Apologies for the Italian parts. If someone wishes to translate them into English, I'll publish that.
This has gone through the mail, so simple text formatting is used, but it should read fairly clearly anyway.
> true). As I progress I keep coming across something called a `phase
> factor` which is applied to wavefunctions. Would you be kind enough to
> explain what the point of a phase factor is? [Am I correct in thinking
> this is also called the Condon-Shortley phase?]. Why do we need it at
> all? This is probably simple question to answer but so far I haven`t
> managed to find an explanation. When I studied associated Legendre
> polynomials a phase factor was applied but now I`m looking at the radial
> wavefunction for hydrogen and it doesn`t seem that a phase factor is
> applied in this case.
OK, the wavefunctions are complex, in general. There are cases in which
you can take them to be real (like the radial wavef. of the H atom), but
often you just have to consider complex functions.
Having said that, consider a wavefunction, any that you like, for example
one of the eigenfunctions of the hydrogen atom you are familiar with. In
quantum mechanics linear operators represent observables (such as the
energy or position or momentum or angular momentum...), and the
"expectation value" (or, if you prefer, mean value) of some physical
observable is the integral:
integrate (psi^conj(x) Operator psi(x) ) dx
Operator acts on the function psi and gives a new function, you multiply
that by the complex conjugate of psi (indicated by psi^conj), and integrate
the resulting function all over space.
Now, LINEAR operator means that Operator (k psi) = k Operator (psi)
where k is any (constant, independent of x) complex number.
Therefore, suppose you decide to multiply your original psi by a constant
number (call it again k) of modulus unity (you want psi to remain
normalized...). You now have a new function, let's call it
chi(x) = k psi(x).
If you apply the same rule as above to compute the expectation value of the
same observable as before on the new wavefunction chi instead of psi, you
can easily convince yourself that you get the same number as before,
essentially because of the linearity of Operator and because
k^conj * k = 1
A complex number of modulus 1 is called a phase factor. We have
demonstrated that any phase factor can multiply any wavefunction without
changing the expectation value of any observable on that wavefunction!!
So phase factors are immaterial.
....almost!
There are subtle cases where you make interfere 2 wavefunctions and get some
effect out of those phase factors (see my page on geometric phases).
The Condon-Shortley phase is actually a phase CONVENTION.
We have seen that phases of quantum states are immaterial.
But if you want to write down on paper a wavefunction you have eventually
to make this choice. Well, the Condon-Shortley phase convention is about
this phase choice for coupled angular momentum states. In standard
textbooks you could look for the Clebsg-Gordan coefficients.
By the way, I strongly advise you to read the first 3 chapters of the
Sakurai book "Modern Quantum Mechanics": it is illuminating.
-------------------------------------------------------------------------------
> Would I only apply a phase factor to a wavefunction which is
> complex or could I also
> apply it to the radial wavefunction (which is not complex)?
> When you have a wavefunction written down in front of you how do you
> decide whether
> it needs a phase factor or not?
You do not "apply" a phase factor.
You just multiply the wavefunction by a number of modulus 1.
Wevefunctions do not "need" phase factors in front.
It is up to you to decide to put one or not.
For example, the radial wavefunction
psi = C*exp(-r/a)
(C is some real normalization factor)
and the other wavefunction
chi = C*exp(-r/a) * exp(i u)
(i is the imaginary unit and u is a FIXED real number, like 0.23 or -12870.2,
called the phase angle)
are perfectly equivalent and represent equally well the SAME state.
It is up to you to decide which one you like best.
(psi, I bet, not chi!)
> Is the phase factor connected to the idea of a simple wave having a `phase`?
No. There is nothing "wavy" in a fixed complex number.
Well, in a very restricted sense yes, but I think this might get you confused.
Basically NO.
> Is the `Condon-Shortley phase` a general phase factor which applies to
> several wavefunctions, or is it special/specific to Legendre functions
> (spherical harmonics)?
To be precise, neither.
But the second option is more close to the truth.
> Are the Berry phase and the Condon-Shortley phase just different `types`
> of phase factor, but still just phase factors?
Yes. There is a difference, though.
The Condon-Shortley phase is just a CONVENTION.
Exactly like you decide to take psi as radial wavefunction and not chi in the
example above, except it applies to angular wavefunctions.
The Berry phase instead is given by some geometric properties of the quantum
system in exam, you cannot change it.
-------------------------------------------------------------------------------
> I have finally managed to get to the ladder operators for angular
> momentum. {I
> looked a these for the QM HO some months ago]. So far so good but I`m a bit
> stuck on understanding how it is that J_+ is proportional to |n, m+1>.
> This does not jump out at me from the page of equations. I have the
> following:
> J_z J_+ |n, m> = (m+1) hbar J_+ |n,m>
> J_z |n,m> = m hbar |n,m>
> J_z |n, m+1> = (m+1) hbar |n, m+1>
> Now, apparently, from these we can see that J_+ is proportional to |n,
> m+1>,
> giving:
> J_+ |n, m> = c_+ |n, m+1>
> This is absolutely key to seeing how the ladder works but I don`t see
> it! Is it possible to break it down so that it becomes more obvious?
The second equation
J_z |n,m> = m hbar |n,m>
actually means that a state |n,m> is an eigenstate of J_z with eigenvalue m hbar
Now, the last equation
J_z |n, m+1> = (m+1) hbar |n, m+1>
tells you the same thing for state |n, m+1>, and since m is any number,
it is equivalent to the previous one.
Given this, the crucial equation is the first one.
J_z J_+ |n, m> = (m+1) hbar J_+ |n,m>
Read it with brackets like this:
J_z (J_+ |n, m>) = (m+1) hbar (J_+ |n,m>)
This tells us that the ket within () is an eigenstate of J_z of eigenvalue
(m+1) hbar.
So, the correct statement is not that "J_+ is proportional to |n, m+1>", but
that J_+ is a m-raising operator, that acts on an eigenstate of J_z with
eigenvalue m hbar and transforms it in a new eigenstate of J_z with eigenvalue
(m+1) hbar.
Once this is clear to you, please note that this does NOT prove that
J_+ |n,m> = |n, m+1> !!
Indeed this last eq. is usually wrong. There is a normalization factor to
account for.
-------------------------------------------------------------------------------
> The other thing is the Heisenberg Uncertainty Principle. I have been
> having another look at the EPR experiment because I`d like, eventually,
> to get to Bell`s theorem, and it seems that it is a good idea to
> understand EPR first! There seem to be several versions/takes on the EPR
> experiment, but this question is more fundamental. Does the HUP mean
> that we cannot know both x and p_x to better than the standard
> uncertainty because the act of measuring means that we disturb the
> system, or does it mean that the actual of existance of x and p_x are
> ruled by the HUP - in other words, it wouldn`t matter even if we could
> measure without disturbing the system, the information we seek is
> inherently affected by the HUP anyway, ie it is not there in the first
> place? If the latter is the case, why do so many books talk about the
> act of measuring disturbing the system (thus giving rise to the HUP)
> when in fact it is irrelevant to the existence of the HUP? So perhaps
> with the act of measuring we get a double effect, a combination of the
> HUP per se plus the `jolt` suffered by the system as a result of taking
> a measurement?
> I hope this makes sense - I`m trying to really get into the mind of the
> HUP, and
> understand exactly what role `measurement` really has in relation to it.
This is one of the deepest points of quantum mechanics, tons of paper has been
inked on this subject.
My attitude is the following: if you believe in Schroedinger (or, say
Dirac) wave description of matter alone, then matter is made of waves that
move in a perfectly deterministic way. The square modulus of the psi is to
be interpreted as a probability distribution. Averages observed properties
are represented by means (expectation values) of some operators.
In this frame, HUP is just a mathematical result that for any two
non-commuting operators, the joint probability distribution is never
factorable, for no psi. Therefore the product of the "spread" of the two
observables around their mean value has a minimum, connected to the value
of the nonzero commutator.
The only problem with all this arises when you actually MAKE measurements.
The crucial postulate of Q.M. is that when you measure some quantum
quantity you do not get its average value, but one of its eigenvalues with
a certain probability. If you measure position, say, then you always find
that your quantum particle is "somewhere", in a well defined (but random)
point in space no matter how spread out is its psi.
This means that the measurements affects deeply your quantum
state no matter how "delicate" is the measurement technique you use.
The x,p case is a slightly complicate and confusing, though.
A much clearer case is that of spin, say spin 1/2. If this subject really
matters to you, I urge you to read through and really understand the first 6
pages of Sakurai's book. Of course, also the interpretation of Sect. 1.4
has to be meditated at length....
-------------------------------------------------------------------------------
> Actually, I am still struggling with the ladder operators for AM, so I`m very pleased
> with your explanation. Hopefully this weekend I can get them finished!
> It is interesting what you say about "this does NOT prove that J_+ |n,m> = |n, m+1> !!"
>
> This is something I have also thought about. I know that the there is a constant of
> proportionality which must be included but it is difficult to understand why it is
> needed, because if you simple follow the algebra then you are led to the conclusion
> that J_+ |n,m> EQUALS |n, m+1> . Yet this is not true. It is difficult to understand
> why the constant of proportionality is required. Maybe if I cannot find an answer to
> this question I will come back to you.
It's easy: take an eigenstate |psi> of some operator, say J_z (any linear
operator, in fact), with some eigenvalue, say (m+1) hbar (whatever number
it may be, in fact):
J_z |psi> = (m+1) hbar |psi>
Then take any constant c times this eigenstate: you get another eigenstate
with the same eigenvalue:
J_z (c |psi>) = (m+1) hbar (c |psi>)
This is true since J_z is a LINEAR operator, thus, you can take a constant
c out of its argument.
This means that eigenstates are always defined up to a constant.
You may fix the modulus of the constant by requiring that the state is
normalized.
However there is no way to fix the phase of this arbitrary constant.
And this takes us back to the origin of all our exchange of mails...
-------------------------------------------------------------------------------
> Can I ask you something about the polarisation of light? I am trying
> to move towards an understanding of Bell`s theorem and many
> explanations use light in their example. Now I know that photons
> are quantum particles and so we are not on solid ground trying to get a
> `picture` of them BUT, perhaps we can try to make some sense of the
> language used to describe the `polarisation` of light. There seems to be
> basically three
> descriptors used for the polarisation: plane polarised, circularly
> polarised and `unpolarised`. The only books I have which try and explain
Yes, you have focused on the 3 main "kinds" of light.
> what these terms mean (with the exception of `unpolarised`) are the
> `popular quantum mechanics` type such as `In Search of Schroedinger`s Cat`
> by John Gribben. `Plane polarised light` is described as like a simple
> wave, sort of two-dimensional (like a wave on a fixed string) such that it
> can pass through an appropriately oriented picket fence. We can have
> vertically polarised and horizontally polarised, or anywhere in between. It
> seems that the distinction here is how these different types of angularly
> polarised light interacts with a polarised filter. This is okay, I can
> understand this description/definition. But how does this compare with
> circularly polarised light and unpolarised light? What are these two
> exactly? If we could pick out one photon of circularly polarised light
> what would it `look` like next to a plane polarised photon or an
> unpolarised photon? Is CP light so called because it is made up of
> angularly polarised light (all angles), or is there something more
> fundamentally different about it? Is each individual photon in CP light
> circularly polarised or is it just a numbers effect whereby if we have
> enough photons all at slightly different angles we might as well call it CP
> light. When CP light passes through a polaroid filter, are the photons
> which emerge just the plane polarised ones which are present in the CP
> light which happen to be at the necessary angle to pass through, or is each
> individual photon in the CP light itself circularly polarised
> and......there is a different process taking place? Well, I hope I have
> managed to give an indication about what confuses me with the these three
> types of light.
First: what is polarization?
It is the direction the electric field points to.
You know that light waves are made up of oscillating electric AND magnetic
fields. In vacuum, you may safely focus on the electric field, since the
magnetic field (which is also there) is completely determined (up to static
fields) by the electric field, due to Maxwell equations.
Now, you remember that the electric field E is a vector quantity
(since the electric force acting on a point charge q is just F = q E, and
any force is obviously a vector, it has to be dragging stuff in some
direction).
Well now, a generic electromagnetic wave travelling to the right along the
x axis is represented by E(x- ct), where E is any vector function of its
argument, except for the constraint, due to Maxwell equations, that E() is
constantly perpendicular to the x axis - the axis of propagation. (So it
is restricted to the y-z plane). c is obviously the speed of light, which
is a fundamental constant appearing in Maxwell equations.
The simplest thing you could take for E is a fixed vector E0, independent of
position and time, multipled by an arbitrary scalar function:
E(x - ct) = E0 f(x-ct)
This would represent linearly polarized light, in the direction of the
vector E0. For example if E0 = yy = (unit vector in y direction): that
would be called y-polarized x-travelling light. Of course E0 could point
in the z direction or in any intermediate direction.
For circularly-polarized light, life is slightly more complicate. First of
all you have to take light of a fixed "color", i.e. of a fixed frequency w.
Given w, you can make up the dimensionless quantity (a phase!)
w(x/c -t)
which is again, as you can check, only a function of x-ct.
Plug this phase into an oscillating function such as sin() or cos() or, even
better, in the complex exponential exp(i* ).
[Remember that taking the exp(i* ) for a physical field means implicitly that
you are assuming it is the real part of the function that representing the
actual physical field. Complex exponential here are just used for algebriac
convenience in place of sin/cos functions. This is completely different from
quantum mechanics!]
The resulting field is an oscillatory field at a fixed frequency. The generic
field f(x-ct) can be written as an infinite superposition of fixed-frequency
fields, according to Fourier's theorem.
Now, take an y-polazed x-travelling wave of angular frequency w:
yy exp[i w(x/c -t)]
Take also a z-polazed x-travelling wave of the same angular frequency w:
zz exp[i w(x/c -t)]
If you sum the two you just get a linearly-polarized wave in the 45 degrees
direction between yy and zz. However, suppose you de-phase the second wave by
ninety degrees, so that the maxima of the yy-polarized component occur exactly
when/where the zz component vanishes:
zz exp[i w(x/c -t) + pi/2]
Try to plot this vector at a given point in space, say x=0. At t=0 it starts
pointing y-ward, but after pi/(2 w) seconds it points z-ward. It actually
rotates around a circle (assume the lengths of yy and yy are the same).
It is a good old circular motion. If you see it in space, at a given time,
as a function of x the electtic field describes a corkscrew spiral...
This is called circularly-polarized light, and it may be right- or
left-polarized according to having a clockwise or counter-clockwise
rotation of the field looked at in the direction of propagation (or
equivalently to having a left or right handed spiral in space).
Note that in the complex formulation you can also take the complex sum:
yy exp[i w(x/c -t)] + i zz exp[i w(x/c -t)]
which gives automatically the right dephasing.
This tells you that a suitable complex combination of linealy polarized states
yields a circularly polarized state.
This is a fundamental fact you'll encouter in your Bell's inequality quest.
By the way, the usual inaccessible Sakurai presents this topics (including
light polarization and Bell's inequality) in some detail....
-------------------------------------------------------------------------------
> One last question about light. With unpolarised light, if I had say 100
> photons of unpolarised light, would all of the photons be identical, or would
> they all be different? Can an individual photon be unpolarised?
No, an individual phonon is always in SOME polarization state.
100 photons may build up unpolarized light by having the same k-vector and
different (random) polarization states.
> Actually, thinking about CP light, I find it a bit difficult to think of
> it as `pure light` in the same way as PP light is `pure`, so it seems a bit
> odd to use the two types of CP light as a basis to construct PP light -
> it`s like using something which is quite complicated to make something simple!
> Can CP light be considered `pure` in the sense of being the most basic
> of building blocks?
Yes it can.
There is nothing more complicate about CP light than linear polarized light.
A CP photon is just a linear combination of two LP photons of the same k,
with a pi/2 dephasing, i.e. with coefficients 1 and i.
This can be inverted to give LP light in terms of CP light spinning
oppositely.
-------------------------------------------------------------------------------
> And now for something completely different......EPR. I wonder if you have
> ever looked into this very deeply? I`m trying to get a feel for it and it`s
> not easy.
I even wrote a paper on EPR.
Meaning Electron Paramagnetic Resonance, which is the standard meaning of this.
On Einstein Podolsky Rosen, I read the paper, and several discussions of it.
> I am trying to understand exactly what Einstein and coworkers were
> driving at when they published their famous 1935 paper. I have read many
> `explanations` of the paper but they all seem to be slightly different so I
> managed to get hold of the original paper. I guess you have access to it
> but I thought I`d attach it as it anyway. Have you ever read it before?
Yes, of course I have access to PR.
> Anyway on page 779 of their famous 1935 paper EPR talk about two
> particles which are correlated particles and separated. They talk about
> `collapse of the wave packet` when a measurement is carried out on one of
> the two particles (so that the opposite particle also instantly assumes an
> eigenvalue). My question is, do the two particles continue to be correlated
> after the wavefunction has collapsed? I ask this because just after
> Equation (8) they say "If now the quantity B is
Your question is a deep one, and the answer can only be given by experiment.
The amount of this correlation was put in a mathematical form by Bell, and
several experiments investigated whether there is correlation between the two
non-interacting states or not, and the experimental answer is YES.
A famous exp in this field was realized by A. Aspect and collaborators.
> measured and is found to have value b_r, we conclude that after the
> measurement the first system is left in the state given by v_r(x_1) and the
> second system is left in the state given by phi_r(x_2). We see therefore
> that, as a consequence of two different measurements performed on the first
> system. the second system may be left in states with two different wave
> functions." But why, after one measurement, do the particles not become
> totally independent? If the wave function has collapsed then it makes sense
> that any correlation which was present originally is destroyed. In other
> words any further measurements on particle A can only give information
> about particle A - if we want to know
No, the collapse of the wavefunctions refers just to the fact that upon
measurement the wavefunction switches from the initial psi to an eigenstate fi_n
of the operator representing the quantity being measured, with probability
||^2
The fact that the two subsystems are no more interacting (and could be even
light years away) does not matter: it is the _total_ wawefunction that
collapses, carrying along the correlation.
> anything more about particle B we have to make measurement on particle B
> itself. This point of view would not get around the problem of particle B
> `knowing` which particular measurement was first carried out on particle A,
> but it would get around EPR`s contention that we can predict simultaneous
> values for the momentum and position of particle B, because any
> measurements on particle B would destroy any information we may have about
> it, thus preserving the Heisenberg Indeterminacy Principle.
> I hope I`m making sense,
What EPR contend is that in principle one could know x and p of particle B.
It is rather a philosophical issue. What is wrong in their paper is the
paragraph after Eq. (8) " We see..." where they imagine that one could
assign two different wavef. to the same reality, simply on the basis that
the two subsystems are noninteracting.
As usual, Sakurai gives a nice explanation of this "paradox".
If you really can't get hold of it I may fax you the relevant pages.
> Thanks for your email. I now understand that once the wavefunction has
> collapsed the correlation is lost - we have now have two separate,
> unconnected particles. But is it the case that any further measurements
> we now make on either of the two particles only gives us information
> about that specific particle?
Yes, the first measurement on particle A shows the correlation that was present
in the total wavefunc. before it was carried out. After that measurement,
assuming that particles A and B are not interacting any more, all correlations
are lost, and systems A and B are completely independent.
> Thus, on page 779, a little way down from equation (8) they say "We
> therefore see that, as a consequence of two different measurements
> performed on the first system, the second system may be left in stsates
> with two different wavefunctions."
>
> Do you mind if we discuss this a little?
> Imagine we have two oppositely moving correlated particles. We measure
> the position of particle 1. We therefore know the position of particle
> 2. The key questions are, a) are the two particles still the same
> distance from the source, and b) do they have the same momentum? If so
Which source? There is no source in this conceptual experiment. If you take
EPR's wavefunction, it represents a wave of two particles uniformly spread out
in space travelling in opposite directions with a phase correlation. There is
no position info for either particle.
After taking this position measure, you know the position of particle A (and
consequently also of particle B). That's it. Those delta functions are the
starting states for the Hamiltonian evolution that continues independently for
each state, according to Schroedinger time-dep. equation after the measure is
compete.
> what if we now measure the momentum of particle 1? Do we discover the
> momentum of particle 2?
>
If, after the position measurement of particle A, you measure the
momentum of particle A, or of particle B or of both A and B, you get for each of
them an uniformely distributed random value between -infinity and +infinity
Also, you find no more correlation between the A measure and the B measure.
In particular, you will in general NOT find pA + pB = 0.
The reason of this non conservation of momentum is that a position measurement
implies interaction of the particles with some device that "knows where it is",
therefore that break the translational invariance of free space.
> I guess that the answer to all these questions is `yes`. But this does
So you see that the answer is basically NO!
> not mean that the position and momentum of particle 2 were `real`
> (predetermined) before any experiments were performed. What`s more,
> according to the Copenhagen, if we first measure the position of
> particle 1, and then measure it`s momentum, we destroy all knowledge of
> it`s position again. But what about particle 2? This leads me to my
> final question which brings up non-locality.
> If we measure the position of particle 1 we discover the position of
> particle 2. If we then measure the momentum of particle 1, we discover
> the momentum of particle 2 (I think this experiment is fair). But by
The exp is faire bu the result is not what you expect.
> doing this we violate the Uncertainty Principle because we will find out
> both the position AND momentum of particle 2. So in order for this not
> to happen, this must mean that the the position of particle 2 becomes
> totally unknown when the momentum of particle 2 is measure. And this is
> `spooky action at a distance`!
Read again the above paragraph. Aren't there too many 2's?
If you actually meant:
"So in order for this not
to happen, this must mean that the the position of particle 2 becomes
totally unknown when the momentum of particle 1 is measured. And this is
`spooky action at a distance`!"
then the answer is given above: the measure of the momentum of particle A does
not affect the motion of particle B, AFTER THE FIRST COLLAPSE, which had occured
as you measure the position of particle A.
However THERE IS some `spooky action at a distance`, but not where you
said. This `spooky action at a distance` occurs at the first (A-position)
measurement. Before this measure you could tell with precision the
momentum of particle B. Immediately after this measure, conducted on
particle A, all knowledge of the momentum of particle B is lost!
-------------------------------------------------------------------------------
> Thanks again for your email. It has certainly got me thinking, and hopefully put me
> back on the right track.
> But....
>
>
>>Which source? There is no source in this conceptual experiment. If you take
>>EPR's wavefunction, it represents a wave of two particles uniformly spread out
>>in space travelling in opposite directions with a phase correlation. There is
>>no position info for either particle.
>>
>
> It`s a bit hard to think about there being no source. What about the place where the
> wavefunction (for the two particles) was formed. Failing this, once we have
> measured, say, the position of one of the particles, can`t we think of the source as
> a point halfway between the two particles?
My idea of source is also the place where the particles were formed.
Its possible existence shouldn't related to measuring any position, right?
It's either there somewhere or not there, regardless of what you do to the
particles once they exists...
So, for a travelling-wave state exp(i p x), which is uniformely distributed
in space, which has just a trivial time dependence (I assume no potential
is acting on the particle), and which has always been there in the past and
always will be in the future, it is impossible to think of a source.
Of course this means that no actual state like that could be realized in
practice. Any actual wavef. representing a physical particle would be a
"wavepacket" of finite momentum and position spread.
Same story for 2 particles of course...
In my opinion you are concentrating on a situation where understangind this
uncertainty problem is the most complicate. Since the 1940's people
prefere to use spin systems and operators instead of position and momentum
as much clearer and basic cases. Once things are clear there, you may
always come back to the position-momentum intricacies...
>> > doing this we violate the Uncertainty Principle because we will find out
>> > both the position AND momentum of particle 2. So in order for this not
>> > to happen, this must mean that the the position of particle 2 becomes
>> > totally unknown when the momentum of particle 2 is measure. And this is
>> > `spooky action at a distance`!
>>
>>Read again the above paragraph. Aren't there too many 2's?
>>
>>If you actually meant:
>>
>>"So in order for this not
>>to happen, this must mean that the the position of particle 2 becomes
>>totally unknown when the momentum of particle 1 is measured. And this is
>>`spooky action at a distance`!"
>>
>
> Yes, too many 2`s! Apologies for that. I agree with your version of the sentence!
> I now see that after the first measurement (whichever it was) the two particles are
> totally independent and any subsequent measurement on either of them has no effect
> on the opposite particle. In other words, if I measure the position of particle A
> causing collapse of the wavefunction, then if I measure the momentum of A I will not
> affect particle B, ie. I will not destroy the position information I have for
> particle B.
> I understand that the `action at a distance` is particle B `knowing` which
> measurement I have chosen to make on particle A in the first place. I think this is
> what you said. Sorry for repeating it, but it helps make things clear. Ofcourse if
> there is something wrong here, please say!
Yes it is correct now.
-------------------------------------------------------------------------------
> Thanks for that. I am now getting to grips with the EPR paper, despite
> its interesting wording. Meanwhile, paddling around the web as I do
> (this time looking for stuff on Bell`s theorem, etc), I often come
> across new things. Just now I saw something which says:
> "Quantum mechanical state:
> state vector |y> (pure state) or density operator r (mixture)."
> [actually that `r` after density operator looks more like a `p` (or
> rho?) in the original]
> I am familiar with the state vector but is it possible to explain what
> this density operator business is, please? When I see `mixture` I tend
> to think of a superposition of states, each with its own coefficient
> related to its probability of being realised when a measurement is made.
> But what has this got to do with `density operator`?
No, the density operator has not much to do with quantum superpositions.
It comes about in the formalism of ensembles.
One assumes to have a beam made of non-interacting quantum objects, and that
inside this beam there is some random distribution of different quantum states.
The case of a pure ensemble is that you have been accustomed so far, all the
"particles" in the same quantum states: For example all spins pointing up
towards z: |z+>.
Or else all pointing up towards x: |x+> = (|z+>+|z->)/sqrt(2)
More general than a pure ensemble, is a mixed ensemble: suppose 70% of the
particles are |z+> and 30% are |x+>.
In general one would define a mixed ensemble by the states |psi_i> composing it
(here |z+> and |x+>: note that they need not be orthogonal, and need not be as
many as the dimension of the space of states), and by the weights w_i (here 0.7
and 0,3) of these states.
Note that it is necessary to require that sum_i w_i = 1
On such an ensemble it is natural to define the ensemble average of some
operator A as
[A] = sum_i w_i <psi_i|A|psi_i>
This is the obvious generalization of the quantum average on a pure state <A>.
With this formula in mind, one defines the density operator:
rho = sum_i w_i |psi_i><psi_i|
where, the operator |psi_i><psi_i| is the projector on the state |psi_>.
With this definition it's easy to convince oneself that the ensemble average of
any operator A is given by:
[A] = tr[rho A]
where tr[O] indicates the trace of the operator O, i.e. the sum of its diagonal
matrix elements on any basis.
This means that the density operator contains all the infos of the ensemble.
In particular, one can recognize a rho for a pure ensemble in that it has
tr(rho^2)=1. For all mixed enesmbles one has 0<tr(rho^2)<1.
Also, the entropy, a measure of disorder in thermodinamics and statistical
mechanics, is:
S= -k_B tr(rho ln rho)
where k_B is the Boltzmann constant and ln rho is the logarithm of the operator,
which is meant in the sense of the spectral decomposition function of operators.
In practice on the basis of eigenstates of rho you take the ln of the
eigenvalues rho and multiply them by the corresponding spectral projector.
You see that rho has a fundamental role in ensemble quantum mechanics.
> In case you would like to check out the origin of this I found it at:
> http://www.phys.tue.nl/ktn/Wim/qm1.htm#standard formalism
Interesting page.
----------------------------------------------------------------------------
> vorrei farle una domanda a proposito di un esercizio svolto venerdì in aula.
>
> C'è un corpo nero con temperatura 3000K con solo un'apertura S
> trasparente a un certo intervallo di lambda centrato su lambda_max.
>
> Si cerca il numero di fotoni e di elettroni nel caso ci sia effetto
> fotoelettrico.
>
> Noi calcoliamo prima Wtot come una superficie per una lunghezza per una
> densità di energia, quindi dovrebbe essere un'energia, ma il risultato è
> in Watt.
>
> Poi dividiamo Wtot per l'energia di un singolo fotone e otteniamo il
> numero di fotoni al secondo. Quindi dimensionalmente Wtot dovrebbe
> essere una potenza.
>
> Ma da dove salta fuori la dipendenza temporale? Qualcosa dovrà essere
> per unità di tempo, ma cosa?
>
> Forse R(lambda,T) è già per unità di tempo e quindi è un energia /
> volume secondo?
No, non bisogna confondere u(nu,T), che e` una densita` di energia per
unita` di frequenza (J/m^3/Hz) (o di lunghezza d'onda se si prende
u(lambda,T), in J/m^3 /m), con R(nu,T) che e` una densita superficiale di
POTENZA (in altri termini e` una densita` di flusso di energia). R(nu,T)
si misurera` in W/m^2/Hz
(Nota che dato che Hz=1/s, le unita` di R(nu,T) si possono anche scrivere, in
maniera piu` compatta ma meno chiara, come J/m^2 !!)
Quindi moltiplicando per una superficie e integrando in d nu si ottiene
proprio una potenza in W, quella che tu chiami Wtot. Il resto segue.
> Alla fine dell'esercizio in cui si cercava la probabilità dell'elettrone
> di stare nella regione conica equatoriale di +o- 10°, lei ha fatto delle
> considerazioni.
>
> In particolare abbiamo visto che una combinazione lineare di soluzioni
> del tipo exp(+i m fi) è soluzione. Quindi del tipo sen fi o cos fi.
>
> Però se calcolo = 0
>
> Come nel caso dell'equazione di Schrodinger nel calcolo del momento
> medio per 2cos kx
>
> Però mi sfugge il significato di questo risultato. Cosa mi porta dire
> questo? E qual è la differenza con l'onda del tipo exp (i k x) o exp (i
> m fi)? Che considerazioni si fanno?
Si possono fare un sacco di considerazioni diverse, ma i punto cruciali
sono 2: - osservabili non compatibili (rappresentate da operatori che non
commutano) e quindi principio di indeterminazione; -postulato della misura
in mecc. quantistica.
Per il primo punto, le coppie di operatori non commutanti sono (x , k_x) (o
equivalentemente (x, p_x) ) e (fi, "m") (o equivalentemente (fi, L_z),
visto che "m" significa appunto L_z/hbar).
Localizzarsi in x significa delocalizzarsi in p ecc.
La differenza tra i 2 casi sta nella condizione al contorno: nessuna
condizione (o meglio funzione limitata) per il moto sulla retta reale,
periodicita` tra 0 e 2 pi per il moto angolare di fi.
Per il secondo punto il fatto e` che su uno stato sin(fi) anche se il valor
medio di L_z e` 0, se uno va a misurare L_z, ha probabilita` nulla di
misurare 0, e invece probabilita` 0.5 di misurare m hbar e 0.5 di misurare
-m hbar.
E` bene leggersi a fondo il primo capitolo del Sakurai e meditare questa
cosa a prima vista curiosa. Se ne potra` parlare ulteriormente a voce, se
ci sono ancora punti oscuri.
> Buongiorno.
>
> Qual è la differenza tra effetto fotoelettrico e ionizzazione?
> Ho in entrambi i casi l'emissione di un elettrone...
Il primo e` un caso speciale del secondo. Un elettrone puo` essere emesso
da un materiale anche senza l'azione di fotoni, ad es. per effetto
termoionico o effetto di campo (tunnelling quantistico con un campo
elettrico). In questi ultimi casi si ha ionizzazione, ma NON
fotoemissione.
Peraltro tradizionalmente si parla di ionizzazione per sistemi microscopici
come atomi o molecole, che diventano appunto ioni. Invece la fotoemissione
e` soprattutto una tecnica di investigazione di solidi (di solito
metallici), in cui l'elettrone che si porta via non cambia radicalmente lo
stato del sistema, perche` puo` venire rimpiazzato "chiudendo il circuito"
su un generatore.
> Nell'esercizio 3-15 dell'Eisberg ottengo un valore di lambda
> significativamente diverso dalla soluzione proposta.
>
> Le spiego la mia risoluzione:
>
> parto dall'equazione (di cui non sono certo) p^2 / 2m = K T
Hai ragione a non esserne certo: per l'equipartizione "ogni grado di liberta`
quadratico nell'Hamiltoniana all'equilibrio contribusice all' en. totale per
1/2 KT", quindi l'en. cinetica media e` 3/2 KT.
> trovo p usando la massa del protone e sostituisco in lambda= h / p
>
> io ottengo 1.8 Angstrom la soluzione è 1.596 Angstrom.
>
> Come mai?
Ho provato con 3/2 KT e viene 1.471 Angstrom.
Chissa` che cosa avevano in testa Eisberg e compare.
> invece nel seguente (3-16) quello che viene chiamato principal planar
> spacing corrisponde a d nella formula di Bragg?
Beh, si`, in questo caso, perche` piu` sotto dice "from these planes".
> Poi l'energia cinetica dell'elettrone è confrontabile con la sua energia
> a riposo, posso usare lo stesso Ec=p^2 / 2m ?
Beh, confrontabile ma insomma meno del 10%.
Comunque se vuoi fare le cose per bene usa la formula relativistica esatta
p=1/c sqrt( (Ec + mc^2)^2 - m^2 c^4 )
Vedi poi di quanto differiscono gli impulsi e quindi gli angoli.
> E infine, assumendo vero quello che ho detto sopra, facendo i conti
> ottengo fi(fotone) = 2.38° e fi(elettrone) = 0.00014°!!!!!
>
> Secondo me c'è qualcosa che non va...
Anche secondo me.
> Mi può spiegare come si deve risolvere questo esercizio?
Prima ci si calcola la lambda. Per l'elettrone (con p^2/2m ) viene 6.1321 pm.
Poi si inverte n lambda =2 d sin(theta)
con n=1, e si ricava
theta = Arcsin(lambda /2d) = 0.00979589 rad = 0.56 gradi
Il pezzo del fotone non ho voglia di farlo, ma e` la stessa minestra, cambia
solo la legge di dipersione per tirarsi fuori il lambda.
> 1) in un esercizio da lei svolto si deve calcolare la separazione dei
> livelli di uno stato 2p dell'idrogeno, usando l'approssimazione di campo
> debole.
>
> so che E = g mub B mj
>
> trovo le due diverse g per i due valori di j 1/2 e 3/2
>
> ma poi per trovare E quale mj devo considerare?
Che vuol dire trovare E? E e` funzione di mj, quindi ogni stato
individuato da j e da mj ha una sua energia E = <j,mj|H|j,mj> , dove
H e` l'Hamiltoniana.
> ho 2 e 4 valori diversi rispettivamente?
Certo, in tutto 6 stati con 6 energie diverse.
Poi, a seconda di cio` che l'esercizio chiede, si combineranno queste energie in
opportune differenze di energia, tipicamente.
> 2) perché in un atomo di carbonio se S=1 L non può essere né 0 né 2? e
> se S=1 L non può essere 1?
Per il principio di Pauli.
Magari ci dedico 5 minuti domani, magari ricordamelo.
> Forse se le scrivo il testo mi può aiutare di più.
> B = 0.1 T
> calcolare la separazione dei livelli di uno stato 2p dell'idrogeno.
> Può spiegarmi velocemente cosa devo fare dopo aver trovato i due valori di g
> e avendo capito che sono nell'approssimazione di campo debole?
> Grazie e scusi se non ero stato molto chiaro.
2 p da` 6 livelli (contando lo spin).
Il grosso della separazione e` data dalla spin-orbita, che e` contenuta nella
formula relativistica (quella che c'era anche nell'ex 4 del compitino): i 6
stati si splittano in 2 (sotto) piu` 4 (sopra), etichettati da j=1/2 e j=3/2
rispettivamente.
Dopodiche` all'energia Erel(j) calcolata dalla formula relativistica di cui
sopra bisogna aggiungere il contributo magnetico, che dipende dal numero
quantico mj che etichetta gli stati all' interno dei due multipletti che
j=1/2 e j=3/2:
Etot(j,mj) = Erel(j) + Emagn(l,s,j,mj)
dove l'en. magnetica
Emagn(j,mj) = g(l,s,j) mu_B B mj
e g(l,s,j) e` il fattore g di Lande' che hai gia` calcolato per i 2 casi.
Direi che tutto cio` corrisponde all' esercizio cui avevo dedicato una
buona ora a lezione, ed e` riassunto anche a
http://materia.fisica.unimi.it/manini/dida/many_el_general.notes.html
A quanto pare non e` ancora particolarmente chiaro...
> Nel secondo esercizio sulle molecole servono l'energia di prima
> ionizzazione del litio e l'affinità elettronica del fluoro. Sulla tavola
NO non servono. E` dato esplicitamente il potenziale U(r)!
> degli elementi ho trovato la prima ma non la seconda. Poi mi sono
> chiesto se la prima dipende dal numero di neutroni. Perché altrimenti
> non capisco cosa potrebbe cambiare tra le due molecole dell'esercizio.
>
> Perché bisogna considerare l'energia di punto zero?
Appunto, la differenza sta li`: la freq. di vibrazione dipende dalla massa,
quindi l'en. di punto zero pure, e dunque anche l'en. di legame.
Vedi Eisberg alla fine del par. 12-6.
> Per risolverlo io ho trovato b imponendo che il minimo sia in R=1.4 A e
> calcolando l'energia di dissociazione come Ediss= -V(R) -Elitio +Ecloro
>
> Giusto?
Dice "dissociare negli _ioni_ componenti", quindi le en. ionizz. e
affinita` el. non c'entrano. Vedi Eisberg fig. 12.1
> Non ho capito perch=E9 esistono le zone di Brillouin, non riesco a =
Considera per esempio i modi normali dei fononi che ho descritto nella
lezione sui fononi 1-dimensionali con 2 atomi per cella
(http://materia.fisica.unimi.it/manini/dida/1D-phonons.notes.html).
Prendi inizialmente un k a piacere nella prima zona di Brillouin, cioe`
nell' intervallo -Pi/a - Pi/a.
Prendi ora un'altro modo normale, stessa formula ma con un diverso k,
uguale a quello di prima piu` diciamo 2 Pi/a (o un qualunque multiplo
intero di 2 Pi/a): se guardi bene questo nuovo modo normale, vedi che e`
identico a quello di prima. Questo di dice che tutti i k fuori dalla prima
zona di Brillouin danno modi normali IDENTICI a opportuni k dentro la prima
zona di B. (ottenuti con opportune traslazioni di 2 Pi/a).
Stessa cosa puoi vedere per le funzioni d'onda elettroniche, per esempio
quelle del tight-binding.
Morale: in una zona di Brillouin, diciamo la prima, ci sono tutti e soli i
k che danno gli stati indipendenti del cristallo.
Chiaro?
In 3D i concetti sono gli stessi, ma la questione di come disegnare le zone
di B. e` piu` complicatuccia (c'e` qualche link dalle mie pages.): leggiti
qualche libro di solid state, se la cosa ti arrapa.
> capire bene la dipendenza dalla riflessione di Bragg che causa le bande =
> permesse e quelle proibite.
> Cosa succede per K vicino a n*pi/a?
Bah, questa e` una pittura perturbativa, in cui il potenziale periodico e`
pensato come una debole perturbazione al potenziale costante che da` gli
stati di elettrone libero. Io non ne ho mai parlato durante il corso, ma
ho messo un piccolo commento alla fine di
http://materia.fisica.unimi.it/manini/dida/tight.binding.notes.html.
In effetti, se uno immagina di poter trattare il potenziale come una debole
perturbazione, conviene pensare le cose in spazio k, cioe` di Fourier.
Chiaramente, un potenziale periodico di periodo a ha picchi a k=n 2Pi/a,
con n intero.
Quindi, l'effetto della perturbazione e` maggiore per stati che
differiscono di k=n 2Pi/a rispetto a stati con differenze di k diverse: li`
si aprono gaps nella banda parabolica della particella libera.
By the way, questo getta luce sul perche` nei libri di testo (es. Eisberg
13-6) si pigli sempre come potenziale periodico una roba orripilante come
il Kronig-Penney, invece che il piu` semplice potenziale che a uno verrebbe
in mente con la periodicita` a del reticolo:
V(x)=cos(2 Pi x/a)
In effetti con un potenziale cosinusoidale come quello qui sopra, l'unica
componente di Fourier non nulla e` quella a ±2Pi/a, quindi nella
banda parabolica si aprirebbe un solo gap, all' ordine piu` basso di teoria
delle perturbazioni, e didatticamente la cosa sarebbe un po' antipatica...
Thanx to David & to Alberto!
Comments and debugging are welcome!
| created: 22 Jan 2002 | last modified: 25 Sept 2022 by Nicola Manini |