Potential barriers, wavepackets, probabilities etc.

[speeding electron]

I've been thinking about quantum particle waves interacting with potential barriers, specifically the case of a particle interacting with a step potential, reading from a website which deals with the time-independent solutions for individual eigenfunctions of the momentum operator.
We can solve Schrodinger's equation for both regions: in region I (before the barrier) we have the solution for a free particle, before it hits the barrier, and can model a 'reflected' wave by one with momentum away from the barrier and a lower amplitude than the 'incident' wave. With the solution beyond the potenial step, we can model a wave that has made it through, and which will have a lower amplitude than the incident wave, and a longer wavelength. The ratio of reflection to transmission, as the site puts it, is calculated using the ratio of the reflected and transmitted particles' amplitudes. But what exactly does this mean?
I've heard quantum tunnelling, whereby there is a finite chance that a particle with energy less than that of a thin potential barrier will make it through, and I suppose this is in some ways analogous. But what I wonder is this: after the collision with the barrier does the wavefuntion evolve so that there are two wave packets, each going away from the barrier, but in opposite directions? This would imply not only that there is a probability that the particle will go through or not, but also that each time we measure it, that probability will be there, and we may detect the particle on one side of the barrier the first time round, on the other the next. But even as I write the idea comes to me: do we disturb the wavefunction the first time we measure it, thereby destroying the wavepacket on the side of the barrier where we haven't detected the particle? Is this how the probability of the two events is described in the theory?
One last thing - how do we specifiy the amplitude of the reflected and transmitted waves? Is this purely experimental? It seems that we cannot do it using Schrodinger's equation. If not, how can the equation claim to be able to describe completely the evolution of a wavefunction? Thanks for taking the time to read all this - I would just like some clarification on the issues raised here. Also, recommendations for any good websites/books on this subject would be appreciated. Thanks again.

 

[Eye_in_the_Sky]

The ratio of reflection to transmission, as the site puts it, is calculated using the ratio of the reflected and transmitted particles' amplitudes. But what exactly does this mean?

The time-independent solution for the step potential with E > V_o has the form:

Ae^ikx + Be^-ikx , for x <= 0 ;

Ce^ik'x , for x >= 0 .

The "A-term" is interpreted as the "incident" wave, the "B-term" is interpreted as the "reflected" wave, and the "C-term" is interpreted as the "transmitted" wave.

"The ratio of reflection to transmission" is given by |B|²/|C|² ** .

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** NOTE: The above expression is incorrect! The correct expression is given by:

(k|B|²) / (k'|C|²) .

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... how do we specifiy the amplitude of the reflected and transmitted waves? Is this purely experimental? It seems that we cannot do it using Schrodinger's equation.

But we can(!) do it using the Schrödinger equation.

Looking closely at the (suppressed) details of the above solution, we find that:

k = sqrt{2mE} / h_bar ;

k' = sqrt{ k² – 2mV_o/h_bar² } ;

B = [ (1 – k'/k) / (1 + k'/k) ] A ;

C = [ 2 / (1 + k'/k) ] A .

As you can see, B and C are just proportional to A. The "incident" amplitude A is, so to speak, just a "dummy" variable - it simply "cancels out" when we compute the desired ratios. These ratios are then just a function of k, the "incident" wave-number (i.e. the "incident" momentum, p = h_bark).

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Next:

... what I wonder is this: after the collision with the barrier does the wavefuntion evolve so that there are two wave packets, each going away from the barrier, but in opposite directions?

Yes (... and note that we are now considering a time-dependent scenario).

... do we disturb the wavefunction the first time we measure it, thereby destroying the wavepacket on the side of the barrier where we haven't detected the particle? Is this how the probability of the two events is described in the theory?

Yes, what you say is essentially correct ... where I am using the word "essentially" only because you used words like "disturb" and "destroy" which may have connotations which are not strictly true.

The phenomenon which you have quite accurately described (modulo certain possible connotations) is referred to as "state reduction" or "reduction of the wavefunction" as a result of the performance a "measurement". Ordinary "textbook" QM captures this notion via the, so called, "Projection Postulate".

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Also, recommendations for any good websites/books on this subject would be appreciated.

At the moment I can't really think of anything that good. ... Well, maybe the one which we used to call "Eisberg and Resnick" ... yeah, it's quite elementary, yet it does a very good job at explaining the concepts and giving enough mathematical details without being too technical. Here are some links concerning that book:

http://engineering-books-online.com/047187373X.html



http://www.reviewcentre.com/reviews26372.html

 

[CJames]

Yes, what you say is essentially correct ... where I am using the word "essentially" only because you used words like "disturb" and "destroy" which may have connotations which are not strictly true.

To jump in, would it be better to say that the wavefunction collapses? That is, not only does the other "half" of the wave collapse, but so does the rest of the wave (in a time dependant scenario).

 

[Eye_in_the_Sky]

To jump in, would it be better to say that the wavefunction collapses? That is, not only does the other "half" of the wave collapse, but so does the rest of the wave (in a time dependant scenario).

Now that you mention it, I think that the term "collapse" is in fact used more frequently than "reduction".

But the heart of the point which you are making is that it is the entire wavefunction which "collapses", and not just the component on the other side of the barrier. However, which of these two scenarios will prevail, in principle, depends upon what "kind" of measurement is being envisioned. The kind of measurement which I had in mind was an "ideal" measurement for which the corresponding observable has three eigenvalues, say -1 (to the left of the barrier), 0 (inside the barrier), and 1 (to the right of the barrier), with corresponding eigenprojections given by:

[P_-1f](x) =
f(x) , for x left of the barrier
0 , for x otherwise ;

[P₀f](x) =
f(x) , for x within the barrier
0 , for x otherwise ;

[P₁f](x) =
f(x) , for x right of the barrier
0 , for x otherwise .

While this observable is perfectly fine for the calculation of the associated probabilities and expectation values, I agree that it is not physically realistic with regard to "collapse" and the "future behavior" of the system. In this sense, the measurement represented by the above observable is said to be "ideal".

As far as I know, the only types of physically real measurements which can exhibit this kind of "ideality" are the so called "null" measurements. For example, imagine that a single particle detector is placed on the right side of the barrier. If we wait long enough for the particle to have been either transmitted or reflected, then in the case where the detector does not register the presence of the particle (a "null" measurement), it follows that the wavefunction has been "collapsed" to the fully intact reflected component alone. (However, now there is a new aspect of "ideality" in that the detector is assumed to be 100% efficient. Nevertheless, a proper accounting for this element will not change, in any essential way, the above picture of the "collapse".)

So, in conclusion, regarding your question "... would it be better to say ...?", I'm not quite sure which would have been better. There are certainly better ways of saying it than the way I had said it. One better way would have been to raise the point which you yourself have raised, and then follow it with an appropriate clarification ... which I hope has been accomplished here.

 

[speeding electron]

Thank you for clearing that up Eye in the Sky, and I understand all of it apart from where you calculate the reflected and transmitted amplitudes. Am I right in thinking that A + B = C ? In checking your results, it seems that that is not true - is that my mistake or can it be explained?
That is however not my main query. It seems perfectly reasonable that A and B should be proportional to C, and that the amplitudes should be dependent on how much great the difference is between E and V_0, but I just can't see mathematically speaking why the amplitudes should have anything to do with the wavefactors - the functions are solutions to the Schrodinger equation no matter what the amplitudes. Is there some sort of boundary condition that I'm missing here? Thank you.

 

[Eye_in_the_Sky]

Check again. In the above solution, A + B does equal C.

 

[Eye_in_the_Sky]

Oh, and the 1^st boundary condition is that the wavefunction must be continuous at x=0. This gives A+B=C as a requirement.

The 2^nd boundary condition is that the deriviative of the wavefunction must be continuous at x=0. This gives k(A-B)=k'C as another requirement.

Solving these two equations for B and C in terms of A, k, and k' then gives the above solution (where, of course, k and k' are related to E and V_o through the Schrödinger equation itself).

 

[speeding electron]

OK, thanks. I failed to see that those boundary conditions would lead to the required relationships between A, B, C, k, and k'. The website hardly made any mention of it, just mentioning that the two wavefunctions must be "matched at the interface". I still don't get that A + B = C, rather that B + C = A + 2A(k-k')/(k+k'), but I'll impose the boundary conditions myself and see what I get. Thanks again for your help.

 

[speeding electron]

Sorry, I meant to say that B + C = A was what my intuition told me, which is what is not the case. That first boundary condition tells us that A + B = C, which is true. I would have thought that one wavepacket of large amplitude would have 'broken up' into two wavepackets, each with smaller amplitudes, as it interacted with the barrier, so that the sum of the amplitudes of the two smaller wavepackets equaled the amplitude of the larger one.

 

[Eye_in_the_Sky]

An important correction.

Yes, speeding electron, on account of your rather scrupulous inquiries, I have found an error in my first post. As far as I can see, everything there is correct except for:

"The ratio of reflection to transmission" is given by |B|²/|C|².

The correct expression for this quantity is:

(k|B|²) / (k'|C|²) .

The explanation for why this is so is given below.

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Your intuition that B+C should equal A is on the right "track". Something a little more on that "track" would have been to assert that |A|²=|B|²+|C|² ... but still, this too is incorrect. What we are after here is a mathematical statement which will accurately capture the notion of "conservation of particle number" in the given context. More precisely, we would like to say that:

The incident "particle flux" is equal to that of reflection plus that of transmission.

For a QM plane wave Ae^(+/-)ikx, k>0, the associated "particle flux" is proportional to the amplitude-squared of the wave multiplied by the particle velocity. Hence, our statement for the conservation of particle number becomes:

[1] v|A|² = v|B|² + v'|C|² ,

where v (v') is the associated particle velocity in the region x<0 (x>0).

But,

v = p/m = h_bark/m , and likewise , v' = h_bark'/m .

Therefore, equation [1] becomes

[2] k|A|² = k|B|² + k'|C|² .

You can easily check that this formula does indeed hold for our solution.

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Finally, let us summarize:

reflection coefficient --> R = |B|² / |A|² ;

transmission coefficient --> T = (k'|C|²) / (k|A|²) ;

ratio of reflection to transmission --> R/T = (k|B|²) / (k'|C|²) .

Observe that equation [2] above is equivalent to R+T=1.

 

[speeding electron]

OK, thanks for clearing all that up.