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Expectation value of a finite well, and superposition of first two states.

  1. Aug 15, 2012 #1
    1. The problem statement, all variables and given/known data
    the first two energy eigenstates of a 1 nm wide finite well of barrier height 8vo have energy eigen values of 0.66ε and 2.6ε. calculate the expectation value of a linear superposition of these states?



    2. Relevant equations
    airy equations


    3. The attempt at a solution
    i have attached a photo with working out..

    would appreciate any help
     

    Attached Files:

  2. jcsd
  3. Aug 15, 2012 #2

    Simon Bridge

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    I was expecting something along the lines of [itex]\langle H \rangle = c_1^2E_1 + c_2^2E_2[/itex] where [itex]c_n = \int \psi_n^\star \psi d\tau[/itex] ... are you not given the superposition?

    But then I realized that while you said you had to find the "expectation value of the superposition" you did not say which property you needed the expectation value of ... with retrospect, the expectation value of energy would seem to be too simple. Even so - you seem to have been intent on calculating a value for energy.

    Anyway - I think your reasoning needs to be more explicit in order to be able to advise you.
    eg - in the first line you seem to want to calculate [itex]E_2-E_1[/itex] but you don't say what this is supposed to achieve ... you also have an expression for [itex]E_n[/itex] being equal to an expression for a difference in energies... which seems to me to be flat wrong.

    You have a notation: "@2=0" ... or is it "@z=0" ? (if so - good habit: cross your "z"'s).
    You then spend a lot of time rearranging an equation.
     
    Last edited: Aug 15, 2012
  4. Aug 15, 2012 #3
    the notation is (@z=0) or "z"=0.
     
  5. Aug 15, 2012 #4
    the expression for En should be just E then.

    so now i can calculate a value of E for a 1nm wide finite well in superposition of the 1st two states where (n22-n12)..

    with that plugged in and calculated i get a value of 1.793e-19 J

    does this value come into the airy function to help resolve the energy expectation value of linear superposition of the states given.(0.66 and 2.6ε)
     
  6. Aug 15, 2012 #5
    my other question is: does the barrier height of 8v0 have an effect(should it be included somewhere in the equation).

    im greatly thankful for your help :)
     
  7. Aug 15, 2012 #6

    Simon Bridge

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    The first En seems fine ... it's just that the rest of the line makes no sense mathematically - you have unequal things being equal: a contradiction! You need to get rid of at least one of those equals signs.

    Also the E2-E1 bits would just give the transition energy between states 1 and 2. How is this related to the expectation value?

    Just to be clear - you are told that the particle is in the superposition of the 1st two energy eigenstates?
    ??? then why do the calculation for the first two eigenstates then?
    ... why? How does that get you the expectation value?

    The barrier height would certainly affect the Airy function - it goes into the V(z) part of the Schodinger equation.

    I think one of us is horribly confused.

    You have a state ψ(z) prepared in a superposition of energy eigenstates thus [itex]\psi=c_a\psi_a + c_b\psi_b[/itex] with eigenvalues Ea=0.66ε and Eb=2.6ε

    If you know the Schodinger equation solutions for the energy eigenstates for your shape well then you should be able to just put these E values in to get the wave-functions. But I am unclear about the exact shape of the potential ... please write out the potential function for me: V(z)= ... and I don't see what V0 and ε are supposed to be.

    The expectation value of energy is:[tex]\langle H \rangle = \int_0^\infty \psi^\star H \psi dz[/tex]It gives the average energy of the particle in the state ψ ... if the two eigenstates were equally likely then this is just [itex]\bar{E}=\frac{1}{2}(E_a+E_b)[/itex], but that may not be the case, which is why the probability amplitudes are needed.
     
  8. Aug 15, 2012 #7
    the question reads:

    the first two energy eigenstates of a 1nm wide finite well of barrier 8vo have energy eigenvalues of 0.66ε and 2.6ε, where the symbols have their usual meaning. calculate the energy expectation value of a linear superposition of these states.
     
  9. Aug 15, 2012 #8
    im very confused now :S

    how would u tackle such question, what are the important equations required, and what assumptions do we make to solve it.
     
  10. Aug 15, 2012 #9
    im thinking to ignore the 1st equation i have written down, and start 1st with airy function (@z=0), get a value for that, which will look like (-2.48e-19εi)..

    then take that value and multiply it by the the values of 0.66ε and 2.6ε,
    this will then give me the expectation value for
    E1=-2.48e-19(0.66)=-1.6368e-19 j
    and E2=-2.48e-19(2.6)= -6.448e-19 j

    is that right?

    or just E=(n2hbar2π2)/(2mlz2)

    lz=1nm
    so i get a value for E. which is 1.793e-19 j

    take that value of E, substitute it into the Airy function equation, and solve for E ( two different E's in the Airy function).

    im then left with E=(number)εi

    then take that value and multiply it by the the values of 0.66ε and 2.6ε,
    this will then give me the expectation value for
    E1=-2.48e-19(0.66)=-1.6368e-19 j
    and E2=-2.48e-19(2.6)= -6.448e-19 j
     
  11. Aug 15, 2012 #10
    Hi solas99

    I'm confused here as to exactly what type of potential well we have here - is it the (infinite) linear well with the nasty Airy functions at the top of your page, or is it a finite square well, or is it a finite square well with a barrier?

    In any case, if you know ε, and you know what linear superposition of energy eigenstates you have, then it's dead easy to find the energy expectation value - see Simon Bridge's first reply.
     
  12. Aug 15, 2012 #11
    that is exactly how the question is given:
    the first two energy eigenstates of a 1nm wide FINITE well of barrier 8vo have energy eigenvalues of 0.66ε and 2.6ε, where the symbols have their usual meaning. calculate the energy expectation value of a linear superposition of these states.

    im as confused, on what should i apply to solve it.

    no other information is given, except for this.

    thanks again for replying
     
  13. Aug 15, 2012 #12
    Maybe it's just a plain old finite square well then? So no Airy functions?
     
  14. Aug 15, 2012 #13

    Simon Bridge

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    ... I'm leaning in that direction - but it is looking more and more like it does not actually matter and "linear superposition" will mean an equal mix of the two states and the question just wants to know if the reader understands what "expectation" means.

    @solas99: the question is leaving a lot to context ... do you know what "the usual meanings" for V0 and ε are? You've got the book, not me - they must tell you! Similarly look for what they usually mean by "linear superposition".

    What does "barrier" and "finite well" mean, according to the book.
    There are many possible finite wells and many possible barriers that could fit that description. Is there any special reason you chose the triangle well?


    Off your efforts I was imagining a potential that looked like:

    V(z)=8V0 : z < 0 (V0 := well depth)
    V(z)=(z-w)V0/w : 0≤z<w (w := width of the well)
    V(z)=0 : z≥w
    ... but I see that you are clearly told you have the first two energy states and I somehow doubt this configuration will give those energies. From your reading of the chapter and the related questions, how would you write out the potential?

    But, like Oxvillian says, if you know ε, then you don't need to know the exact potential, and the calculation becomes trivial.
     
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