is actually an abuse of terminology; we can only meaningfully normal-order certain particular expressions for some operators. Is this correct? If not, how does one define the normal-ordering of a linear combination of products of ladder operators?

Answer

The main point is that the normal ordering procedure $:~:$ does not take operators to operators, but symbols/functions to operators. This important point resolves various paradoxes created by abuse of language. For a full explanation, see e.g. this Phys.SE post and links therein.

newtonian mechanics - Is it possible to determine the outcome of any impact knowing only the ratio of masses?

In elastic collisions in 2-D if two balls $A$, $B$ ($m_A = m_B$, $R = 1$) have equal mass we can determine in advance the outcome of the collision.

If cue-ball $A$ impacts object-ball $B$ (at rest) at an angle of +60° (with the x-axis), we can say in advance that ball $B$ will move at an angle of -30° with same axis, we do not need other equations or data.

How do you determine the angle of ball $B$ if the ratio $R: m_A/ m_B \neq 1$? How do you determine the angle of ball $B$ in the above-cited case if, for example, $R = 1/3 \rightarrow \lambda = -46.1$ ?

If you think that impossible, considering the current state of the art, please state that clearly, even in a simple comment. This will help future readers looking for an answer.

I am adding a picture to visualize the problem:

Answer

I'll set up the problem in the following way. Let's assume the line of centers is along the x-axis. Initial speed of $m_1$ is $u_1$ at an angle $\theta$ to the x-axis. Final speed of $m_{1}$ is $v_1$ at an angle $\alpha$ to the x-axis. The final speed of $m_2$ is $v_2$ and is along the line of centers (the x-axis), such that $\alpha$ is the angle between the final velocities of the two balls.

Two conservation of momentum equations $$m_1 u_1 \cos \theta = m_1 v_1 \cos \alpha + m_2 v_2\ \ \ \ (1)$$
$$m_1 u_1 \sin \theta = m_1 v_1 \sin \alpha\ \ \ \ \ \ \ \ \ \ \ (2)$$

If $R= m_1/m_2$ then I can manipulate these to give $$ v_2 = R u_1 \left(\cos \theta - \cos\alpha \frac{\sin\theta}{\sin \alpha}\right)\ \ \ \ \ \ (3)$$

Now using conservation of kinetic energy in an elastic collision $$ m_1 u_1^{2} = m_1 v_1^{2} + m_2 v_2^{2}\ \ \ \ \ \ \ \ (4)$$

Substituting for $v_2$ from (3) and using $v_1 = u_1 \sin\theta /\sin \alpha$ from (2), after a bit of algebra, I get the following horrible expression

$$ R = \frac{\sin^2 \alpha - \sin^2 \theta}{\cos^2 \theta \sin^2 \alpha - 2 \sin\theta \cos\theta \sin \alpha \cos\alpha + \sin^2\theta\cos^2\alpha}\ \ \ \ (5)$$

If $\alpha=\pi/2$, then indeed the solution is $R=1$ for any value of $\theta$.

From formula (5) above. Substitute $1/(1+\tan^2)$ for $\cos^2$, substitute $1 - 1/(1+\tan^2)$ for $\sin^2$ and the corresponding square roots for $\cos$ and $\sin$. This will give an equation exclusively in terms of $\tan\theta$ and $\tan \alpha$. $$ R = \frac{\tan^{2}\alpha(1+\tan^2\theta) -\tan^2\theta(1+\tan^2\alpha)}{\tan^2\alpha - 2\tan\theta\tan\alpha + \tan^2\theta}$$ Rearrange this into a quadratic in $\tan^2\alpha$ thus: $$(R-1)\tan^2\alpha -2R\tan\theta\tan\alpha +(R+1)\tan^2\theta = 0$$ Thus $$\tan\alpha = \tan\theta \left(\frac{R \pm 1}{R-1}\right)$$

Only one of these roots makes sense $$\tan\alpha = \tan\theta \left(\frac{R+1}{R-1}\right),$$ and of course a negative value of $\tan\alpha$ just means that $\alpha>90^{\circ}$.

EDIT: As a test I plugged this into a graphical tool with $\theta=60^{\circ}$ (red curve). The plot below shows $R$ vs $\alpha$ (recall this is the angle between the final velocity vectors). Notice that the minimum value of $\alpha$ is $60^{\circ}$ when $R$ is very large. As $R$ diminishes, $\alpha$ gets larger, reaching about $73^{\circ}$ (i.e. $13^{\circ}$ from the x-axis on your diagram) for $R=3$, then $90^{\circ}$ for $R=1$, and about $106^{\circ}$ ($46^{\circ}$ to the x-axis in your diagram) for $R=1/3$. There is a unique curve for each $\theta$ (the blue curve shows $\theta=30^{\circ}$ for comparison). Note that the red and blue curves only cross at $R=1$. This value of $R$ is the only one that uniquely determines the separation angle, for all other values of $R$ it also depends on $\theta$.

$R$ vs $\alpha$ for two values of $\theta$.

quantum mechanics - Queries about rotational groups $mathrm{SO}(3)$ and $mathrm{SU}(2)$ in QM

In a QM text I am using (Sakurai 2nd edition 'Modern Quantum Mechanics'), he describes two rotation groups, namely the $\mathrm{SO}(3)$ rotation group and $\mathrm{SU}(2)$ rotation group (unitary unimodular group).

He defines $\mathrm{SO}(3)$ as a group with matrix multiplication on a set of orthogonal matrices (which are matrices which satisfy $R^TR = 1 = RR^T$), he then states that this group only includes rotational operators (and not also inverse operators which would be the group $\mathrm{O}(3)$). He does not ever rigorously define 'rotational operation'.

1. How you would distinguish between rotational operators and inverse operators, would a sufficient definition be that rotational operators is a transformation with one fixed point?

He also defines the the group $\mathrm{SU}(2)$ which consists of unitary unimodular matrices, and states that the most general unitary matrix in two dimensions has four independent parameters and it is defined as $$U = e^{i \gamma} \left( {\begin{array}{cc} a & b \\ -b^* & a^* \\ \end{array} } \right) $$ where $|a|^2 + |b|^2 = 1,~~~\gamma^* = \gamma.$

2. Am I right to assume that the $\mathrm{SO}(3)$ rotation group does not have much of application in quantum mechanics but is rather used more in classical mechanics whereas $\mathrm{SU}(2)$ is used more in quantum mechanics, particularly for $s =\frac{1}{2}$ spin systems where we work in a two dimensional Hilbert space?


3. How does it follow that there are four independent parameters for the general unitary matrix, the way I see it there are three independent parameters, namely, $a$, $b$ and $\gamma$?

Could M-Theory explain dark matter as well as dark energy?

Is it possible that M-Theory provides a solution to the mystery of dark matter and dark energy?

The idea of dark matter and dark energy is that there is some inexplicable source affecting our universe.

M-Theory suggests that there are other universes as well as our own. If there were some degree of interaction between the universes, this could explain both the unaccounted for acceleration of our universe's expansion in which dark energy is suggested as well as the unidentified mass of our universe which is termed dark matter.

If the other universes are orthogonal to some or all of our universe's dimensions, then both effects seem very possible. Could the other universes' mass have some affect within our universe causing the gravitational effects on visible matter, radiation, and the large-scale structure of the universe? Could those universes also be causing the accelerated expansion of our universe by literally pulling it apart through similar gravitational effects.

All of the effects and oddities seem like they would fall out of M-Theory easily. The extra dimensions of M-Theory seem well-suited for explaining both effects. Some dimensions could be effected by some alternate universes causing anomalies such as have been detected and relegated to dark matter. While, maybe, all of the alternate universes could act as a gravitational pull on the very fabric of the universe through the more pedestrian three dimensions, as is suggested for dark energy.