homework and exercises - Deriving the Hamiltonian density for a free scalar field

I'm working through my old notes on QFT (cf. Ref 1) and I'm not quite sure how to approach the derivation of the Hamiltonian density for a free scalar field (question 2.3 on page 19) and the subsequent derivation of the components of the energy-momentum four-vector in terms of the new field operators. Could you point me in the right direction?

$${\mathcal{H}~=~\frac{1}{2}\left[\left(\partial_t\phi\right)^{2}+\left(\nabla\phi\right)^{2}+m^{2}\phi^{2}\right]\textrm{.}}$$

References:

1. M Dasgupta, AN INTRODUCTION TO QUANTUM FIELD THEORY, Lecture presented at the School for Experimental High Energy Physics Students Somerville College, Oxford, September 2009.

Answer

First question.

Let's have Lagrangian $$ L = \frac{1}{2}\left( (\partial \varphi )^{2} - m^{2}\varphi^{2}\right) . $$ Stress-energy tensor is equal to $$ T^{\alpha \beta} = \frac{\partial L}{\partial_{\alpha }\varphi }\partial^{\beta}\varphi - g^{\alpha \beta}L = \partial^{\alpha}\varphi \partial^{\beta}\varphi - \frac{g^{\alpha \beta }}{2}((\partial \varphi )^{2} - m^{2}\varphi^{2}). $$ Then $$ H = \int T^{00}d^{3}\mathbf r = ..., \quad \mathbf P^{i} = \int T^{0i} d^{3}\mathbf r = ..., \qquad (.1) $$ and you have an answer.

Second question.

Let's have the solution for scalar real field: $$ \varphi (x) = \int \left( b_{\mathbf k}^{*}e^{ikx} + b_{\mathbf k}e^{-ikx}\right)\frac{d^{3}\mathbf k}{\sqrt{(2 \pi )^{3}2\epsilon_{\mathbf k}}}, \quad kx = \varepsilon_{\mathbf k}x^{0} - (\mathbf k \cdot \mathbf x). \qquad (.2) $$ So you can rewrite $(.1)$ as $$ \mathbf P = \int \mathbf k b_{\mathbf k}b_{\mathbf k}^{*}d^{3}\mathbf k , \quad H = \int \epsilon_{\mathbf k} b_{\mathbf k}b_{\mathbf k}^{*}d^{3}\mathbf k \qquad (.3) $$ (where I ignored the infinite constant summand as we can do for the free field).

Then you want to rewrite operators of $(.3)$ expressions in terms of

$$ \hat {\varphi} , \quad \hat {\frac{\partial L}{\partial \dot {\varphi }}} = \hat {\dot {\varphi }} = \hat {\pi} . $$ You can do next steps.

1. To multiply $(.2)$ by $e^{-i(\mathbf q \cdot \mathbf x )}$ and then take an integration over $\frac{d^{3}\mathbf r}{\sqrt{(2 \pi )^{3}}}$. You'll get $$ \int \varphi (\mathbf r , t) e^{-i (\mathbf q \cdot \mathbf r )}\frac{d^{3}\mathbf r }{\sqrt{(2 \pi )^{3}}} = ... = \frac{1}{\sqrt{2 \epsilon_{\mathbf q}}}\left( b_{-\mathbf q}^{*}e^{i \epsilon_{\mathbf q} t} + b_{\mathbf q}e^{-i \epsilon_{\mathbf q} t}\right) . \qquad (.4) $$


2. To take t-derivation of $(.4)$. You'll get $$ \int \pi (\mathbf r , t) e^{-i (\mathbf q \cdot \mathbf r )}\frac{d^{3}\mathbf r }{\sqrt{(2 \pi )^{3}}} = i\sqrt{\frac{\epsilon_{\mathbf q} }{2}}\left( -b_{\mathbf q}e^{- i\epsilon_{\mathbf q} t} + b_{-\mathbf q}^{*}e^{i \epsilon_{\mathbf q} t} \right) \qquad (.5) $$


3. Now you have from $(.4), (.5)$ a system of linear equations. You can express $b_{\mathbf q}$ from these equations as function of $\varphi , \pi$.


4. You can use the hermitian nature of $\hat {b}_{\mathbf q}, \hat {b}^{+}_{\mathbf q}$. By assuming the $\hat {b}_{\mathbf q}$ expression you can get $\hat {b}^{+}_{\mathbf q}$ expression. Finally, you'll get $$ \hat {b}_{\mathbf q } = \int \left( \epsilon_{\mathbf q} \hat {\varphi} + i \hat {\pi} \right)e^{i q x}\frac{d^{3}\mathbf r }{\sqrt{2 \epsilon_{\mathbf q} (2 \pi )^{3}}} , \quad \hat {b}_{\mathbf q }^{+} = \int \left( \epsilon_{\mathbf q} \hat {\varphi} - i \hat {\pi} \right)e^{-i q x}\frac{d^{3}\mathbf r }{\sqrt{2 \epsilon_{\mathbf q} (2 \pi )^{3}}}. $$


5. Use these expressions for the transformation of $(.3)$ expression.

If you want some details of derivations which were used in steps above, I'll help you.

electromagnetism - A puzzler in induction and special relativity

Some reading I was doing jogged my memory about a puzzle in E&M that hit me back in my undergrad days, but I just let it go at the time and never found an answer.

A pretty common undergrad E&M problem goes something like, "We manage, by way of external coils, to produce a magnetic field going through the center of a wire loop. The field occupies and area of A within the loop and we make it change at x Tesla/sec. What is the EMF induced around the loop?"

And of course we solve this with Faraday's equation

$\displaystyle \nabla \times E = -\frac{\partial B}{\partial t}$

Integrate it over the surface contained by our wire loop and get

$\displaystyle EMF = -\frac{d \Phi}{dt} = -A\frac{dB}{dt}$

Straightforward. Except ....

The EMF, according to this equation, is directly related to the instantaneous rate of change of $B$. And this is true even if $B$ is confined to a very small area within the current loop. It is true even if, for instance, $B$ is confined to, say, one cm square, and the current loop is one million miles in radius. Again, the relationship is instantaneous. So, if we turn on our coils and generate $B$, immediately there is an EMF 1 million miles away. But information cannot travel faster than the speed of light, so this makes no sense. Where have we gone wrong?

I am pretty sure the answer to this question has something to do with the "slipperiness" of the concept of simultaneity in special relativity. So taking an integral across the entire area of the current loop "simultaneously" is probably a tricky thing to do. So maybe the question to be asked is, given the strictures of special relativity, can we cast the Maxwell's equations, particularly induction, in meaningful integral forms, and what do those forms look like?

Answer

Suppose, as you said, we have a thin solenoid pointing upward, going through the middle of a loop one light-year wide. Then the three equations $$\nabla \times E = - \frac{\partial B}{\partial t}, \quad \mathcal{E} = -\frac{d\Phi_B}{dt}, \quad B = \mu_0 n I$$ together imply violation of causality, as shown in your question.

Maxwell's equations are already fully relativistic, so the first two equations are always true. In particular, the path from equation (1) to equation (2) is pure math and always holds; there's no need to modify that for relativity either.

The mistake is in equation (3). This is the field of a solenoid that has been on for a long time, but you want us to consider the case where it's "suddenly" switched on. If we switch the solenoid on quickly, $\partial B / \partial t$ is large, so we suddenly get a large electric field circulating just outside the solenoid. But then we have to account for the other induction effect, $$\nabla \times B \sim \frac{\partial E}{\partial t}.$$ If you draw a picture, you'll see this produces both an upward and downward additional magnetic field. A detailed calculation shows that, if you take a light-year-wide loop, the total magnetic flux is zero! Then $\mathcal{E}$ is zero, as expected by causality.

This additional magnetic field will induce additional electric fields, which induce more magnetic fields, so that after a year, the fields propagate to the loop and $\mathcal{E}$ becomes nonzero. You might already know this propagation mechanism by a different name; it's called light.

quantum mechanics - "Reversed" Clebsch Gordan coefficients

Consider the sum of angular momenta $J=J_1+J_2$.

When one has a state $|J,M\rangle$ of an eigenbasis of a common eigenspace of $J^2$ and $J_z$, one can write it in terms of the elements $|j_1,m_1;j_2,m_2\rangle$ of an eigenbasis of a common eigenspace of $J_1^2,\,J_2^2,\,J_{1z},\,J_{2z}$: $$|J,M\rangle=\underset{m_1,m_2}{\sum} C^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}\ \ |j_1,m_1;j_2,m_2\rangle$$ Where $C^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}$ are the Clebsch Gordan coefficients.

I can do this easily using a Clebsch Gordan Coefficient's table. What if I want to express the kets $|j_1,m_1;j_2,m_2\rangle$ in terms of the kets $|J,M\rangle$? In this case, I can write all the $|J,M\rangle$ in terms of the states $|j_1,m_1;j_2,m_2\rangle$ and combine them in such a way that I get one of the $|j_1,m_1;j_2,m_2\rangle$ states. This can lead to long calculations.

I wonder if there is a table like the one for Clebsch Gordan coefficients but for a kind of "reversed" Clebsch Gordan coefficients $B^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}$ such that $$|j_1,m_1;j_2,m_2\rangle=\underset{J,M}{\sum} B^{j_1,\,j_2,\,J}_{m_1,\,m_2,\,M}\ \ |J,M\rangle$$

Answer

If I understand well, the coefficients $B$ are just in fact CG's: to be explicit $$ \vert j_1m_1; j_2m_2\rangle = \sum_{J(M)} C^{j_1j_2J}_{m_1m_2M} \vert JM\rangle $$ Note the sum of $M$ is not really a sum as $M$ must satisfy $M=m_1+m_2$.

The best way to see this is to start with $\vert j_1m_1; j_2m_2\rangle$ and just insert the unit $I=\sum_{JM}\vert JM\rangle\langle JM\vert$. One then has $$ \vert j_1m_1; j_2m_2\rangle=\sum_{JM}\vert JM\rangle\langle JM\vert j_1m_1;j_2m_2\rangle $$ with $$ \langle JM\vert j_1m_1;j_2m_2\rangle = \langle j_1m_1;j_2m_2\vert JM\rangle = C^{j_1j_2J}_{m_1m_2M} $$ since the CGs are real.

Is the center of mass in general relativity equal to the center of mass in newtonian gravity?

Consider 2 point masses $A,B$ a distance $d$ away from each other without velocity or rotation spin. Is the center of mass in general relativity equal to the center of mass in newtonian gravity?

In newtonian gravity the center of mass is where the particles collide, if they collide. Is the center of mass in general relativity also the place where the particles collide, if they collide? Or maybe the center of mass in general relativity is the place where the particles collide if they had no initial velocity or rotation spin?

Answer

There is generally no center of mass in general relativity.

The notion of center of mass becomes useful in Newtonian (and special relativistic) mechanics since because of simplifications we obtain by using it as a reference point. For instance, we could decouple the motion of COM from the relative motion of various parts of the system.

But in general relativity there are generally no conservation of momentum. Moreover the notion of a point particle also has some problems.

In certain special cases (such as for an isolated system with an asymptotically flat metric) in general relativity we could still define the notion of center of mass. For instance in the paper:

Huisken, G., & Yau, S. T. (1996). Definition of center of mass for isolated physical systems and unique foliations by stable spheres with constant mean curvature. Inventiones mathematicae, 124(1), 281-311. doi:10.1007/s002220050054, (online pdf).

the COM is defined not as a 'inner' point of a spacetime, but rather through foliation of asymptotic exterior, that is by considering the behavior of metric in the almost flat region. Such definition is essentially Newtonian.

In general spacetime such expansion of the metric is impossible, so the notion of COM does not exist. Its nonexistence is illustrated by the process of relativistic swimming in a curved spacetime. This effect, suggested by J. Wisdom in the paper:

Wisdom, J. (2003). Swimming in spacetime: Motion by cyclic changes in body shape. Science, 299(5614), 1865-1869. (free pdf).

essentially allows motion of an initially at rest system through internal deformations. (Like Baron Münchhausen pulling himself by his hair).

Are all instances of quantum non-locality problem artifacts of the use of classical concepts in quantum physics?

Consider experiments involving entangled spins, say two-spin 1/2 particles in the singlet state:

$$\left|\psi\right> =\frac{1}{\sqrt{2}}\left[\left|\uparrow\downarrow\right> - \left|\downarrow\uparrow\right>\right]$$

where Alice is going to measure one spin in the $z$-direction and Bob the other spin in the same $z$-direction. Alice and Bob are space-like separated at the time they perform their measurements. There then seems to be a (benign) locality paradox. Suppose that in some frame, Alice performs her measurement first. She then knows what Bob will find, but how can the information of Bob's result already exist before Bob makes his measurement?

The problem here is that we know that there are no local hidden variables underlying quantum mechanics. John Bell demonstrated this by showing that a local hidden variables theory gives rise to certain inequalities (now know as the Bell inequalities) for certain correlations of spin measurements of entangled spins, while quantum mechanics implies that these inequalities can be violated in certain cases (the same singlet state we're considering here, can show correlations that do not satisfy the Bell inequalities). later the violations of these inequalities was verified in an experiment conducted by Alan Aspect.

So, the non-existence of local hidden variables means that the information that Alice has about what Bob will find cannot also be present locally at Bob's place. Now, in principle, there is no problem in principle with information about this popping up non-locally at Alice's place. One cannot exploit this to allow for faster than light communications. Also we can note that non-local hidden variable theories can be compatible with quantum mechanics. But this does go against the spirit of locality. It then seems that there is an inherent benign non-local aspect to quantum theory.

However, in the MWI, there is no non-locality at all. After Alice measures her spin, there are two copies of her and both copies are equally for for Bob, so it's not true that in the frame where Alice has made her measurement and Bob hasn't, that the information about what Bob will find already exists in Alice's place.

Now, the whole point of the MWI is to treat observers and measurement devices in a quantum mechanical way too instead of in an ad-hoc classical way and thereby to get rid of the collapse of the wavefunction. But if we do treat observers in a classical way and stick to a single universe theory, then we do get a non-locality problem, albeit a benign one.

Then having resolved the source of the apparent non-locality in this case as a classical spanner thrown into the quantum world, we can ask if this is always the case. E.g. we can read here that Lev Vaidman claims that this is also true for the famous Aharonov-Bohm effect, but also that this is in dispute. Vaidman's argument is rather technical, but he argues that ultimately the whole non-locality issue there is artifact of using classical potentials.

Is there an example of a non-local aspect of quantum mechanics that cannot explained away as an artifact of a classical concept?

mathematical physics - What's the point of having an einbein in your action?

One often comes across actions written with an extra auxiliary field, with respect to which if you vary the action, you get the equation of motion of the auxiliary field, which when plugged into the original action lets you retrieve a more familiar looking action without the auxiliary field. An example is

$\begin{equation*} S=\int d\tau e^{-1}(\tau)\eta_{\mu\nu}\frac{dx^{\mu}}{d\tau}\frac{dx^{\nu}}{d\tau} \end{equation*}$

which when solved for the einbein $e$ gives you the familiar square root form of the action. Why would you want this extra auxiliary field apart from having an action which is not in an annoying square root form? Why, for example, does GSW mention on Page 18 of "Superstring Theory" that "the role of $e(\tau)$ is to ensure the action is invariant under reparametrizations of $\tau$" when you can have reparametrization invariance even in the square root form of the action?

Answer

first, you and Moshe may have somewhat misunderstood the comment that "the purpose of introducing $e$ is to guarantee the reparameterization symmetry of the action". This wasn't meant to express the circular argument that we need to add a redundant field in order to have a redundancy.

Instead, I think that the authors simply wanted to write that "in order for this action - which we already chose to study - to be equivalent to the einbein-frei form, the new field $e$ has to be unphysical, so the reparameterization has to be insured, so the factor of $e$ has to be included." If the $e$ were simply omitted in this form of the action, the action wouldn't be reparameterization-invariant, so it wouldn't be equivalent to the "proper length" of the world line we want to describe: the proper length of a world line is independent of the way how we parameterize the world line to calculate the integral (the proper length), so any equivalent description has to have the same property.

Motivation for gauge redundancies

In this particular case, the main technical reason is that by introducing the new auxiliary variable, the einbein, you get rid of the square root. Note that without this variable, the action would depend on the standard Lorentz gamma factor, $1/\sqrt{1-v^2/c^2}$. From the Lagrangian, it would imprint itself into the Hamiltonian as well. Hamiltonians that depend on square roots of functions of observables are hard - relatively to what you get with $e$. In the world sheet generalization, the Hamiltonian would even include an integral of a square root which is really bad - especially if you want to Fourier-transform things.

On the other hand, the underlying physics actually doesn't need the messiness as can be seen by introducing clever redundancies. There is no square root in the action including the einbein - but it's still equivalent. By introducing this fake degree of freedom, or a redundancy, we actually convert the equations of motion for $x$ to a simple wave equation, $\nabla_\mu\nabla^\mu x^\alpha=0$. Well, it's a wave equation in at least $1+1$ dimensions. In $0+1$ dimensions, it's just an equation for a uniform motion of $x^\alpha$.

Case of other gauge symmetries

The reparametrization of $\tau$ is a redundancy of the system, but so is any gauge symmetry, for example the colorful $SU(3)$ in QCD or the diffeomorphism group in general relativity (which is nothing else a higher-dimensional generalization of this einbein example: $e(\tau)\equiv \sqrt{g_{00}}$ is the redundant field, and gravity has no physical components in $0+1$ dimensions). You could also work without it - and the twistor description of $N=4$ $d=4$ $SU(N)$ gauge theory actually has no locality so it has no local gauge symmetry in spacetime. Other descriptions that don't have any gauge redundancy include the unitary gauge in spontaneously broken gauge theories; AdS/CFT that replaces the redundancy by a totally different one in the dual description; the S-dual description that replaces the gauge group with the magnetic monopoles' dual gauge group; and many others.

The purpose of having local symmetries - or gauge redundancies - is always either to simplify the equations or to make some of their symmetries manifest.

In particular, if you want elementary particles with spin equal to one or higher, and if you want a manifest Lorentz symmetry in spacetime and manifest locality, you need local vector- or tensor-valued fields such as $A_\mu$ and $g_{\mu\nu}$ that also have time-like components. In a quantum theory, the time-like components would produce negative-norm states: the theory could predict negative probabilities if these polarizations would stay in the spectrum.

Of course, gauge invariance is the only consistent way to remove them. Because gauge symmetry is a "symmetry" in the sense that it commutes with the Hamiltonian, the gauge invariance of the initial state guarantees the gauge invariance of the final state, so unphysical, gauge-dependent states are not produced by the evolution.

At any rate, the choice of a redundant equivalent description is always made in order to make some or several of the properties below manifest:

1. Unitarity combined with manifest Lorentz invariance: gauge symmetry is needed to remove the ghosts


2. Locality - one needs local tensor fields


3. Simple, harmonic-oscillator-like underlying character of the dynamics that would be masked by square roots in the einbein-frei formulation (typical in the case of diffeomorphisms of the world volumes)

Cheers, LM

particle physics - What is the smallest observable structure in the universe?

I've been wondering about the Planck length recently, but it is not observable. What is the smallest actually observable structure in the universe?