GR9677 #97


Problem


This problem is still being typed. 
Quantum Mechanics}Probability Density Current
If one has forgotten the expression for the probability density current, then one needs not despair! If one remembers the vaguest definition of a ``probability current," one can solve this problem without the usage of the forgotten formula:
Recall that the probability current density J is the difference between incoming P(in) and outgoing P(out) probability densities. This is just the difference between the probability densities of a rightwards moving plane wave and a leftwards moving plane wave, since the probability density is related to the wave function by
The given wave function can be written in terms of plane waves,
gives the coefficient for the leftwards (negativedirection) wave, while gives the coefficient for the rightwards traveling wave. The probability density for each wave is given by,
Plugging these quantities into the formula for the probability current density above (J), one gets,
which is choice (E).
Alternatively:
The formal expression for the probability current density can be effortlessly derived from recalling the definition of probability and Schrodinger's Equationboth of which every physics (or engineering) major should know by heart.
Probability is defined (in the Born Interpretation) as . One should recall that in general (to wit: the absolute value squared of a complex expression is itself times its complex conjugate).
The timedependent Schrodinger's Equation is
,
where has the form of the familiar timeindependent Hamiltonian. From this, one finds that .
Generalizing the idea of a current from classical physics to the idea of a probability current, one takes the time derivative of the probability to get , where the productrule for babymath derivatives has been used and the derivative has been taken inside the integral because the integral and derivative are with respect to different variables.
Plugging in the expression for from the Schrodinger's Equation, one gets
,
where the terms involving V's cancel out, and thus,
Rewriting , one can eliminate the integral in the probability current by applying the fundamental theorem of calculus (to wit: ),
. But, since the probability current is usually define as , one has
.
(Aside:) One can printout a cool poster or decent Tshirt ironon to remember the Schrodinger's Equation (among other miscellanai) at a site the current author made several years ago,
\begin{quote}
http://anequationisforever.com/ds.php
\end{quote}
One can remember the general form of the probability current by recalling that it has to do with the difference of times its conjugate.
Right, so onwards with the problem:
The problem gives the wave function, so one needs just chunk out the math to arrive at the final answer,
Thus,
and,
, where one notes that the imaginary terms go to unity from the complex conjugate.
Plugging this into the probability current, one arrives at the expression for choice (E).


Alternate Solutions 
physics_guy 20180815 02:45:41  I think ETS really does want you to know the definition of current density. Otherwise, you should know that the probability current vanishes for nondegenerate stationary states, i.e. wavefunctions of the form (possibly timedependent phase factor) x (realvalued function). In this problem, if we let a and b be real, then the probability current should vanish. Only choices A and E satisfy this, and choice A does seem suspiciously stringent. To see why A is false, let b = i * a. Then we have the wavefunction of a free particle of wavenumber k, whose probability current is not zero (even though its probability density is constant everywhere in space), but is instead equal to the probability density times the velocity of the the particle.   wavicle 20111015 14:23:55  Kudos on the trick, but how do we KNOW that the PCD is not just zero for all values of and ?
It seems to me, that we can consider cos(kx) = 0 and calculate the PCD since the PCD should be valid for all values of kx. Then the PCD is
might be off by a sign, but who cares!  

Comments 
q 20191016 00:55:56  Bob Nightengale (BNightengale) Février 18, 2015The \'Los Angeles Times\' explique comment Rodriguez serait un ajustement parfait parce qu\'il ne peut statistiquement remplacer Jansen, et il en co?terait beaucoup moins que de négociation pour Jonathan Papelbon des Phillies de Philadelphie.\r\nq http://computechnic.dk/tobuyen26.asp   physics_guy 20180815 02:45:41  I think ETS really does want you to know the definition of current density. Otherwise, you should know that the probability current vanishes for nondegenerate stationary states, i.e. wavefunctions of the form (possibly timedependent phase factor) x (realvalued function). In this problem, if we let a and b be real, then the probability current should vanish. Only choices A and E satisfy this, and choice A does seem suspiciously stringent. To see why A is false, let b = i * a. Then we have the wavefunction of a free particle of wavenumber k, whose probability current is not zero (even though its probability density is constant everywhere in space), but is instead equal to the probability density times the velocity of the the particle.
physics_guy 20180815 03:03:11 
Alternatively, letting a or b = 0 gives you a particleinabox stationary state times a phase factor. Since bound stationary states in 1d are nondegenerate, the probability current must vanish. Again the only solutions that work are A and E, and A is not true in general.

  dragore 20120816 11:58:34  If by chance you can vaguely remember that:
Then all choices but (A) & (E) can be eliminated since (E) is the only one that has an imaginary unit in the denominator, as required. (A) sounds implausible given the freedom in the choice of and .
QuantumCat 20140918 14:51:32 
I think because and are given to be complex constants, E is the only correct answers, since A is a special case of E. When the imaginary components of and are zero we have A, however this is given not to be the case in the problem.

  wavicle 20111015 14:23:55  Kudos on the trick, but how do we KNOW that the PCD is not just zero for all values of and ?
It seems to me, that we can consider cos(kx) = 0 and calculate the PCD since the PCD should be valid for all values of kx. Then the PCD is
might be off by a sign, but who cares!
wavicle 20111015 14:25:46 
btw I don't bother with the terms since they just "cancel"

  itorsics 20110909 19:42:46  Here's a quick heuristic way to derive the probability current.
If we assume the wavefunction looks locally like a plane wave , then the "local value" of the velocity is . The probability current is then , where is the probability density. This gives
. However, this is not necessarily real, so we take the real part, which gives the standard formula for the probability current.   scrabble 20110322 07:50:55  Note that the wave function can be reduced to that of a plane wave travelling in the +x or x direction by setting and . In that case we should get either a positive or a negative probability current (either the particle is travelling to the left or to the right). The only answer that admits this change of sign is (E).   The_Duck 20100705 19:52:12  Here's an instant solution if you can think of it: if alpha and beta are real (or if either equals zero, as ramparts mentioned), then I believe psi describes a stationary state and there should be no probability current. This immediately gives E, because C and D don't care whether alpha and beta are real or not (and A can't be right).
Another approach is to notice that if alpha = 1 and beta = i then you can simplify to the familiar form psi = exp(i[wt + kx]) which is a state with momentum eigenvalue hbar*k. For a normalized momentum eigenstate like this I believe you can equate the probability current with the velocity, p/m = hbar*k/m. Plugging these values of alpha and beta into the answers gives E.
Of course, I didn't think of either of these under time pressure. :/
flyboy621 20101106 00:31:08 
This has to be the way to do it.
If or is zero, or if , we have a stationary state and the probability current must be zero. Otherwise the current must be nonzero.
(A) gives zero only if one of the coefficients actually is zero, so is not right in general. (B) doesn't depend on the coefficients at all, so can't be right. (C) gives zero only if BOTH coefficients are zero, which means there is no wave function at all. (D) would not give zero if only one coefficient is zero (though it would if ), so it must be (E).

physics_guy 20180815 03:00:04 
If b = i*a then you have the wavefunction of a free particle with wavevector k, which does have nonzero probability current. It is true that if a and b are real  or if a is any complex number and b is simply zero  then the probability current is zero. But it is not simply because psi is a stationary state. The condition for a state to be stationary is that the divergence of the probability current must vanish, not the probability current itself. However, if a stationary state is also nondegenerate, then the probability current vanishes. It is not clear that psi is a bound state here, indeed it looks more like a superposition of free particle wavefunctions, which are degenerate even in one dimension. However, if you are really itching for a shortcut, it follows from the definition of probability current that if psi is of the form (possibly timedependent phase factor) x (real function of position) then the probability current vanishes. This is why letting a and b be real (or just letting b=0) gives you an instant solution. It\'s a nice trick for this problem, but I would argue it\'s hard to convince yourself it\'s true unless you knew the definition of J in the first place.

  ramparts 20091002 12:11:54  Oh, I think the probability current should go to 0 if , no? That would only leave A and E (and A is just wrong ;) ).
niux 20091104 19:23:14 
nice

  ramparts 20091002 12:10:48  1.8 minutes per problem.
I have yet to see a satisfactory answer to this problem for people who *don't* have the probability current memorized.   poop 20051208 15:44:00  Umm, I think there's an error in your sweet ironon tshirt. On the blackboard, doesn't the Schrodinger Equation have a second derivative in x, rather than a first derivative??
:)   keflavich 20051111 19:41:37  That's a sweet 'effortless' derivation =P  




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