Minimum uncertainty wave packets energy interpretation

[ESmithy]

I am having trouble understanding (not for homework) what a wave packet is in terms of the correspondence of the idea of a wave packet to a "point like" particle. I'd like to focus on the 1d wave packet ultimately, but in order to describe my consternation -- let me detour to a well defined idea first;

In describing atomic states, for example, a stationary state is associated with an electron by solving for an energy "E" or a set of energies that an electron may have. eg: S1, P1,P2,P3, etc. An electron may be in "one" of these states, or in a combination of them (i suppose?) and then one has to add the necessary energies from the possible states that an electron is into get the total. Eg: if it is *only* S1, then it has the energy of an "S1" state -- well defined. If it is *only* in S2, then it would have the energy of S2, etc. If it is somehow in the combination of S1 and P1 (uncertain of which it is for a moment) I am not certain of how one determines the "energy" that ought to be assigned to it.

Just so, I find the same issue becoming very acute in the wave packet idea. Now, instead of just two energy states -- there are infinite energy states in the distribution of a gaussian. ?!
How much energy does a particle described by the Gaussian wave packet actually have according to the summation of the energies of the individual frequencies (plane waves) -- I know I could "assign" the mass of an electron to a wave packet, take it's group velocity (instantaneous), and come up with an energy -- but that is rather like pulling a rabbit out of a hat trick;

What is the more formal way of computing the total energy of an item in a superposition of solutions (plane waves) such as the Gaussian which allow normalization?

If you know of a book title which addresses this well let me know, I am not shy to purchase something that goes into more detail than the college texts (undergraduate) that I have.

 

[chromosome24]

Keep in mind the wave packet is a probability distribution. The energy of a spatial Gaussian wave packet is itself Gaussian in form. That said, despite the energy wave packet being composed of an infinite set of energy eigenvectors, it will have a non infinite expectation value.

In other words the value that should be assigned to energy, position, angular momentum, or any kind of observable, is the expectation value, and the expectation values is dependent on the probability distribution.

When you have a system that's in a combination of energy eigenstates, then the time dependency of the wave function becomes more apparent. Time only alters the phase of the eigenstates. With only one eigenstate describing the state vector, when taking the expectation value, the phases cancel and there is no time dependency in the expectation value. However, when you have a superposition of energy eigenstates, the state vector equation becomes a summation of multiple eigenstates having phases that change with time at rates corresponding to their energy eigenvalue. This means that one will find a time dependency in its expectation value.

This is how the gaussian distribution is linked to our classical interpretation of a point particle moving in space. Think of it this way what happens to the probability distribution, in postion space, of a particle (state vector) if it is only described by ONE energy eigenvector (whats the inverse Fourier transform of a delta function).

hope this helps.

 

[The_Duck]

Stationary states have a definite energy. If a particle is in a stationary state and you measure it's energy, you get that definite energy with 100% probability.

As you say, it is possible for a particle to be in a superposition of multiple stationary states. In this case if you measure the energy of the particle the outcome is not determined, but somewhat random! Your energy measurement could return the energy associated with anyone of the component stationary states. The probability of measuring a given energy is determined by how much of the associated stationary state there is in the superposition. If the state is "mostly" S1 but "some" S2 then you will likely measure the energy to be E1 (energy of S1) but with some probability you will measure it to be E2.

With wave packets, the state is an infinite superposition of plane waves of definite momentum and energy. If you measure the momentum or energy you will get a number in some continuous range. You can construct probability density functions that give the expected distribution of energy and momentum if you make repeated measurements of such states. So wave packets don't have definite values of energy and momentum; there is some uncertainty in both. But these uncertainties tend to be very tiny considered from a human scale, which is why it usually makes sense to talk about e.g. an electron with definite momentum and energy. The wave packet of such an electron is simply composed of plane waves that all have very similar momentum and energy. The uncertainty principle is the statement that if you want a wave packet that has low uncertainty in energy, the price you pay is that the wave packet spreads out: you have an increased uncertainty in the position of your particle.

Any introductory textbook on quantum mechanics ought to cover this stuff.

 

[ESmithy]

Keep in mind the wave packet is a probability distribution. The energy of a spatial Gaussian wave packet is itself Gaussian in form. That said, despite the energy wave packet being composed of an infinite set of energy eigenvectors, it will have a non infinite expectation value.

Good. That was bothering me -- for I know infinite energies are a problem in QCD, QED, with virtual photons (etc) and I wasn't sure if this was another instance of that just swept under the rug...

When you have a system that's in a combination of energy eigenstates, then the time dependency of the wave function becomes more apparent. Time only alters the phase of the eigenstates. With only one eigenstate describing the state vector, when taking the expectation value, the phases cancel and there is no time dependency in the expectation value. However, when you have a superposition of energy eigenstates, the state vector equation becomes a summation of multiple eigenstates having phases that change with time at rates corresponding to their energy eigenvalue.

Let me restate that to see if I am understanding you:
Schrodinger's equation's solutions are dispersive in nature; so that a wave packet is composed of plane waves with differing phase velocities and therefore, with time, the relative phases will change the overall shape of the wave packet. Thus, for any single experiment, the probability of finding the object described by the wave packet will change with time and be spread in space instead of linearly related to the original position as a classical particle would be.

This means that one will find a time dependency in its expectation value.

I don't quite understand this last sentence: Expectation values <> are single numbers describing the average (I don't have my book here, this is just from memory) result of doing an experiment many times.
In my understanding, an expectation value typically tracks the classical value -- in the example I am interested in, the 1d minimum uncertainty wave packet, then -- are you saying the "average" energy would change with time? eg: If I take a snapshot of the wave-packet at time (t=t1 >0), calculate the wavepackets "average" energy by doing some kind of integral over space of an energy operator multiplied by that evolved wave-packet shape -- this new "average" energy will differ from that at "t=0"?

This is how the gaussian distribution is linked to our classical interpretation of a point particle moving in space. Think of it this way what happens to the probability distribution, in postion space, of a particle (state vector) if it is only described by ONE energy eigenvector (whats the inverse Fourier transform of a delta function).

hope this helps.

It helps some. Thank you.

 

[ESmithy]

Stationary states have a definite energy. If a particle is in a stationary state and you measure it's energy, you get that definite energy with 100% probability.

Yes.

As you say, it is possible for a particle to be in a superposition of multiple stationary states. In this case if you measure the energy of the particle the outcome is not determined, but somewhat random! Your energy measurement could return the energy associated with anyone of the component stationary states. The probability of measuring a given energy is determined by how much of the associated stationary state there is in the superposition. If the state is "mostly" S1 but "some" S2 then you will likely measure the energy to be E1 (energy of S1) but with some probability you will measure it to be E2.

OK. That is extremely interesting; There is a coefficient of each of the energy states -- say "a" and "b" which then are added as in: phi = a*S1 + b*S2; that determine the relative "amounts" of each state that an item is in. In none of my books is the "sampling" of a combined energy state really discussed.
You're saying, that in an experiment -- one will not measure the energy that is intermediate between the stationary states -- but only the energy of one or the other -- and in some proportionality to the coefficients?
(perhaps a**2 vs b**2) etc? What prevents an electron from having enough energy to be in both S1 and P1 at the same time? eg: E(S1)+E(P1); I don't expect it to "stay" in this combined energy state -- but that's different from it ever being in that combined state ever (at all).

With wave packets, the state is an infinite superposition of plane waves of definite momentum and energy. If you measure the momentum or energy you will get a number in some continuous range. You can construct probability density functions that give the expected distribution of energy and momentum if you make repeated measurements of such states. So wave packets don't have definite values of energy and momentum; there is some uncertainty in both.

OK. I thought that a wave packet still defined an object with a definite energy; just an uncertainty in position and momentum. In the book I have, the uncertainty in energy is related to time -- and the author makes a comment concerning this indicating that it is *not* the uncertainty in the measurement of energy for a period of time; He was emphatically denying that tunneling, for example, is because a particle has an uncertainty in energy that can "borrow" energy and then return it.. etc.

But these uncertainties tend to be very tiny considered from a human scale, which is why it usually makes sense to talk about e.g. an electron with definite momentum and energy. The wave packet of such an electron is simply composed of plane waves that all have very similar momentum and energy. The uncertainty principle is the statement that if you want a wave packet that has low uncertainty in energy, the price you pay is that the wave packet spreads out: you have an increased uncertainty in the position of your particle.

Any introductory textbook on quantum mechanics ought to cover this stuff.

They don't really -- They generally show that a superposition of plane waves satisfy the S.E; that such a superposition of waves is therefore a solution; that the superposition of waves can be normalised over space, whereas a single plane wave (free particle) can not be normalised. They then will simply give explanations regarding HUP; some explain the Gaussian packet is the best one can do -- but of the books I have read, none go into detail of how much energy such a packet "represents" -- rather they say this is a close as one can get to a "classical" particle, and then abandon it. Very little is done in the way of examples of calculations on a wave packet for various experiments somebody could try -- and the physical phenomena such experiments actually represent. I'm sure there are probably better books, but after searching through about four -- I decided it was probably better to just ask for directions...

 

[The_Duck]

There is a coefficient of each of the energy states -- say "a" and "b" which then are added as in: phi = a*S1 + b*S2; that determine the relative "amounts" of each state that an item is in. In none of my books is the "sampling" of a combined energy state really discussed.
You're saying, that in an experiment -- one will not measure the energy that is intermediate between the stationary states -- but only the energy of one or the other -- and in some proportionality to the coefficients?

Yes, as you suggested: if S1 and S2 are properly normalized stationary states with energy E1 and E2, then a*S1 + b*S2 is a properly normalized state if |a|^2 + |b|^2 = 1. If you measure the energy of this state you get E1 with probability |a|^2 and E2 with probability |b|^2.

What prevents an electron from having enough energy to be in both S1 and P1 at the same time? eg: E(S1)+E(P1); I don't expect it to "stay" in this combined energy state -- but that's different from it ever being in that combined state ever (at all).

That measurements will come out as described above is simply one of the postulates of quantum mechanics. The idea is that you will only observe energies which are eigenvalues of the energy operator (i.e., the Hamiltonian). Further when you measure a given state you can only observe energies corresponding to the stationary states that appear when you express your state as a sum of stationary states, and the probabilities of observing each possible value of energy are calculated as above.

OK. I thought that a wave packet still defined an object with a definite energy; just an uncertainty in position and momentum.

Since a free particle's energy is related to its momentum by E = p^2/2m, uncertainty in momentum means uncertainty in energy.

In the book I have, the uncertainty in energy is related to time -- and the author makes a comment concerning this indicating that it is *not* the uncertainty in the measurement of energy for a period of time; He was emphatically denying that tunneling, for example, is because a particle has an uncertainty in energy that can "borrow" energy and then return it.. etc.

Yeah, there is an energy-time uncertainty principle much like the position-momentum uncertainty principle, but a bit harder to define rigorously. In quantum mechanics, energy is the same thing as temporal frequency (e.g. a stationary state with energy E has a frequency that is E/h or something). For seeing how this leads to an uncertainty principle, I really like http://scienceblogs.com/builtonfacts/2010/03/hearing_the_uncertainty_princi.php" .

The position-momentum uncertainty principle gives a lower bound of (Δx)(Δp) ~ h. Similarly you can say there is a lower bound of (ΔE)(ΔT) ~ h if you are careful about what that means. In the case of a wave packet, we can say that ΔE is the spread in energies in the wave packet due to the fact that it is composed of a range of plane waves of different momenta (with spread Δp) and thus different energies. We can have ΔT be the length of time it takes for the wave packet to pass a point (which is nonzero because it is spread out over a length Δx). You can calculate ΔE ~ Δ(p^2/2m) = (p/m)(Δp) and ΔT = (Δx)/v = (m/p)(Δx). Then (ΔE)(ΔT) = (Δp)(Δx) with a lower bound around h.

of the books I have read, none go into detail of how much energy such a packet "represents"

You want to read about the "posulates of quantum mechanics." These are introduced early but often don't mean much until you've done some calculations. http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html" is a possible source, you can Google more.

 

[chromosome24]

In my understanding, an expectation value typically tracks the classical value

Spot on. In our case, the thing we are observing (the free particle) is moving. This means that its expectation value must also be moving, which means that the probability distribution describing the position of the particle, this is the minimum uncertainty Gaussian wave packet, is also moving with time.

If I take a snapshot of the wave-packet at time (t=t1 >0), calculate the wavepackets "average" energy by doing some kind of integral over space of an energy operator multiplied by that evolved wave-packet shape -- this new "average" energy will differ from that at "t=0"?

I may have been unclear with this. The total energy (PE + KE) of our free moving particle system, or any system there of, is described by the Hamiltonian. In our case Hamiltonian is just the particles KE. Furthermore the expectation value of the KE, or momentum, of the free particle is not time dependent. However time dependency in the expectation value of the momentum space prob. dist. is very much so possible when the systems Hamiltonian has a non zero potential.

So you can think of a free particle as being described by two Gaussian distributions; one being a position space time dependent Gaussian and the other being a momentum space time independent Gaussian.

check out these simulations http://www.falstad.com/qm1d/
it helps build some intuition

 

[ESmithy]

Thank you both, very much!

Yes, I am starting to see now. :) That pretty much solves my problem.