electromagnetic radiation - Polarization and mirrors

When a light beam reaches a dielectric surface, the incident and reflected beams have different intensities depending on polarization. For the so-called Brewster's angle, the reflected light is lineary polarized.

My question is: how does this law work in case of mirror-like surface, when (ideally) all the light is reflected?

Answer

The easy answer is to say that Brewster's law only applies to reflection from the interface with a transparent medium, and a mirror isn't transparent. Indeed for an ideal perfect mirror, all light of both polarizations is reflected perfectly, so there is nothing to say.

For an actual real-world mirror, the metal mirror surface will have a finite skin depth, and can be considered a dielectric medium with a very large, complex index of refraction. This does lead to a small polarization dependence of the reflection coefficient for near grazing incidence angles. The analogue of Brewster's angle occurs at an angle given by 2*pi*(skin depth) / wavelength above grazing, where the parallel polarized reflection coefficient will reach a minimum ( but not zero, still close to 1). For more details, see for example Landau and Lifshitz, volume 8, section 87.

"Inverted" quantum oscillator

I'm trying to understand the problem of the "inverted" oscillator, which has the following Hamiltonian: $$ \hat{H}=\frac{\hat{p}^{2}}{2m}-\frac{k\hat{x}^{2}}{2} $$

Suppose that a particle at the initial time is in a state

$$ \Psi(x)=\frac{1}{(\pi b^{2})^{\frac{1}{4}}}\exp^{-x^{2}/2b^{2}} $$

We want to find $\langle\Psi\mid\hat{x}(t)\mid\Psi\rangle$ and $\langle \Psi \mid \hat{x}^{2}(t) \mid \Psi \rangle$ and their asymptotic when $t \rightarrow +\infty$.

Heisenberg equation for operator $\hat{x}$ (we use that $\hat{x}$ does not depend explicitly on time, $[\hat{x},\hat{x}]=0$, $[\hat{x},\hat{p}]=i\hbar$ and Leibniz rule:$[A,BC]=[A,B]C+B[A,C]$):

$$ i\hbar \dot{\hat{x}}=\left[\hat{x},\hat{H}\right]=\left[\hat{x},\frac{\hat{p}^{2}}{2m}-\frac{k\hat{x}^{2}}{2}\right]=\left[\hat{x},\frac{\hat{p}^{2}}{2m}\right]=\frac{\hat{p}}{m}\left[\hat{x},\hat{p}\right]=i\hbar\frac{\hat{p}}{m} $$

$$ \dot{\hat{x}}=\hat{p}/m $$

Similarly, Heisenberg equation for operator $\hat{p}$:

$$ i\hbar\dot{\hat{p}}=\left[\hat{p},\hat{H}\right]=\left[\hat{p},\frac{\hat{p}^{2}}{2m}-\frac{k\hat{x}^{2}}{2}\right]=-\left[\hat{p},\frac{k\hat{x}^{2}}{2}\right]=-k\hat{x}\left[\hat{p},\hat{x}\right]=i\hbar k\hat{x} $$

$$ \dot{\hat{p}}=k\hat{x} $$

So, we have following set of equations: $\dot{\hat{x}}=\hat{p}/m$; $\dot{\hat{p}}=k\hat{x}$.

Using them, we can write: $$ m\ddot{\hat{x}}=k\hat{x} $$

Solution of this differential equation is $\hat{x}=A\cosh(\omega t)+B\sinh(\omega t)$, where $\omega=\sqrt{\frac{k}{m}}$.

Initial conditions: $A=\hat{x}(0)$; $B=\frac{\dot{\hat{x}}(0)}{\omega}$. Remember that $\dot{\hat{x}}=\hat{p}/m$. Then $B=\frac{\hat{p}(0)}{m\omega}$.

So we have $$ \hat{x}=\hat{x}(0)\cosh(\omega t)+\frac{\hat{p}(0)}{m\omega}\sinh(\omega t) $$

Then $$ \hat{x}^{2}=\hat{x}^{2}(0)\cosh^{2}(0)+\frac{\hat{p}^{2}(0)}{(m\omega)^{2}}\sinh^{2}(0)+\frac{1}{m\omega}(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\sinh(\omega t)\cosh(\omega t) $$

Now we want to calculate $\langle\Psi\mid\hat{x}\mid\Psi\rangle$ and $\langle \Psi \mid \hat{x}^{2} \mid \Psi \rangle$:

$$ \langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=\langle\Psi\mid\hat{x}(0)\cosh(\omega t)+\frac{\hat{p}(0)}{m\omega}\sinh(\omega t)\mid\Psi\rangle=\langle\Psi\mid\hat{x}(0)\mid\Psi\rangle\cosh(\omega t)+\langle\Psi\mid\hat{p}(0)\mid\Psi\rangle\frac{\sinh(\omega t)}{m\omega} $$

Similarly

$$ \langle \Psi \mid \hat{x}^{2}(t) \mid \Psi \rangle=\langle\Psi\mid\hat{x}^{2}(0)\mid\Psi\rangle\cosh^{2}(\omega t)+\langle\Psi\mid\hat{p}^2(0)\mid\Psi\rangle\frac{\sinh^{2}(\omega t)}{(m\omega)^{2}}+\frac{1}{m\omega}\langle\Psi\mid(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\mid\Psi\rangle\sinh(\omega t)\cosh(\omega t) $$

Lets introduce the notation:$x_{0}=\langle\Psi\mid\hat{x}(0)\mid\Psi\rangle$, $p_{0}=\langle\Psi\mid\hat{p}(0)\mid\Psi\rangle$,$x_{0}^{2}=\langle\Psi\mid\hat{x}^{2}(0)\mid\Psi\rangle$, $p_{0}^{2}=\langle\Psi\mid\hat{p}^{2}(0)\mid\Psi\rangle$. Then:

$$ \langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=x_{0}\cosh(\omega t)+\frac{p_{0}}{m\omega}\sinh(\omega t) $$

$$ \langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle=x_{0}^{2}\cosh^{2}(\omega t)+p_{0}^{2}\frac{\sinh^{2}(\omega t)}{(m\omega)^{2}}+\frac{1}{m\omega}\langle\Psi\mid(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\mid\Psi\rangle\sinh(\omega t)\cosh(\omega t) $$

Here is the first big trouble: i've found some lecture notes in the internet, where they say that

$\langle\Psi\mid(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\mid\Psi\rangle=0$ for real $\Psi(x)$. I tried to prove this statement, but i can't. Can you explain why it is equal to zero?

Well, suppose that this statement is true. Then: $$ \langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=x_{0}\cosh(\omega t)+\frac{p_{0}}{m\omega}\sinh(\omega t) $$

$$ \langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle=x_{0}^{2}\cosh^{2}(\omega t)+p_{0}^{2}\frac{\sinh^{2}(\omega t)}{(m\omega)^{2}} $$

Thus, we expressed values through their meanings at the initial time. So, lets find $x_{0}$, $p_{0}$, $x_{0}^{2}$, $p_{0}^{2}$. From this point we will work in the coordinate representation.

$$ x_{0}=\langle\Psi\mid\hat{x}\mid\Psi\rangle=\int\Psi^{\ast}x\Psi dx=\int\Psi x\Psi dx=\int\Psi^{2}xdx $$

Remember that $\Psi(x)=\frac{1}{(\pi b^{2})^{\frac{1}{4}}}\exp^{-x^{2}/2b^{2}}$ :

$$ x_{0}=\frac{1}{(\pi b^{2})^{\frac{1}{2}}}\int x\exp^{-x^{2}/b^{2}}dx $$ We integrate in domains from minus infinity to plus infinity. But this integral is equal to zero! So $$ x_{0}=0 $$ Similarly: $$ p_{0}=\langle\Psi\mid\hat{p}\mid\Psi\rangle=-i\hbar\int\Psi^{\ast}\frac{\partial}{\partial x}\Psi dx=\frac{2}{b^{2}}i\hbar\frac{1}{(\pi b^{2})^{\frac{1}{2}}}\int x\exp^{-x^{2}/b^{2}}dx $$ The domains of integration are the same:from minus infinity to plus infinity. But this is also is equal to zero! So, we have : $$ \langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=0\cosh(\omega t)+\frac{0}{m\omega}\sinh(\omega t)=0 $$ I'm confused by this result! Is there a mistake? If not, can you explain me this result?

Now we are going to calculate $x_{0}^{2}$ and $p_{0}^{2}$: $$ x_{0}^{2}=\langle\Psi\mid\hat{x}^{2}\mid\Psi\rangle=\int\Psi^{\ast}x\Psi dx=\frac{1}{(\pi b^{2})^{\frac{1}{2}}}\int x^{2}\exp^{-x^{2}/b^{2}}dx=\frac{1}{(\pi b^{2})^{\frac{1}{2}}}\frac{\sqrt{\pi} (b^{2})^{\frac{3}{2}}}{2}=\frac{b^2}{2} $$

$$ p_{0}^{2}=\langle\Psi\mid\hat{p}^{2}\mid\Psi\rangle=-\hbar^{2}\int\Psi^{\ast}\frac{\partial^{2}}{\partial x^{2}}\Psi dx=\frac{\hbar^{2}}{2b^{2}} $$

Finally: $$ \langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle=\frac{b^2}{2}\cosh^{2}(\omega t)+\frac{\hbar^2}{2b^{2}}\frac{\sinh^{2}(\omega t)}{(m\omega)^{2}} $$

Using the equation $\cosh^{2}(x)-\sinh^{2}(x)=1$ we get:

$$ \langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle=\frac{1}{2}\left(b^{2}+\frac{\hbar^{2}}{b^{2}(m\omega)^{2}}\right)\cosh^{2}(\omega t)-\frac{\hbar^{2}}{2b^{2}(m\omega)^{2}} $$

Thus we see that when $t \rightarrow +\infty$ dispersion of coordinate $\langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle$ tends to infinity.

To sum up, I have these several difficulties:

1. why $\langle\Psi\mid(\hat{x}(0)\hat{p}(0)+\hat{p}(0)\hat{x}(0))\mid\Psi\rangle=0$ for real $\Psi(x)$?


2. why $\langle\Psi\mid\hat{x}(t)\mid\Psi\rangle=0$?



3. why when $t \rightarrow +\infty$ $\langle\Psi\mid\hat{x^{2}}(t)\mid\Psi\rangle$ tends to infinity? Also i would be happy if you provide some articles(or books) where i can read about it.

thermodynamics - How does cooling scale with volume?

What equation would give me the answer to the question, "If i have a cup of water at a tempature of say boiling, how long would that cup of water take to cool off compared to say half that size of a cup of water." So the volume is in half. Its a general question I am just looking for where to start.

Answer

An actual cup is slightly complicated because there are 2–3 distinct types of surfaces. Let's deal with free-floating cubes of water instead, both at an initial temperature T_i in an environment with temperature T_e .

A cube with half the volume will have 50% the thermal mass C, but 63% of the surface area A. Newton's Law of Cooling implies $\frac{dT}{dt}=-h\frac{A}{C}(T-T_0\!)$ , where h is a property of the environment. So the smaller cube will cool 26% faster initially, when both cubes are at T_i .

If you want to know the temperature of a cube at any given time, the solution to the differential equation above is $\frac{T-T_e}{T_i-T_e}=\exp\left(-ht\frac{A}C\right)$. If you solve for t, it follows that when the small cube is at a given temperature, it will take the large cube 26% more time to reach it.

quantum mechanics - Complete set and Klein-Gordon equation

In http://www.physics.ucdavis.edu/~cheng/teaching/230A-s07/rqm2_rev.pdf, it says that when there is some external potential, the Klein-Gordon equation is altered, and it says the following:

The solution $\Psi$ can always be expressed as a superposition of free particle solutions, provided that the latter form a complete set. They from a complete set only if the negative energy components are retained, so they cannot be simply discarded.

Can anyone explain what this is talking about? First of all, what is a complete set? And why does forming a complete set require negative energy components being retained? Why is the equation (solutions, I guess) need to be a complete set?

Answer

This statement is false. The positive energy solutions of the free Klein Gordon equation form a complete set. What this means is that any wavefunction of the form $\psi(x)$ where x ranges over space not time, can be written as a sum of free-particle solutions.

In this case, the way you do it is by Fourier transform: you write $\psi(x)$ as

$$ \psi(x) = \int e^{ik\cdot x} \psi(k) dk $$

and this expands an arbitrary function, and then you make it time dependent by replacing $e^{ikcdot x}$ with $e^{ik\cdot x - i E_k t}$ where $E_k = \sqrt{k^2 + m^2}$. The result gives the positive energy time evolution of any initial wavefunction data.

This is no good not because it isn't complete, but because it isn't local-- the single particle positive energy solutions go places faster than light. This requires making a multiparticle theory to fix the formalism.

The completeness is not completeness in the sense of expressing every solution of the Klein Gordon equation in terms of initial data, because the initial data for KG is the function $\psi$ and it's first time derivative. This is not the data you specify for a quantum mechanical wavefunction, it's the data for a classical field.

The book is confused on this issue, and you should skip this part.

classical mechanics - What is the function type of the generalized momentum?

Let

$$L:{\mathbb R}^n\times {\mathbb R}^n\times {\mathbb R}\to {\mathbb R}$$

denote the Lagrangian (it should be differentiable) of a classical system with $n$ spatial coordinates. In the action

$$S[q]=\int_{t_1}^{t_2} L(q(t),q'(t),t)\, \mathrm{d}t,$$

the first $n$ slots are evaluates at a path $q:{\mathbb R}\to {\mathbb R}^n$, the second $n$ at $q'$ and the last one is for possible explicit time dependencies.

Are generalized momenta defined as functions of the generalized coordinates, i.e.

$$p_j=\frac{\partial L(x^1,\dots,x^n,v^1,\dots,v^n,t)}{\partial v^j},$$

or as associated with a curve $q:{\mathbb R}\to {\mathbb R}^n$, i.e.

$$p_j=\left.\frac{\partial L(q^1,\dots,q^n,v^1,\dots,v^n,t)}{\partial v^j}\right\rvert_{\large{q=q(t),\ v=q'(t)}},$$?

In the latter case it's a function of time only, and $q$ is buried somewhere within it.

A question that is codependent with this might be: What is the type of the total derivative and the term $\frac{\partial \mathcal{L}}{\partial q_i}$ in differential equation in $q:{\mathbb R}\to {\mathbb R}^n$, which is usually expressed as

$$\frac{\mathrm{d}}{\mathrm{d}t} \dfrac{\partial \mathcal{L}}{\partial {\dot q_i}} - \frac{\partial \mathcal{L}}{\partial q_i} = 0.$$

newtonian mechanics - Why don't we resolve $mg$ i.e. weight while solving problems related to banking of road in Circular motion?

It requires concept from Newton's laws of motion and circular motion. While solving for net force on a body on an inclined plane , we always resolve mg (weight) parallel and perpendicular to the plane but in the banking of road concept, we resolve normal reaction. Why don't we simply resolve weight and then equate it with normal reaction?

Answer

As a general rule, resolve force vectors into components that are parallel to and perpendicular to the acceleration of the object. In the case of a vehicle driving around a banked curve the acceleration is horizontal, towards the center so it's the normal force (and friction if any) that you would resolve into components.

quantum mechanics - What are phase conventions in angular momentum and rotation calculations?

I work with complicated angular momentum calculations related to atomic physics; nevertheless, I never need to use anything related to a phase convention (apparently because it's taken care of in a consistent way everywhere with me not noticing it), but I keep reading this term within subjects related to Clebsch-Gordan coefficients, Wigner D-matrices, 3j and 6j symbols, and Racah W-Coefficients, and more.

Can someone please explain what a phase convention is and how to use/utilize it?

Or guide me to some good book that would explain that?