quantum field theory - Why must the Dirac equation multiplied by its complex conjugate give the KG equation?

This may be a simple question. I can show this is the case mathematically but cannot explain why it happens. It was only when asked why this happens when I realised I couldn't explain it intuitively/physically.

So I suppose there must be some way in which the Klein-Gordan equation can be recovered from the Dirac equation, just like a lot of things in Physics. But why does this $\textit{have}$ to be done by multiplying by the complex conjugate and not some other operation?

Answer

The reason is as follows.

The scalar Klein-Gordon equation is (and has to be) a second-order equation because the box $$\square = \partial_\mu \partial^\mu $$ is the simplest Lorentz-invariant differential operator that may act on scalar fields. However, Dirac wanted to find a first-order equation and it is indeed possible for spinors because $$ \partial \!\!\!\! / = \gamma^\mu\partial_\mu $$ is also Lorentz-invariant (doesn't change the transformation properties of the object it acts upon), it is a first-order operator, and it is able to act on spinors (spinors are needed because the operator contains the Dirac matrices which must contract their indices with a spinor).

By special relativity, the wave equations must enforce the correct invariant mass. So the de Broglie waves $$ \exp(ip_\mu x^\mu ) u$$ multiplied by a special form of the spinor (only in the Dirac case) must be a solution for $p_\mu p^\mu = m^2$. In other words, special relativity implies that a solution to the mass $m$ Dirac equation must be a solution to the mass $m$ Klein-Gordon equation, too.

However, the Dirac equation is stronger because it constrains the first derivatives; the Klein-Gordon equation allows you to choose the wave function and its first derivative(s) and it only constrains the second derivatives.

In the momentum basis, the Dirac equation says $$ (p\!\!\!/ - m) \Psi = 0$$ which says that the field $\Psi$ must be an eigenstate of "p-slash" with the eigenvalue $+m$. The condition $$ p\!\!\!/ -m = 0 $$ implies that $p^2=m^2$ because $$p^2 \equiv p_\mu p^\mu = p\!\!\!/ \cdot p\!\!\!/ $$ due to the anticommutator $$\{\gamma_\mu,\gamma_\nu\} = 2g_{\mu\nu}\cdot {\bf 1}$$ However, $ p\!\!\!/ =+m $ (when acting on $\Psi$) is not equivalent to $p^2=m^2$ (when acting on $\Psi$). The former condition is stronger, as I said. We may very well have $ p\!\!\!/ =-m $ and this will still imply $p^2=m^2$ (when acting on $\Psi$). This is nothing else than the claim that there are two solutions to the equation $x^2=m^2$, namely $x=+m$ and $x=-m$.

So the "Klein-Gordon" condition $(p^2-m^2)\Psi=0$ acting on $\Psi$ is simply equivalent to $$(p\!\!\!/ +m) (p\!\!\!/ -m)\Psi = 0$$ which isn't more controversial than the formula $a^2-b^2=(a-b)(a+b)$ and this product may be written as "p-slash plus m" times the Dirac equation.

In other, more physical words, the Klein-Gordon equation allows the positive-energy vector $p^\mu$ as well as $-p^\mu$, its negative-energy opposite. However, $(p\!\!\!/ -m)\Psi=0$, the Dirac equation, only allows one of the signs of the energy and this sign is correlated with the up/down polarization of the electron/positron.

The "other" or "reverted" Dirac equation, $(p\!\!\!/ +m)\Psi=0$, imposes the opposite sign of the energy as the function of the polarization. It's clear that the Klein-Gordon equation (acting on the spinor) which doesn't care about the sign of energy is equivalent to saying that the energy's sign is either what the Dirac equation demands (as a function of the polarization); or the opposite sign. The opposite sign is obtained by the solutions of the "reverted" Dirac equation.

So $\Psi$ obeys the Klein-Gordon equation iff it obeys the Dirac equation; or the reverted Dirac equation; or if it is a combination of these two types of solutions. This is mathematically equivalent to saying that the solution to the Klein-Gordon equation is annihilated by the product of the "Dirac operator" and the "reverted Dirac operator".

The reversal of the Dirac operator is also linked to some kind of complex conjugation, more precisely the charge conjugation (C), because the C-conjugated spinor simply obeys the reverted Dirac equation. This may be checked by defining C more carefully and counting the signs.

quantum mechanics - Expectation of momentum in the bound state

Is it logically correct to assert that the expectation of the momentum $$\langle \hat p \rangle=0$$ for any bound state because it is bound to some finite region? What is the physical interpretation of the fact that $$\langle \hat p \rangle=0$$ in an energy eigenstate $\psi_n(x,t)$ but $$\langle \hat p \rangle\neq0$$ in some superposition state $$\psi(x,t)=c_m\psi_m(x,t)+c_n\psi_n(x,t)~?$$ Here $\psi_n(x,t)$ the eigenstates of the Hamiltonian, for example, in the problem of particle in a box (say).

Answer

Is it logically correct to assert that the expectation of the momentum $\langle p \rangle=0$ for any bound state because it is bound to some finite region?

Bound state means the particles are bounded somewhere. Its wavefunction will vanish at the asymptotic limit. A bound state could be a superposition of a finite number of bound eigenstates. For instance, the superposition of the ground and first excited-state wavefunction of particle-in-box will still vanish at far limit.

I think one can only conclude for non-relativistic, bound, eigenstate (not any bound state) $\langle \hat{p} \rangle=0$. Since $$\langle n | p | n \rangle \sim \langle n | [H,x] | n \rangle = \langle n | Hx-xH | n \rangle = E_n ( \langle n| x | n \rangle - \langle n| x | n \rangle) =0 $$. If we relax the state into any bound state $| \rangle$, we have $$\langle | p | \rangle \sim \langle | [H,x] | \rangle = \sum_n c^*_n E_n \langle n | x | \rangle - c_n E_n\langle |x | n \rangle \neq0 $$ in general.

quantum mechanics - States in QM and in the algebraic approach

In QM states are vectors in a Hilbert space $\mathscr{H}$. These are often denoted like $|\psi\rangle$.

On the other hand, in the algebraic approach, we have one $\ast$-algebra $\mathscr{A}$ and states are linear functionals $\omega : \mathscr{A}\to \mathbb{C}$ such that $\omega(a^\ast a)\in [0,\infty)$ and $\omega(1)=1$.

It is not at all clear how these two things are related.

For a first step, we have the GNS construction. The GNS construction is the following:

GNS Construction: Given a $\ast$-algebra $\mathscr{A}$ and a state $\omega : \mathscr{A}\to \mathbb{C}$ we can construct one Hilbert space $\mathscr{H}_\omega$, one $\ast$-representation $\pi_{\omega} : \mathscr{A}\to \mathscr{A}(\mathscr{H}_\omega)$ and one $\Omega \in \mathscr{H}$ such that $\pi_{\omega}(\mathscr{A})\Omega$ is dense and $$\omega(a)=\langle \Omega ,\pi_\omega(a)\Omega\rangle.$$

Now we have some interesting things:

1. 
   

   Every algebraic state $\omega$ gives rise to a whole Hilbert space on which $\omega$ becomes the distinguished $\Omega$ and it produces one mean value on the usual QM sense.

   
   


2. 
   

   The other unit vectors on the Hilbert space give rise to algebraic states. Actually, if $\Phi\in \mathscr{H}$ we have that $$\omega_\Phi(a)=\langle \Phi, \pi_{\omega}(a)\Phi\rangle$$ is one algebraic state. It is obviously a linear functional and certainly satisfies $\omega_{\Phi}(1)=1$ and $$\omega_{\Phi}(a^\ast a)=\langle \Phi,\pi_{\omega}(a^\ast a)\Phi\rangle=\langle \Phi,\pi_{\omega}(a^\ast)\pi_{\omega}(a)\Phi\rangle=\langle \Phi,\pi_{\omega}(a)^\ast \pi_\omega(a)\Phi\rangle=\langle \pi_\omega(a)\Phi,\pi_{\omega}( a)\Phi\rangle=|\Phi|^2\in [0,+\infty)$$

   
   



3. 
   

   On the other hand it doesn't seem that every algebraic state gives rise to one usual state in $\mathscr{H}_\omega$. In truth, because of the Riesz representation theorem it would suffice that to every algebraic state $\phi$ there was one algebraic state $\tilde{\Phi}$ on $\mathscr{A}(\mathscr{H}_\omega)$. This in turn requires that $\tilde{\Phi}(\pi_{\omega}(a))=\phi(a)$ thus for this to be true we would need $\pi_\omega$ to be invertible. In other words, the representation must be faithful.

   
   

These points show that although related to the usual state vectors from QM, the algebraic states aren't equivalent to them. In fact, it seems we have more algebraic states than state vectors.

Furthermore, GNS allows us to indeed represent each state as a state vector, but on different Hilbert spaces. The point (2) I made then guarantee that each such state vector can be identified with one algebraic states, but there are other apart from it which do not belong on this Hilbert space. Even, though, if we pick one of those states in (2) and perform the GNS construction with them it seems we get an entirely different Hibert space.

It seems that the role of algebraic states is to generate a representation only and this is quite strange, considering the usual QM point of view on states.

So what is the correct way to understand algebraic states? How they relate to the usual notion of states from QM? To work with them in practice do we always need to use the GNS construction?

How do we deal with the fact that it appears that there are more algebraic states than QM vector states, in the sense that when we perform the GNS construction some algebraic states appear to be "left out"?

Answer

The algebraic formulation is more general and takes into account many subtleties that arise in QFT and that are hidden in quantum mechanics.

In fact, in quantum mechanics the Stone-von Neumann theorem tells us that the irreducible representation of the algebra of quantum observables (more precisely, of the algebra of canonical commutation relations) is essentially unique (i.e. it is unique up to unitary transformations). So the only relevant representation is the usual one (called Schrödinger representation), and the physically relevant states are the ones that are normal with respect to such representation (i.e. that can be written as density matrices on the corresponding Hilbert space $L^2(\mathbb{R}^d)$).

In quantum field theories, on the other hand, there are infinitely many inequivalent irreducible representations of the canonical (anti)commutation relations. Therefore, there are indeed states that can be represented as density matrices (or vectors) in one representation, but not in another (it is said that they are not normal with respect to the latter).

In addition, the so-called Haag's theorem explains that inequivalent representations, or more precisely disjoint states (states that are not normal w.r.t. the GNS irrepresentation of each other), play a very important role in QFT. In fact, given a group $G$ acting on the C*-algebra of observables, and two $G$-invariant states $\omega_1,\omega_2$ (with an additional technical condition that is not important here), then either $\omega_1=\omega_2$, or $\omega_1$ and $\omega_2$ are disjoint. In a relativistic theory, the ground state (or vacuum) is invariant w.r.t. the restricted Poincaré group. In addition, it is easy to see that in general the vacua of a free and an interacting theory must be different (and both invariant). Therefore, by Haag's theorem they are disjoint, and so they cannot both be represented as density matrices in a single representation.

This is just one example of why non-normal states (w.r.t. the free or Fock irrepresentation) are very important in QFT, and of why the algebraic description of quantum theories is so often used for relativistic quantum mechanics.

Why is current the same in a series circuit?

So I am a 10th grade student and my teacher told me that the current is the same at every point in a series circuit. It does split up in parallel circuit but it then recombines and the current flowing out of the battery is the same as the current flowing back into it.

My question is - Why does the current remain the same?

I mean let's say that there is a light bulb somewhere in a series circuit. Now, current(or electrical energy) will flow into it and then convert into light energy.

But if the amount of current flowing into the filament of the bulb = the amount of current flowing out of the filament and at the same time it is producing photons(light energy) [and some heat energy too] then aren't we creating energy ? Which is not possible.

I am really confused and can't seem to grasp this idea. Please help.

radioactivity - Is sub critical plutonium "safe" to handle?

Apparently, in Los Alamos scientists handled sub critical masses of plutonium (for example the demon core) with little or no protection. Richard Feynman and others mentioned that plutonium spheres were warm to the touch, which seems to imply that they actually touched them. Slotin (one of the victims of the demon core) performed experiments wearing blue jeans and cowboy boots.

Are certain plutonium isotopes safe to handle (as long as they don't go critical) or is it just that people back then didn't know any better?

Answer

Please be aware that plutonium cores are supposed to be plated with another metal (nickel or silver, if my memory serves me right). Machining plutonium is very hazardous and is done with remote manipulators, since it increases risk of inhalation.

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Source: http://toxnet.nlm.nih.gov/cgi-bin/sis/search/r?dbs+hsdb:@term+@na+@rel+plutonium,+radioactive :

Absorption through the skin can occur through occupational exposure. Experiments show that the skin is an effective barrier and the percentage absorbed /seldom/ exceeds 0.05% for intact skin. [Seiler, H.G., H. Sigel and A. Sigel (eds.). Handbook on the Toxicity of Inorganic Compounds. New York, NY: Marcel Dekker, Inc. 1988., p. 724] PEER REVIEWED

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Source: Plutonium ANL FactSheet Oct 2001

Plutonium metal. Plutonium isotopes are primarily alpha-emitters so they pose little risk outside the body. Here the plastic bag, gloves, and outer (dead) layer of skin would each alone stop the emitted alpha particles from getting into the body.

What Happens to It in the Body? When plutonium is inhaled, a significant fraction can move from the lungs through the blood to other organs, depending on the solubility of the compound. Little plutonium (about 0.05%) is absorbed from the gastrointestinal tract after ingestion, and little is absorbed through the skin following dermal contact. After leaving the intestine or lung, about 10% clears the body. The rest of what enters the bloodstream deposits about equally in the liver and skeleton where it remains for long periods of time, with biological retention half-lives of about 20 and 50 years, respectively, per simplified models that do not reflect intermediate redistribution. The amount deposited in the liver and skeleton depends on the age of the individual, with fractional uptake in the liver increasing with age. Plutonium in the skeleton deposits on the cortical and trabecular surfaces of bones and slowly redistributes throughout the volume of mineral bone with time.

What Is the Primary Health Effect? Plutonium poses a health hazard only if it is taken into the body because all isotopes but plutonium-241 decay by emitting an alpha particle, and the beta particle emitted by plutonium-241 is of low energy. Minimal gamma radiation is associated with any of these radioactive decays. Inhaling airborne plutonium is the primary concern for all isotopes, and cancer resulting from the ionizing radiation is the health effect of concern. The ingestion hazard associated with common forms of plutonium is much lower than the inhalation hazard because absorption into the body after ingestion is quite low. Laboratory studies with experimental animals have shown that exposure to high levels of plutonium can cause decreased life spans, diseases of the respiratory tract, and cancer. The target tissues in those animals were the lungs and associated lymph nodes, liver, and bones. However, these observations in experimental animals have not been corroborated by epidemiological investigations in humans exposed to lower levels of plutonium.

As a note, the common myth that plutonium is the “deadliest substance known to man” is not supported by the scientific literature. It poses a hazard but is not as immediately harmful to health as many chemicals. For example, for inhalation – the exposure of highest risk – breathing in 5,000 respirable plutonium particles, about 3 microns each, is estimated to increase an individual’s risk of incurring a fatal cancer about 1% above the U.S. average “background” rate for all causes combined.)

EDIT: As an aside: I recommend reading Eileen Welsome's The Plutonium Files: America's Secret Medical Experiments in the Cold War to get some idea of what early plutonium health safety experiments really meant (e.g. injecting a solution of plutonium salt into a patient's leg).

general relativity - Speed of light when accelerating

I'm studying special relativity and saw the 2 postulates of Einstein. The most remarkable one for me is the universal speed of light. Einstein postulated that the speed of light in vacuum is the same for all inertial observers, regardless of the motion of the source. But what if we are accelerating and we are changing from one inertial frame to another, how will we observe the speed of light? Is it still the same? I suggest it's not? Because I alread saw a small introduction to general relativity where the light will bend in a gravitational field and therefore also when you accelerate. So does the speed of light change because it bends?

Answer

To understand this you need to understand what we mean by speed.

If I want to measure positions and times I need to set up a coordinate system. For example I can take my stopwatch and my metre rule and construct some Cartesian axes $t$, $x$, $y$ and $z$, then I can describe every point in spacetime by its position in my coordinates $(t,x,y,z)$. Once I have done this I can calculate the velocity of some object by watching how its position measured using my coordinates changes with the time measured using my coordinates. So for example if something is moving along my $x$ axis the speed is just:

$$ v = \frac{dx}{dt} $$

All very well, but why did I repeatedly use the phrase measured using my coordinates in the paragraph above? Well, it's because my coordinate system is just one way of measuring out spacetime and it doesn't necessarily have any fundamental physical significance. That means the speeds determined by my coordinates don't necessarily have any fundamental physical significance either.

To look into this a bit farther let's stick to flat spacetime so we don't have the complications introduced by general relativity. The obvious coordinates to choose are those of an inertial frame, and you'll meet the phrase inertial frame over and over again in studying relativity. In these coordinates the speed of light is always $c$.

But now suppose I'm accelerating with some acceleration $a$. There's nothing to stop me choosing coordinates where I remain at the origin i.e. I measure all distances and times relative to me. This would be a non-inertial frame, and as you'd expect if I use a non-inertial frame all sorts of weird things can happen e.g. Newton's laws no longer apply.

To make this concrete suppose I am accelerating along the $x$ axis and I measure the velocity of a light beam travelling along the $x$ axis using my non-inertial coordinates. I get the result:

$$ v_\text{light} = c\,\left(1 + \frac{a}{c^2}x \right) \tag{1} $$

I won't go thorough the derivation, but the accelerating coordinates are known as Rindler coordinates and the speed of light is calculated using an equation called the Rindler metric.

Anyhow, what I find is that the velocity of light now changes with the distance along the $x$ axis away from me. When the product $ax$ is positive (i.e. ahead of me) the velocity of light is greater than $c$ and when $ax$ is negative (behind me) the velocity of light is than than $c$. In fact when $ax = -c^2$ the velocity of the light slows to zero and there is an event horizon there (called the Rindler horizon).

But it's the same light in the same spacetime, just described using two different sets of coordinates. So what then is the real speed of the light. And here we reach the key point: there is no real speed. Relativity tells us that any set of coordinates is as valid as any other set of coordinates - you cannot say the inertial coordinates are right and the accelerating coordinates are wrong because they are equally valid ways of describing the same spacetime.

There is one last point to make. Suppose we take my equation (1) for the speed in the accelerating coordinates and calculate the speed at my position i.e. at $x=0$. The value is:

$$ v_\text{light} = c\,\left(1 + \frac{a}{c^2}\,0 \right) = c $$

So even though the speed of light is variable, at my position the speed is equal to $c$. And this is another absolutely key point: although the speed of light may vary in some coordinate systems, if you measure the speed of light at your position you will always get the value $c$. So the speed of light is always locally constant, it's just not globally constant.

momentum - How to escape the center of a room without gravity?

Imagine you're an astronaut on the International Space Station and your fellow astronauts played a prank on you by taking all your clothes and putting you in the center of a module so that you cannot reach anything with either your hands or your feet.

What would be the most effective way to escape that situation if you're reluctant to start peeing?

Answer

It is worth calculating (rather than just speculating) some of the methods described.

1. Breathe in one way, and out the other. Your resting tidal volume is about 0.5 liter; take a deep breath and it can be 3-5 liters (thanks @Aaganrmu). That is about 4 gram of material. If you purse your lips to increase the velocity with which you expel the air, you can get it to around 10 m/s (I estimate this from the data in this paper which measured the velocity of air in a cough - 15 m/s - and spoken word - 4 m/s.). This gives a net momentum of 0.04 kg m/s, which means a 70 kg astronaut will get a reaction velocity of 0.05 mm/s, moving 1 cm every 20 seconds. But if you do this 10 times per minute, your acceleration will be about $10^{-4}~\rm{ m/s^2}$ and you will travel 2 m (to the nearest wall) in about 200 seconds. Feeling somewhat dizzy from the hyperventilation... Note that it's not necessary to turn your head: it's enough to breathe in slowly, and out rapidly.


2. "swim" with your arms. If you can change the area of your arms by 20 cm$^2$ between the "forward" part of the stroke, and the "back" part of the stroke, and you can move the hand with a peak velocity of about 2 m/s for 50 cm, the approximate drag force will be $F = \frac12 \rho v^2 A C_D = 0.5*1*4*0.02*1.0 = 0.04~\rm{ N}$ - four times more than breathing out. And of course you can probably move your arms a great deal faster - let's say one complete stroke (two arms) per second, for an acceleration of $6\cdot 10^{-4}~\rm{m/s^2}$ and a time across the capsule of 80 seconds.

Combining the two techniques, the exercise with the arms will allow you to breathe more rapidly (at peak exercise, an adult male can move about 100 liters of air per minute - that is 10 times higher than the value I used). This should comfortably get you to the side of the capsule in under a minute.

If you can increase the mass you accelerate, you can greatly improve on these numbers. I briefly considered that spitting might be the answer, but your mouth will run dry pretty quickly. Other bodily fluids would greatly improve on the time - but given that you still have to live in that space after the fact, I think that spending a bit more time waving your arms about, then laugh at the videos your crew mates made, is the best approach here.