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Physical meaning of solution to step potential with E = V0

  1. May 7, 2014 #1
    I am able calculate the solution for the TISE relating to the interval to the right of a step potential; however, I am unclear as to what this solutions represents physically for this scattering problem.

    So, starting with

    (-ħ2/2m)d2/dx2Ψ2(x) + V(x)Ψ(x)2 = EΨ(x)2

    where V(x)Ψ(x)2 = EΨ(x)2

    So, d2/dx2Ψ2(x) = 0

    In order for d2/dx2Ψ2(x) to equal 0

    d/dx Ψ(x)2 = A2 and Ψ(x)2 = A2x + B2

    Again, was does this tell us about what happens to the incident wave function as it hits the a step potential exactly equal to the enegy of the incident wave function?
     
  2. jcsd
  3. May 7, 2014 #2

    Simon Bridge

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    Your solution is: ##\psi_2(x)=A_2x+B_2##
    What sort of thing is this the equation of? (i.e. a circle, a parabola? what?)
    What does this say about the amplitude as x increases?
    Does this make sense?

    Your approach looks funny to me.

    1. why not start with plane-wave solutions?
    2. you should not give the eigenvalue E (associated with the wavefunction) the same symbol as the height of the step (associated with the potential). i.e. Do you mean that V(x>0)=V and V(x<0)=0 ... solve equation then put E=V?

    Please show your reasoning.
     
  4. May 8, 2014 #3
    ψ2(x)=A2x+B2 is a series of line equations.

    As x increases the amplitude increases. That does not make sense to me in this context.

    I do mean that V(x>0)=V and V(x<0)=0.

    Thank you for the response.
     
  5. May 9, 2014 #4
    Why do you assume that A2 is positive?
     
  6. May 9, 2014 #5

    Simon Bridge

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    What's wrong with that?

    Express what it means in terms of the probability of finding the particle at different positions of x>0. What should be happening to the probability density with position?
    What does that tell you about your answer?

    How would making A2 negative make the wavefunction, and the probabilities, behave better?
     
  7. May 9, 2014 #6
    It wouldn't. Could it be complex instead?
     
  8. May 9, 2014 #7

    Simon Bridge

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    How would that help?

    Is there another possibility?
    How would you expect the wavefunction to behave inside the potential?
    Wait a while before answering... OP need to try this too: if the result of the calculation does not make physical sense, what does it usually mean?
     
  9. May 9, 2014 #8
    Based on the a calculation of the reflection and transmission probabilites derived from the probability current density, we find that for E > V0 a percentage of the incident wave function is reflected and a percentage of the wave function is transmittted.

    Based on the a calculation of the reflection and transmission probabilites derived from the probability current density, we find that for E < V0 all of the incident wave is reflected.

    We therefore assume that for E= V0 the incident wave function is definitely reflected, but also might have a portion that is transmittted.

    So this is what one might expect.

    Instead by solving the TISE (-ħ2/2m)d2/dx2Ψ2(x) + V(x)Ψ(x)2 = EΨ(x)2

    where V(x)Ψ(x)2 = EΨ(x)2

    So, d2/dx2Ψ2(x) = 0

    In order for d2/dx2Ψ2(x) to equal 0

    d/dx Ψ(x)2 = A2 and we find Ψ(x)2 = A2x + B2.

    A line exquation where the amplitude increases with increasing x, is like an "exploding" exponential.

    Further, if we try to calculate the reflection and transmission probabilites derived from the probability current density, we find that for E = V0

    Wave equation for interval 2: Ψ2 = A2eik2x + B2e-ik2x 0 < x < ∞

    Wave number for interval 2

    k2= [2m/ħ2 (E - V2)]1/2 Interval 2; x > 0 V2 = E

    k2= [2m/ħ2 (0)]1/2 = 0

    The angular frequency in this eigenfunction corresponds to the wave number k2 and is:

    ω2 = 0 = 0

    Thus, the terms in the eigenfunction correspond to the following:

    A2 – incident wave (momentum 0)
    B2 – reflected wave (momentum 0)

    Producing the following wave equation for interval 2: Ψ = A + B

    When the result of the calculation does not make physical sense, it usually means that there is no solution for that interval which produces the nonsensical result.
     
  10. May 9, 2014 #9

    Simon Bridge

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    Why do you have a superposition of two plane waves for x>0? Is the wave incident from the right?
    Did you apply the boundary conditions to get A and B as a proportion of the incoming wave amplitude?

    Notice that your revised equation for ##\psi_2## is different from before?
    What shape is the new function? Have you seen that for before?

    Other reasons why a calculation may not make physical sense is if there is something wrong with the model or if there is something wrong with the calculation or if the underlying assumptions don't match the model used.

    I think You have reached this place in the standard lecture:
     
    Last edited by a moderator: Sep 25, 2014
  11. May 10, 2014 #10
    I believe the terms in the eigenfunction correspond to the following:

    A1eik1x – incident wave (momentum +ħk1)

    B1e-ik1x – reflected wave (momentum -ħk1)

    A2 – incident wave (momentum 0)

    B2 – reflected wave (momentum 0)

    Thus there is no directionality to the B2 term - you do not get a wave traveling to the right or to the left.

    Wave equation for interval 1 becomes: Ψ1 = A1eik1x + B1e-ik1x (-∞ < x < 0)

    Wave equation for interval 2 becomes: Ψ2 = A2 + B2 (0 < x < ∞)

    I did notice the revised equation for ψ2 is different from Ψ2 = A2x + B2 (0 < x < ∞)

    That difference, and my inability to see what it represents, forms the crux of my confusion.
     
  12. May 10, 2014 #11

    Simon Bridge

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    If E > V would you still have the reflected wave in region 2?
    If yes: then where does it reflect from?
    If no: then why is there one for E=V?

    Your revised form for the wavefunction is different because you did the analysis properly that time.

    Your first form was \psi = Ax+B (dropping subscripts).
    That was simply wrong since it is not a solution to the schrodinger equation for x>0.
    You can easily check by substituting it in.

    You are still puzzled about what it all means?

    You have not answered all the questions. If you don't answer questions, it is difficult to help you.
    What is the shape of the x>0 wavefunction? (It has a name.)
    Where have you seen this shape before?

    You have not completed the calculations.
    Use the boundary conditions to find values for B1, A2, and B2, in terms of A1.

    Did you watch the video I linked to?
     
  13. May 10, 2014 #12

    Yes. I've watched all his videos. The gentlemen talks very fast. :)
     
  14. May 10, 2014 #13

    Simon Bridge

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    Does this really make sense to you?

    Does this mean that you would also have a reflected wave in region 2 for E<V?

    QM may not be entirely like Classical mechanics, but it does have to contain classical mechanics.
    The calculation you just did should also work for a gaussian wavepacket travelling left to right ... you may need to have a go doing it for this case before you get it. In the picture you just described, because of the presence of a barrier, creating a localized wavepacket at x=-x0 at t=0 must simultaniously create a mirror image wavepacket at x=+x0. This means that it is possible to detect the particle whose behaviour is modelled by the wavepacket in a small volume about x=+x0 at time t=0. We can test this startling physical teleportation by putting a detector at x=+x0 and then switching the apparatus on.

    OK so leaving that aside for now.

    ... No it doesn't - x is free to vary along the whole number line.

    The previous wavefunction was a straight line with slope A.
    This wavefunction is a special case of a straight line - what is it's slope?
    Where have you seen that before.

    You are mistaken here ... aside from not being asked to do this for E=0, but E=V, A1 is the common term. Use the boundary conditions to relate the two sides.

    i.e.

    psi2=A2+B2 is just a constant (more usual to put B2=0 but you don't believe that for some reason.)
    ... so you can just put psi2=A2.

    All amplitudes are to be in terms of A1, so put A1=1.

    There are two boundary conditions - which gives you two equations.

    There are two remaining unknowns: B1 and A2.

    Simultaneous equations.

    The video series shows you this - you tell me you watched it? Did you have trouble following it?
    3:00-4:00mins into the linked video, Donny puts B2=0 (he uses D) ... why does he do that?

    Is your situation different from his in some way?
     
    Last edited: May 10, 2014
  15. May 12, 2014 #14
    The previous wavefunction was a straight line with slope A.
    This wavefunction is a special case of a straight line - what is it's slope?



    Slope of a vertical line - infinity.




    Where have you seen that before.




    Not sure that I have.





    Use the boundary conditions to relate the two sides.

    i.e.

    psi2=A2+B2 is just a constant (more usual to put B2=0 but you don't believe that for some reason.)
    ... so you can just put psi2=A2.

    All amplitudes are to be in terms of A1, so put A1=1.

    There are two boundary conditions - which gives you two equations.

    There are two remaining unknowns: B1 and A2.

    Simultaneous equations.

    The video series shows you this - you tell me you watched it? Did you have trouble following it?3:00-4:00mins into the linked video, Donny puts B2=0 (he uses D) ... why does he do that?

    Is your situation different from his in some way?[/QUOTE]





    This is where I am having the problem. In solving the two equations with three
    variables in the case of E > V I get

    A1 + B1 = A2 and

    ik1A1 - ik1B1 = ik2A2, i.e. equations that relate A1, B1 and A2.

    But with

    Ψ1 = A1eik1x + B1e-ik1x

    Ψ2 = A2 + B2

    this does not appear to be the case. Is there some other way to associate the
    terms in Ψ1 and Ψ2 for E = V that I am not seeing?
     
  16. May 12, 2014 #15
    Your first two equations are still fine to work with:

    A1 + B1 = A2 and

    ik1A1 - ik1B1 = ik2A2.

    As E tends to V, k2 tends to zero so your second equation gives you A1 = B1 and your first equation gives you A2 = 2A1.

    I don't think there should be a B2 in there and you cannot have a left moving wave inside the barrier.

    Physically the solution Ψ2 = A2 = 2A1 inside the barrier means that you have an equal chance to find it anywhere within the barrier (at any distance from x=0), but as the transmission coefficient is still zero it will still be squeezed back out.
     
  17. May 12, 2014 #16

    Simon Bridge

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    Jilang's post is what I was trying to ease you towards.

    You have shown some major confusions about the math as well as the physics.

    1. there being no reflected wave for x>0
    ... don't understand how you did not catch on to that since every lecture series stresses that point.
    perhaps you are having trouble understanding or following the lecture material?

    (when people speak too fast on a video - you know you can just pause after each sentence and write it down right?)

    2. you say that ##\psi(x)=C## is a vertical line - slope infinity - but this is plain not true: you can see if you only make a graph of ##\psi(x)## vs ##x##: the line is horizontal. (You appear to have a peculiar blindness to the shapes of functions.)

    The place you have seen this before is the wavefunction of a free particle.
    This is the solution to the Schrodinger equation where V(x)=0 for all x.

    3. the boundary condition (you should actually write it down as part of your working) are the relationships between ##\psi_1## and ##\psi_2## that you were looking for. They give you the first two equations.
    The boundary conditions are:
    $$\psi_1(0)=\psi_2(0)\\
    \left.\frac{d}{dx}\psi_1(x)\right|_{x=0}
    =\left.\frac{d}{dx}\psi_2(x)
    \right|_{x=0}$$ ... which just to say that the wavefunction has to be "continuous" across the barrier.

    4. you identified this one yourself: you have a very hard time relating the wavefunctions to probabilities.
    you should try sentences like: "when x approaches <some point/location> the probability approaches <some value>", for different locations - and let your results tell you the numbers.

    In your original wavefunction, the probabilities blow up as x gets large ... which is nonsense and a good clue you got it wrong. I couldn't answer your question until you got that. An easy check would have been to see if the solution you got was actually a solution to the schrodinger equation (it wasn't). You do this by substituting it back.

    Conclusion:
    If I were your teacher I would be very concerned for your future.
    I think that if you do not address these issues you will find yourself getting more and more stuck until you won't be able to continue.

    My general recommendation to you is to get a tutor who can work on these things in real-time with you.
     
    Last edited: May 12, 2014
  18. May 12, 2014 #17
    Not to worry. I'm not a student. Just an interested amateur. :smile:

    Thanks for all your help.
     
  19. May 12, 2014 #18

    Simon Bridge

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    That would explain things, yes.
    I stand by the recommendations - you need to address those issues if you are serious about wanting to understand this stuff.
    Where does your education go up to in maths?
     
  20. May 13, 2014 #19
    I have a mediocre understanding of undergraduate calculus and differential equations.
     
  21. May 13, 2014 #20

    Simon Bridge

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    How about vectors and linear algebra?

    Introductory QM is usually a second-year semester-long paper - taught just after or concurrently with algebra and differential equations math papers.

    You do need the math to follow the lectures at the level you are attempting.
     
    Last edited: May 13, 2014
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