electromagnetism - Gauge Invariance of the Hamiltonian of the electromagnetic field

The Hamiltonian for an electron of mass $m$ and charge $e$ in an exterior electromagnetic field is $$H=\frac{1}{2m}(p-(e/c)A)^2+e\varphi.$$ The corresponding (via canonical quantization) quantum mechanical Hamiltonian is not invariant under Gauge transformations (for $A$ and $\varphi$). What's the physical meaning of this? And what is the physical meaning of the fact that the classical Hamiltonian is not invariant under gauge transformations?

Answer

I will try to slightly elaborate on @VladimirKalitvianski answer.

From Maxwell's equations, we can derive that the following combination of gauge transformations on $\mathbf{A}$ and $\Phi$ leave both $\mathbf{B}$ and $\mathbf{E}$ invariant: \begin{align} & \mathbf{A}'=\mathbf{A}-\mathbf{\nabla} \alpha \\& \Phi'=\Phi+\frac{\partial \alpha}{\partial t} \end{align} where $\alpha=\alpha(\mathbf{x},t)$. This means that all the field configurations of $\mathbf{B}$ and $\mathbf{E}$ related by a gauge transformation are physically equivalent. Note that this has nothing to do with the Hamiltonian operator in QM.

Now in QM, we know that a wave function can always be multiplied by a phase factor: $$ \psi'=e^{-iq\alpha}\psi, $$ where $\alpha \neq \alpha(\mathbf{x},t)$, because the probability of finding the particle at a particular position is unaffected by the above transformation, and also the Schrodinger equation and the probability current are unaffected by the above transformation. If we now demand that the above also holds for when $\alpha = \alpha(\mathbf{x},t)$ (i.e. a gauge transformation), then the Schrodinger equation must be made gauge invariant: \begin{equation} i\frac{\partial\psi}{\partial t}=-\frac{1}{2m}(\mathbf{\nabla}-iq\mathbf{A})^2\psi+(V+q\Phi)\psi \end{equation} such that the Schrodinger equation is invariant under the simultaneous gauge transformations: \begin{align} &\mathbf{A}'=\mathbf{A}-\mathbf{\nabla} \alpha \\& \Phi'=\Phi+\frac{\partial \alpha}{\partial t} \\& \psi'=e^{-iq\alpha}\psi \tag{1} \end{align} Note that we can say that we have adjusted the "normal" Hamiltonian by replacing the ordinary (partial) derivatives by: \begin{equation} \begin{array}{cc} \displaystyle \mathbf{\nabla} \rightarrow \mathbf{D}\equiv \mathbf{\nabla} - i q \mathbf{A} \; ,& \displaystyle \frac{\partial}{\partial t} \rightarrow D^0 \equiv\frac{\partial}{\partial t} + iq \end{array} \end{equation} To sum up, by demanding that our theory is invariant under the gauge transformation expressed by equation $(1)$, we are forced to change the Hamiltonian operator as we have done above. However, by doing this, the new Hamiltonian describes a particle interacting with the potentials $\mathbf{A}$ and $\Phi$. If you're not convinced by this argument, I strongly recommend you to read up on the Aharonov–Bohm effect (http://en.wikipedia.org/wiki/Aharonov%E2%80%93Bohm_effect).

Furthermore, note that we require that a gauge transformation does not affect any observables. This means that we must demand that the probability current is also unaffected. You can show (although it is quite tedious) that the current is made gauge invariant by making the replacement: $\displaystyle \mathbf{\nabla} \rightarrow \mathbf{D}$.

universe - Atoms pop out of nothing/vacuum/pre-big-bang?

I saw a great documentary last night about 'nothing'. It's about vacuums, and how if you have a total vacuum atoms will pop out of nowhere! Pretty crazy stuff. Atoms literally coming out of nowhere, then disappearing quickly.

It got me thinking, before the big bang if there was nothing, then presumably atoms would be popping out of nowhere still. If enough atoms suddenly popped simultaneously into the same space could it create the big bang? The chance of this is 1 in trillions on trillions but if time is infinite then surely the chance of this happening is 1?

Also if time is infinite and atoms really do pop out of nowhere, if you go back in time far enough was there ever a moment when a donkey appeared out of nowhere in a giant vacuum simply as a result of a complex pattern of atoms appearing out of nowhere?

I'm a Physics noob but love thinking about all this stuff, is this something people have thought about before? And what other theories of how the big bang happened in the first place are there?

Do human bodies give off a consistent but unique radiation/electromagnetic/energy signature?

Is there any facet of the energy emitted by a human body that is consistent and unique - like a fingerprint, but a signal that could be detected by a remote device?

thermodynamics - Is energy the ability to do work?

Here was my argument against this, the second law of thermodynamics, in effect says that, there is no heat engine that can take all of some energy that was transferred to it by heat and do work on some object. So, if we can not take a 100% the thermal energy of an object, and use it to do work, what about the thermal energy that is rejected to the environment, can we use all of that energy to do work on an object? No, if energy is supposed to be the ability to do work, well that’s a contradiction.

Answer

"The ability to do work" is certainly a lousy definition of energy.

Is it "merely" a lousy definition, or is it actually an incorrect definition? I think it could be either, depending on precisely how the word "ability" is interpreted. But if the words are interpreted as they would be in everyday speech and everyday life, I would say it's an incorrect definition.

UPDATE -- What is a definition of energy that is not lousy?

This is a tricky issue. Defining a thing that exists in the real world (like you do in physics) is quite different than defining a concept within an axiomatic framework (like you do in math).

For example, how do you "define" Mount Everest? Well, you don't exactly define it, you merely describe it! You describe where it is, you describe what it looks like, you describe how tall it is, etc. Since there is only one mountain that has all these properties, you wind up with a "definition".

Likewise, if I start describing energy (i.e. listing out various properties of energy), I will eventually wind up with a definition of energy (because nothing except energy has all these properties). Here goes:

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  The following are examples of energy: Kinetic energy, electric potential energy, gravitational potential energy, ...

  
  


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  The fundamental laws of physics are the same at every moment in time -- they were the same yesterday as they are today. This fact implies, by Noether's theorem, that there is a conserved quantity in our universe... This quantity is energy.

  
  
  


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  Special relativity relates energy to mass / inertia.

  
  


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  General Relativity relates energy to the curvature of spacetime.

  
  


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  In quantum mechanics, the energy of a system is its eigenvalue with respect to the Hamiltonian operator.

  
  



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  Whatever other things I'm forgetting or haven't learned...

  
  

All these properties are interrelated, and out of them bubbles a completely precise and unambiguous understanding of what energy is.

(I'm sure that some people will claim that one bullet point is the fundamental definition of energy, while the other bullet points are "merely" derived consequences. But you should know that this is a somewhat arbitrary decision. The same thing is true even in mathematics. What aspects of "differentiable manifold" are part of its definition, and what aspects are proven by theorems? Different textbooks will disagree.)

But can you boil that understanding of energy down into a one sentence "definition" that is technically correct and easy to understand? Well, I can't, and I doubt anyone on earth can.

photons - How is wavelength actually related to space (/distance)?

Is a photon of 400nm in wavelength "smaller" than a photon of 1km in wavelegth?

Answer

It can be, but a photon's "size" does not primarily depend on its wavelength: it depends on its bandwidth.

Saying "I have a photon with a well-defined wavelength of 400 nm" means that you have one quantum of energy in an electromagnetic mode with well-defined wavelength. That mode is necessarily a plane wave* and therefore occupies all of space. Well-defined 10 nm photons are as spatially large as 1 km photons - infinitely big.

Of course, this is simply hiding the issue under the carpet, as we never deal with physically infinite systems. In reality, all light pulses will have some 'main' frequency but also a finite extent. That means that their wavelength is not well defined (it cannot be measured to infinite precision, as this requires an infinite number of cycles), and they can be thought of as superpositions of plane waves. As it happens, the shorter the pulse is spatially, the greater the spread in wavelength it will have - the greater its bandwidth.

To put this on a slightly sounder foundation, you can relate the spatial spread $\Delta x$ and the spread in wavenumbers $\Delta k$ with a Heisenberg-style uncertainty relation, $$\Delta x\,\Delta k\gtrsim1,$$ though it is important to note that this is not (yet) the Heisenberg Uncertainty Principle - it is simply a fundamental property of waves.

Of course, in real life, if you manage to make a source of 400 nm photons, you will likely achieve a spread in wavelength of about that order, or maybe a few times that, but it's unlikely to be as big as 1 km. (Why? it would require your light source to be frequency-stable to the ratio 1 km/400 nm.) Thus your photons will probably be of the order of a few times 400 nm in size.

(Another more complex possibility is that your photons be very large but lose coherence over a length scale of a few times 400 nm, so that they're only useful single photons over that range.)

Thus, in general, yes, 400 nm photons are likely to be 'shorter' than 1 km photons, but you need to be quite careful with how you define terms and what you are assuming your light sources to be.

_{*or possibly in some other geometry, without affecting the conclusions.}

general relativity - Factorization of the energy/mass relation (as used to obtain the Dirac equation) applied to the invariant interval

In considering the Klein-Gordon equation Dirac was examining the energy equation:

$$m^{2}c^{2}=g_{\mu\nu}P^{\mu}P^{\nu}$$

(though he only was considering the flat space case $g_{\mu\nu}=\eta_{\mu\nu}$. Rewriting the above as:

$$0=g_{\mu\nu}P^{\mu}P^{\nu}-m^{2}c^{2}$$

Proceeding to factor this Dirac obtained an expression like:

$$0=\left(\gamma_{\mu}P^{\mu}-mc\right)\left(\gamma_{\nu}P^{\nu}+mc\right)$$

Where $\left\{ \gamma_{\mu},\gamma_{\nu}\right\} =\gamma_{\mu}\gamma_{\nu}+\gamma_{\nu}\gamma_{\mu}=2g_{\mu\nu}$. This is entirely general, though in quantum theory, the $P^{\mu}$ would be replaced with their quantum operators acting on a wavefunction $\Psi$.

Continuing with the general case, choosing one factor or another leads one to the Dirac equation for either positive or negative mass solutions:

$$0=\left(\gamma_{\mu}P^{\mu}\pm mc\right)$$

In choosing either + or - I would think one would have to make the same choice (for consistency) in all other areas of spacetime geometry.

For example, the infinitesimal invariant interval is generally expressed as:

$$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

The mathematical parallel to the first equation seems pretty straight forward: $$0=g_{\mu\nu}dx^{\mu}dx^{\nu}-ds^{2}$$

$$0=\left(\gamma_{\mu}dx^{\mu}-ds\right)\left(\gamma_{\nu}dx^{\nu}+ds\right)$$ however, in attempting to do this in another question of mine: (Is the Invariant interval S between two points independent of the path taken?) I was literally browbeaten by users. (This may have been due to my poor choice of wording)

My question is:

If the factorization of the first equation is soo completely accepted, then what is wrong with factoring the latter equation (i.e the interval)?

The answers I got didn't address any proper issues that I'm aware of, especially when compared to Dirac's factorization as above (If I took them at face value then the Dirac equation would be invalid ha!).

Answer

You are mixing up notation and actual algebraic operations. The metric is a section of the symmetric square of the cotangent bundle, which in turn is a factor of the symmetric algebra on this bundle. If you extend the scalars from the real numbers to the Clifford algebra of the scalars product in each point, you indeed can factorise the expression

$$g_{\mu\nu}dx^{\mu}dx^{\nu} =\left(\gamma_{\mu}dx^{\mu}\right)\left(\gamma_{\nu}dx^{\nu}\right)$$

as you did. Note that the $\gamma^\mu$ take values in a different Clifford algebra at every point. I guess one can find a parameterized matrix representation so that they become smooth matrix valued functions.

The problem is with your factorisation of $ds^2$. Unlike $m^2c^2$, which actually is a square, this is just notation. It seems to suggest that it is the square of some quantity $ds$, and moreover that this $ds$ actually is the differential of some function, but neither is true in general.

Where

$$m^{2}c^{2}=g_{\mu\nu}P^{\mu}P^{\nu}$$

Is an equation (with physical content),

$$ds^{2}=g_{\mu\nu}dx^{\mu}dx^{\nu}$$

is the expansion of $ds^2$, which could just as well (or better) have been written $g$ (for example), in a local basis derived from a coordinate basis.

quantum mechanics - Questions about Berry Phase

I'm learning about the Berry Phase from the original paper, and from the TIFR Infosys Lectures The Quantum Hall Effect by David Tong (2016).

I have some questions regarding the original derivation of the phase, and an example.

1. 
   

   In the derivation, Berry derives the phase as $$\gamma_n(C) = i \oint_C \langle n (\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle \cdot d\mathbf{R} $$ He then proceeds to say that the normalization of $|n(\mathbf{R})\rangle$ implies that $\langle n (\mathbf{R})|\nabla_{\mathbf{R}}n(\mathbf{R})\rangle$ is imaginary. Why does this implication arise?

   
   



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   In p. 33 of the notes, the example of a spin-1/2 particle in a magnetic field is used. Here the magnitude of the field remains constant but the direction can be changed. First, the Berry connection is calculated using spherical coordinates as $$\mathcal{F}_{\theta \phi} = -\sin{\theta} $$ And when the coordinates are reverted to Cartesian, it becomes $$\mathcal{F}_{ij} = -\epsilon_{ijk} \frac{B^k}{2|\mathbf{B}|^3} $$ This is called a magnetic monopole in parameter space. However when this quantity is integrated over the sphere of possible configurations (i.e. the sphere of radius $|\mathbf{B}|$ in parameter space), the following result is presented: $$\int \mathcal{F}_{ij}dS^{ij} = 4\pi g$$ where $g$ is the 'charge' of the monopole. How does this result come about?

   
   

   One idea I had for this one is that if we consider the Berry connection in Cartesian coordinates are considered in analogy to Coulomb's Law, then the field produced by the monopole should obey a kind of Gauss' Law. And if we take the quantities' analogues as follows: $$\frac{1}{4\pi\epsilon} \rightarrow 1 \text{ and } Q \rightarrow -\frac{1}{2}$$ Then we get the result. However I am not happy/convinced with this argument. Can someone shed some light on this?

   
   

Answer

The argument about the Gauss law is correct. The monopole magnetic field has exactly the same form of the electric field of a point charge, thus the Gauss law can be applied.

However, the integral can be perfectly performed in Cartesian coordinates. In fact, it is a basic exercise in the integration of differential forms. We have:

$$F = -\frac{g}{2} \frac{\epsilon_{ijk}}{|\mathbf{B}|^3}B^k dB^i \wedge dB^j$$ The integration is performed on the surface of the $2$-sphere: ${B^1}^2+{B^2}^2+{B^3}^2 = |\mathbf{B}|^2$

$$\int_{S^2} F = \int_{{B^1}^2+{B^2}^2+{B^3}^2 = |\mathbf{B}|^2} -\frac{g}{2} \frac{\epsilon_{ijk}}{|\mathbf{B}|^3}B^k dB^i \wedge dB^j$$

The integrand has 6 terms corresponding to the different permutations of $i,j,k=1,2,3$. The antisymmetry of the Levi-Civita tensor is countered by the antisymmetry of the wedge products. Thus due to the rotational symmetry of the sphere, all 6 terms have the same contribution and we can integrate one term and multiply by 6:

$$ \begin{align*} \int_{S^2} F &= \int_{{B^1}^2+{B^2}^2+{B^3}^2 = B^2} -6 \times \frac{g}{2} \frac{B^1}{|\mathbf {B}|^3}dB^2 dB^3\\ &= - \frac{3g}{|\mathbf {B}|^3} \int_{-|\mathbf {B}|}^{|\mathbf {B}|} dB^3 \int_{-\sqrt{|\mathbf {B}|^2-{B^3}^2}}^{{\sqrt{|\mathbf {B}|^2-{B^3}^2}}} dB^2 2 \sqrt{|\mathbf {B}|^2-{B^2}^2-{B^3}^2} \end{align*} $$ The integral on the $B^2-B^3$ plane is an integral over the equatorial disc. The factor of 2 in this integral takes into account the integration over the upper and the lower hemisphere. This integral can be performed using the substitution: $$B^2 = \sqrt{|\mathbf {B}|^2-{B^3}^2} \sin\phi$$ The $\phi$ limits will be $\frac{3\pi}{2}$ and $\frac{\pi}{2}$ respectively. Thus: $$ \begin{align*} \int_{S^2} F &= - \frac{3g}{|\mathbf {B}|^3} \int_{-B}^{B} dB^3 (|\mathbf {B}|^2-{B^3}^2)\int_{\frac{3\pi}{2}}^{\frac{\pi}{2}} d\phi 2 \cos^2\phi\\ &= - \frac{3g}{|\mathbf {B}|^3} \int_{-B}^{B} dB^3 (|\mathbf {B}|^2-{B^3}^2)\times -\pi\\ &= \frac{3\pi g}{| \mathbf {B}|^3} \int_{-|\mathbf {B}|}^{|\mathbf {B}|} dB^3 (|\mathbf {B}|^2-{B^3}^2)\\ &= \frac{3\pi g}{| \mathbf {B}|^3} \left (|\mathbf {B}|^2 B^3-\frac{{B^3}^3}{3} \right ) \bigg\rvert_{-|\mathbf {B}|}^{|\mathbf {B}|}\\ &= \frac{3\pi g}{| \mathbf {B}|^3} \frac{4 \pi}{3}|\mathbf {B}|^3 \\ &= 4 \pi g \end{align*} $$