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Homework Help: Wavefunctions and boundaries of a 1D-potential well

  1. Apr 13, 2010 #1
    the attachment here shows a potential well and i am practicing writing down wave functions in each region and boundary conditions (for this problem I suppose the boundaries would be x= -a, 0, b.

    i think that for this problem i need 4 functions and i thought that they may be:

    when
    x<-a (1) = 0
    -a<x<0 (2) = Asin(k1x) + Bcos(k1x)
    0<x<b (3) = Csin(k2x) + Dcos(k2x)
    b>x (4) = 0

    where
    k1 = ([tex]\sqrt{}2mE[/tex])/[tex]\hbar[/tex]
    k2 = ([tex]\sqrt{}2m(E-Vo))/\hbar[/tex]

    and i think at the boundary x = -a and b the wave function is continuous and at x=0 the wave function and the first derivative is continuous.

    Is this right?
    thanks
     

    Attached Files:

    • qu7.jpg
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  2. jcsd
  3. Apr 13, 2010 #2
    |......................................|
    |......................................| Vo < E
    |......................................|
    |....................__________|
    |....................|
    |....................|
    |....................|
    |....................|
    -a...................0.................b

    V(x)=
    infinity --- x<-a --- region 1
    0 --- -a<x<0 --- region 2
    Vo --- 0<x<b --- region 3
    infinity --- b<x --- region 4

    if you cant see the pic
     
  4. Apr 13, 2010 #3
    first thing:

    maybe you want to think again about what you stated here when you said "b>x (4) = 0", i believe you have done simple mistake since you wrote the first three equations correctly, it actually should be "x>b (4) = 0"

    second thing:

    dont you think that you should make your wavefunction continuous by appliying the fact that the wavefunction and its derivative (1st) in region(1) = the wavefunction and its derivative (1st) in region(2) and that when x = -a , and the same goes with region(2) and (3) when x = 0 , also region(3) and (4) when x = b ..

    hmm, well .. I think so .. what do you think?
     
  5. Apr 13, 2010 #4
    for the first thing your right i did write it down wrong t=it was meant to be b<x.

    but with the the reason i wrote that the first derivative of the wavefunction is not continuous at the boundary x=-a, and b was cause i thought that when on the otherside of a boundary the potential->infinity, so KE->-infinty at boundaries so first derivative->infinity also. this is cause it seems to be infinite potential and so not physically realistic.
     
  6. Apr 13, 2010 #5
    for the first thing your right i did write it down wrong t=it was meant to be b<x.

    but with the the reason i wrote that the first derivative of the wavefunction is not continuous at the boundary x=-a, and b was cause i thought that when on the otherside of a boundary the potential->infinity, so KE->-infinty at boundaries so first derivative->infinity also. this is cause it seems to be infinite potential and so not physically realistic.
     
  7. Apr 13, 2010 #6
    thats mainly the reason behind choosing the wavefunction = 0 in both regions 1 and 4..

    lets first consider the situation when we say that the wavefunction and its derivative should be continuos in region (1) and (2) when x = -a ..

    you have in region (2) that the wavefunction = A sin(k1x) + B cos(k1x) .. so when x = -a

    in region(2) the wavefunction = - A sin(k1a) + B cos(k1a) & you know that the wavefunction in region(1) at x = -a is zero >>> then 0 = -A sin(k1a) + B cos(k1a)
    im not allowed to proceed more than this, but i think you got the point (I hope!) .. continue with the derivatives .. and dont forget the other regions ..

    good luck..
     
  8. Apr 13, 2010 #7
    for the first thing your right i did write it down wrong t=it was meant to be b<x.

    but with the the reason i wrote that the first derivative of the wavefunction is not continuous at the boundary x=-a, and b was cause i thought that when on the otherside of a boundary the potential->infinity, so KE->-infinty at boundaries so first derivative->infinity also. this is cause it seems to be infinite potential and so not physically realistic.
     
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