LaTeX Issue

For the last few months I have been very busy and my laptop frequently gave me trouble.This was the reason I did not post anything in last few months...Anyway,I found that in between that the LaTeX symbols were not working properly...So,today I fixed it.

Active transformation & Passive transformation

These are famous terms in all math physics and QM books.Generally,they do not bother about explicitly writing the mathematical expressions which make them clearer.And it is also true,at least for a skeptic like me,to confuse the mathematical expressions as they look fairly similar.In a way,they should be;still I think it is worthwhile to match the meaning(told in every book) to those mathematical expressions.Today I do not have time to write on them.However,I am referring the interested reader to two books:Leslie Ballentine p-176 to 178 & Franz Schwabl (Advanced QM) section 7.1.These two should convey the idea to the reader.If I get time,I will write on them.A good pictorial sense is offered by Greiner p-19 in his book on symmetries.

Two Particle Interferometry

The present post will deal with the penultimate section of Gottfried's book.Here the author examines a two particle quantum state to illustrate something that has no classical counterpart.In this way,we will be able to appreciate the "full import" of the superposition principle.

Let us start from the classical double slit experiment.A classical double slit interference involves superposition of two coherent electric fields (secondary waves emitted from the different parts of the wavefront of the primary wave in conjunction with optical devices).Here,intensity is

$\ I_{total}\ =\ I_1\ +\ I_2\ +\ I_{12}\$

and this is equal (apart from a constant factor) to

$[\ <\vec{E}_1^2\ >_{T}\ +\ <\vec{E}_2^2\ >_{T}\ + 2\ <\vec{E}_1\cdot\vec{E}_2\ >_{T}]\$

Clearly,the first two terms are constant.The interference effect comes from the last term,namely the interference term.the interference term arises because of dotting of $\vec{E}_1\$ and $\vec{E}_2\$.

However,if the intensity of the source is so reduced that only a single photon is allowed to fall on the screen at a time,even then, the interference is observed after waiting sufficiently long time,so that our eyes can see the intensity pattern.This means,even a single photon can produce interference pattern, though our eyes may not perceive it.

This raises the question,how can a single photon produce interference...as it is believed to pass through any one of the two slits.This question was resolved in QM saying that a single photon is to be considered as linear superposition of the two states: one passing through the upper (1) pinhole,the other passing through the lower pinhole(2).A single photon is written as

$\ a\ |1>\ +\ b\ |2>\$

When there exists no evidence of which path the photon has followed,we can see the interference pattern.This interference pattern will consist of $\ <1|1>\$,$\ <2|2>\$, and $\ <1|2>\$ terms. The cross term is the interference term arising from the interaction between the two different parts of the "same photon",leaked through two holes.

Needless to say,this experiment has been successfully done with material particles (electrons) also,thus establishing the wave nature of matter.In all such cases,"the possible paths can be combined coherently (in this case the experiment does not reveal which path a photon/electron follows) resulting in an interference pattern.If the experiment reveals the path of the photon/electron,the interference pattern destroys.The meaning of "coherence" is sheerly different than in classical optics(coherence is no way an easy thing to understand,either in optics or in quantum mechanics(occurs frequently in many body QM;refer to Ballentine ch19.Later I will write something on quantum coherence when myself will be in a position to write).

Thus far,we have seen interference with the one-particle systems.These experiments have more or less direct counterpart to phenomena in classical optics.They do not show how enigmatic observations superposition principle can lead to.However,if we investigate two particle systems,because they are the next to the simplest,we have examples which have no classical counterpart and thus, the novel nature of quantum mechanics becomes clear.

A two particle quantum state will be most conveniently described by 3N=3X2=6 co-ordinates. We assign a six-dimensional configuration space to our system unlike the classical wave phenomenon.The multidimensional wavelike character of the quantum states gives coherence a richer meaning in quantum mechanics,than in classical optics.

If the two bodies in the system do not interact with one another,then,the Hamiltonian is H=H1+H2.Each body may have internal degrees of freedom;our demand is-they do not interact with that of the other body.Then,two distinct types of Schrodinger's equation comes into play: a two-body equation and two one-body equation.The consequence is that the product of one body wave functions is also a solution of the two body equation.

However, if the two particles in the system are mutually interacting,then,Hamiltonian will have an additional interaction term: H_12 and the above separation of variable method will not work.

We are now in a position to investigate the experiment described in the book:p-14.



The heart of the experiment is that "the two particle states can show interference effects if both particles are detected in co-incidence while not showing the conventional one-particle interference effects".Our experiment observes the daughter particles of the decay:

$\ A\rightarrow\ a\ +\ b\$

Our experimental set-up,we will make it possible to determine which path one particle takes by doing some measurement on the second particle;in this case neither of the two particles show interference $\ P_a\$ or $\ P_b\$,but there may be interference effect in the co-incidence rate $\ P_{ab}\$

It is important to appreciate that whether the observation is made or not does not matter;what matters is whether such an observation is possible or not.The strength of this statement will be clear shortly.

From the figure,we see the detector screens $\ D_a\$ and $\ D_b\$ are sensitive to a and b particles only.They record the co-ordinates of the particles striking them in coincidence.This is the joint probability distribution $\ P_{ab}\$.We agree to the convention that the upper pin holes would be designated (+) and the lower ones (-).

First of all,if A is at rest $\Delta\ p_A\rightarrow\ 0\$,its wave function resides across whole of the region within the two double-slit screens.In this case,the daughter particles are emitted back to back so as to conserve linear momentum.

Since we are assured that our set up enables us to determine the path followed by one state by performing some measurement on the other state,there will be no interference effect for each of them.Hence,the daughter states are to be seen as particles,not as waves.If the daughter particle behaved like waves,then, it were possible to write each of them in the form $\ c_{1}\ |+>\ +\ c_{2}\ |->\$ and interference pattern would have been obtained (provided there had existed no means to determine which path a state follows)---Our experiment does not entertain this facility.

We will make sure if $\ a\$ passes through the upper hole in the right side(+R),$\ b\$ must pass through the lower hole (-L) in the left side---i.e.$\ b\$ follows diametrically opposite path.Two jets of a and b particles rush in the opposite directions.If we name the pinholes as $\ +\ R\$,$\ -\ R\$,$\ +\ L\$ and $\ -\ L\$ then,there might be four such back to back jets of our concern: (+,+),(-,-).(+,-),(-,+).One such case(+,-) is shown below:



If the position of A were prescisely known to be at the centre of the set up,its momentum would be totally uncertain,and there will be no correlation of the directions of a and b,and the observation on any one of them cannot reveal the path taken by the other.In this case $\ I_1\$,$\ I_2\$ and $\ I_{12}\$ are all present.


All these being said,we must clarify how we do ensure that the two daughters can pass through one pair of diametrically opposite pin-holes.If particle $\ a\$ comes out from the upper hole in the right side,the corresponding particle $\ b\$ can either pass through (+L) or (-L) hole.It turns out that it depends on the size of the source.

We can look upon it this way: if the two daughter states are to pass diametrically opposite holes,i.e. A is to illuminate one or the other of the two opposed holes,then angle $\Theta\$ between their momenta (remembering $\Delta\ p_A\rightarrow\ 0\$ but $\Delta\ p_A\neq\ 0\$) must be very small compared to $\phi\$ where the latter is defined in this figure:



Note that when I drew the jets(2nd figure of the post),I drew the angle of the cone enough small.
Clearly,$\Theta\sim\frac{\Delta\ p_A}{\hbar\ k}\ll\phi\$.Here $\Delta\ p_A\$ is the uncertainty in the y component of A.This manifests as the difference angle $\Theta\$ between the momenta of the daughter particles.Again,from uncertainty of A's vertical position,$\Delta\ p_A\ge\frac{\hbar}{s}\$.From all these,we have $\ s\gg\frac{\ 1}{\phi\ k}\$

This condition being satisfied,the two daughter particles will be tracing opposite path and will reach diametrically opposite pin holes.In that case,measurement on one will definitely ensure the path traced by the other.

If this condition is not satisfied,i.e.the source size is small enough,then,$\Delta\ p_A\$ must be much bigger.This means,$\Theta\sim\frac{\Delta\ p_A}{\hbar\ k}\$ is much bigger compared to $\phi\$.In this limit,(+,+) or (-,-) decay cannot be neglected.In that case,one cannot definitely say the path traced by a particle by doing measurement on the other.



It is clear from the above figure,as the angle $\Theta\$,with the same $\Delta\ p_A\$,A may also illuminate two back to back pin-holes.Clearly, angle AOB > angle AO'B

Beyond the pin-holes,$\psi_{out}\$ is a linear combination of four terms:

$\psi_{out}\ =\ C_1\psi(\ L_a^+)\psi(\ L_b^-)\ +\ C_2\psi(\ L_a^+)\psi(\ L_b^+)\$

$\ +\ C_3\psi(\ L_a^-)\psi(\ L_b^+)\ C_4\psi(\ L_a^-)\psi(\ L_b^-)\$

Here $\psi\$ is a general spherical wave of the form $\psi\ =\frac{\ e^{\ i\ k\ r}}{r}\$

Matching the above with $\psi_{in}\$ at the boundary,we can evaluate the constants.Gottfried assumed a spherical wavefunction and the interaction to be rotation-invariant.This makes the state $\psi_A\ =\psi(\vec{r}_a\vec{r}_b)\$.From the symmetry argument in the book,we have

$\psi(\vec{r}_{\ a^+}\vec{r}_{\ b^-})\ =\psi(\vec{r}_{\ a^+}\vec{r}_{\ b^-})\ =\alpha\$ &

$\psi(\vec{r}_{\ a^+}\vec{r}_{\ b^+})\ =\psi(\vec{r}_{\ a^-}\vec{r}_{\ b^-})\ =\beta\$

And assuming that the screen is quite far off from the pin-holes,we arrive at eq (32) in the book:

$\psi_{out}\ =\alpha\ (\ e^{\ i\ k\ L_{a^+}}\ e^{\ i\ k\ L_{b^-}}\ + \ e^{\ i\ k\ L_{a^-}}\ e^{\ i\ k\ L_{b^+}}\ )\$

$\ +\beta\ (\ e^{\ i\ k\ L_{a^+}}\ e^{\ i\ k\ L_{b^+}}\ +\ e^{\ i\ k\ L_{a^-}}\ e^{\ i\ k\ L_{b^-}}\ )\$


Now,with proper substitutions,like $\ L_{\ a^+}^2\ =\ L_0^2\ +\ (\ a\ -\ y_a)^2\$ etc,we can get $\psi_{out}\$ in terms of other variables.

Since,in our experiment (+,+) and (-,-) decays are prohibited by adjusting the vertical height of the source,$\alpha\gg\beta\$ and we may safely neglect the $\beta\$ term in the expression of $\psi_{out}\$

The interference term comes out ~$\ P_{ab}\doteq\ |\ cos(\ k\theta\ [\ y_a\ -\ y_b])|^2\$



There will be a little more discussion on eqn (36) and (37) that shed more light on the experiment.I will not write right now as I am enough tired with this experiment.May be some time later I will be back at this.

This post will be related to the discussion on uncertainty in simultaneous measurement of electro-magnetic fields,as sketched out in Gottfried p.10-11.

Personally,I did not find earlier the notion of uncertainty in the simultaneous measurement of EM fields up to arbitrary accuracy.The author claims the simultaneous measurement of E-M fields up to arbitrary accuracy will tell us the position and momentum of the charge emanating those fields simultaneously with arbitrary accuracy.

Now,from classical electrodynamics,the most general electric and magnetic fields are given by Griffths equation (11.62) & (11.63).

$\vec{E}\ =\frac{\ q}{4\pi\epsilon_0}\frac{\ R}{(\vec{\ R}\cdot\vec{\ u})^3}\ [(\ c^2\ -\ v^2)\vec{u}\ +\vec{R}\times(\vec{u}\times\vec{a}\)]$

where $\vec{R}\ =(\vec{r}\ -\vec{r"})\$

$\vec{u}\ =(\ c\hat{R}\ -\vec{v})\$

And $\vec{B}\ =\frac{\hat{R}\times\vec{E}}{\ c}\$

They are complicated,as expected.But what does the present author mean by saying that "position and momentum" can be simultaneously measured? I hope it is difficult to invert the formulae above,if not impossible.

The answer is not that straightforward.After a second thought,I found that given $\vec{E}\$ and $\vec{B}\$,there might be a way to solve for the momentum and velocity.But,we must refer to relativistic results.

In relativity,the law of motion is expressed in terms of three force and three velocities etc. as

$\vec{F}\ =\frac{\ m}{\sqrt(\ 1\ -\frac{\ v^2}{\ c^2}) }[\vec{a}\ +\frac{\vec{v}(\vec{v}\cdot\vec{a})}{\ c^2\ -\ v^2}]\$

[Griffiths,eqn-12.73;this is not difficult to prove]

Since the electric and magnetic fields are known to us,the force on the charged particle or currents may also assumed to be known(I must check this point).Then,the above becomes a differential equation in the first time derivative of $\vec{v}\$.It is not possible to say that this equaton is solvable.However,under certain condition on the net force,velocity can be found as a function of time;and hence,the position.One such example is worked out in Griffiths 12.10,though the context is entirely different.

That the speed can be found as a function of time is also clear from the result of problem 12.
36.It uses the formula written above and says

$\vec{a}\ =(\frac{\ q}{\ m})\sqrt(\ 1\ -\frac{\ v^2}{\ c^2}) [\vec{E}\ +(\vec{v}\times\vec{B})\ -\frac{\vec{v}(\vec{v}\cdot\vec{E}}{\ c^2}]\$

Again,the form of the equation is not so illuminating to solve.But under certain condition (I do not anticipate what the conditions could be at present),$\vec{a}\ =\frac{\ d\vec{v}}{\ dt}\$ can be solved for $\vec{v}\$.
This would give the momentum and hence,position...

Again,the uncertainty product for electromagnetic fields as written there is somewhat strange.The uncertainty is said to evolve from the interference of one test body in one space-time volume $\Omega_1\$ with another test body (how does it come into picture?) in space-time volume $\Omega_2\$ (the space-time regions can connected by light signal).This is unlike the uncertainty principle we are familiar with...position and momentum cannot be measured simultaneously with arbitrary accuracy.There we need not worry about the presence of any other body anywhere.

However,we will be in a position to justify the claim later when we read the relevant chapter.

"Definiteness" in the Position of photon

The above does not mean that photon's position can never be specified;consider the case where one tries to find out which path a photon follows in the double slit experiment.In fact,the position measure-ment gives $\ x\ +\delta\ x\$ where $\delta\ x\sim\lambda\$ and cannot be made further less anyway whereas for massive particles,we just saw $\delta\ x\$ can be made $\ll\lambda_{deB}\$.If it were possible,$\delta\ p \gg\frac{\hbar}{\lambda}\ =\frac{\hbar\omega}{\ c}\$; i.e. $\delta\ p\sim\ p\gg\frac{\hbar\omega}{\ c}\$ which must be wrong.

Less excerpts from the books

This is to say that from now onwards,I will be more economical in posting parts from the text surely for copyright reasons.

What Is the "Inability" he refers to?

This discussion should be compared with the first topic in this blog.There we said,following Gottfried,that a massive particle's position can best be estimated as $\delta\ x\sim\lambda_C\$.This is the lower limit of the scale where $\lambda_C\$ is the Compton wavelength of the particle.

The error in the non-relativistic length scale goes down to the order of $\lambda_C\$ which satisfies $\lambda_C\rightarrow\ 0\$ and this means in the non-relativistic limit,we are able to specify the particle's position with fantastic accuracy.

After saying all these,Gottfried says the inability to specify a massive particle's position to ARBITRARY accuracy is acceptable in non-relativistic Quantum Mechanics.I do not understand what does he mean by saying inability...

It is true that $\lambda_C\$ is very very small while being finite yet.Hence,the word "inability" is not wrong in that sense...But then why should he has written it is zero in the non-relativistic limit!

An Experiment Demonstrating A Fundamental Result

The LASER beam is producing linearly polarized light which can be thought of as to be produced by equal number of R state photons and L state photons.Therefore,the Hg+ ion has equal probability of interacting with +h spin photons or -h spin photons[Hecht,p-330:discussion on angular momentum picture of photon].Therefore, as a whole,the LASER light cannot possibly impart angular momentum to any of the ions and flip its spin.Hence,the probability of (resonant) scattering is small.This makes the experiment a succession of individual photon collisions by one OR the other of the two ions---ONE AT A TIME.The observed scattering is non-resonant scattering.

If the ions were point scatterers,they would have scattered the light (may be their unique P state would have changed---but as far as interference is concerned,state of polarization does not matter).Light scattered from the trap material would not interfere for lack of coherence. Whereas,scattered light from the two scondary point sources will maintain constant phase relationship and will interfere.This way it is equivalent to Young's double slit experiment.

Even though we started with linearly polarized light,it's not ideal.A few trace of elliptically polarized light will invole unequal count of R and L state.As a ressult,net angular momentum CAN BE imparted on the Hg+ ions (though probability is extremely small).This corresponds to resonant scattering where spin-flip occurs.More generally,where there is no spin-flip,non-resonant scattering occurs and scattered wave produce interference patterns.

LaTeX

This is just to announce that I have enabled LaTeX in this blog.

$\sum_{i=1}^nx_i$

$\ x^2\ -\frac{a}{b}\$

So,it's working nice...It took some time to understand that the black colour of the LaTeX was overlapping with the black background of my blog.So,I changed it.

Unfamiliar Aspects of Uncertainty Principle & "Instantaneousness" of Position and Momentum Probability Distributions

Gottfried introduces a series of questions to the betterment of the understanding of the process of position and momentum measurement.We consider them as below:






Consider a position measurement is being done on a particle.Say,the clock shows time t at the instant we measure the particle at x; if the uncertainty in the position measurement is δx and the corresponding uncertainty in the time-count is δt,then we are interested in the minimum of δt as t+δt is the degree of definiteness of the time measurement.The less is δt,the more definite is the time.

If we take a closer look,the photon which comes from x hits the detector at time t.But the photon coming from x+δx must hit the detector at time different from t; say,at t+δt.So,we are assigning δt to the scattered photon coming late or before from x+δx.
δx is some finite distance multiple of the photon wavelength λ' (wavelength of the scattered photon is the unit distance in this problem).So,the time error is δt~λ'/c ~>(ћ/mc2)





In Compton scattering the photon is scattered in a variety of angles (empirically produced probability distribution).So,it is a stochastic (random and only statistically predictable) drift in various angles.If we wish to reproduce the position measurement accurately,we must perform the measurememnt very quick so that the wave-function does not spread appreciably.As a result,it is found where it was found the last time.

Momentum is [mass*(L/T)] where L=|x1 - x2| and T=(t2 -t1).Measurement of both x and t involves uncertainty or error.With large L,δx1 and δx2 are small compared to L=|x1 - x2|.Similarly, δt1 and δt2 which we found ~[ћ/mc2] are also very small compared to (t2-t1).Hence,we are more or less justified with momentum=mass*|x1 – x2|/T.




A free particle state with well defined position will not persist in general.With time,it will evolve.If we remember the instance of Compton effect,the state will lose its localization by stochastic drift.As time increases more and more,τ(p+δp)/m [=distance]of different particles will increase more and more and the accuracy of position determination decreases.Clearly,as the time shrinks,the accuracy of position determination increases.






As we found earlier,momentum measurement is done by measuring the position of the particle at two different points: x1 and x2.So,again we wish to find the minimum δt to see how quickly momentum measurement can be done.

Let x1 be the point where 1st Compton scattering event occurs, and say we want to produce a state of so and so (momentum) magnitude and direction.Then we set x2 lie in that direction.Then,we select the photon scattered from x2.This will prepare (and hence,determine) a particle with desired momentum.Clearly,in this case, δt=λ/c can be made arbitrarily small.

Since, δt can be made indeed very small,(t+ δt) approaches t, the assumption of non-relativistic quantum mechanics that position and momentum probability distributions exist instantaneously is justified.