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Eigenfuntions QM problem

  1. Sep 4, 2005 #1
    Ok so Im revising for an exam thats in a day or so and i'm totally stuck on some revison questions, needless to say i need to understand whats going on not just get an answer so if somone can help me do this I would be grateful. Should be pretty easy it's basic stuff from last yr but my mind is going at about 1000miles an hour and i can't think straight.

    Hamiltonian operator for particular system has energy eigenvalues En (n = 1,2,3) and corresponding normalized eigenfunctions Un. the eigenfuntions corresponding to different eigenvalues are orthogonal. The wavefuntion of the electron is given by.

    Si = K(3U1 +4U2)

    (a)calc real normalisation const, k
    (b)what is probablity P1 that measurement of energy would give E1?
    (c)...................................................................................E2?
    (d)...................................................................................E3?
    (e)Using p1 and P2 determine the average value for energy of the electron.
    (f)Evaluate expectation vaule of hamiltionian operator for an electron in state Si and show that this is the answer obtained in (e)

    numerical answers:
    a) 1/5
    b)9/25
    c)16/25
    d)0
    e) (9E1+16E2)/25

    I'm supposed to use what in our lectures were postlates 4 and 6.

    4. something about si (x,t) = SUM (j) cj SIj (x) exp(-iEjt/hbar)

    where Ej is energy associated with SIj

    mmm cant write rest as with limited keyboard.

    and postulate 6.

    Pl = |al|^2/ SUM |an|^2 where SI = SUM (n) an SIn

    Sorry I realise the type is confusing will clarify anything needed, thanks for help
     
  2. jcsd
  3. Sep 4, 2005 #2

    HallsofIvy

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    Science Advisor

    You vector is k(3U1+ 4U2). Assuming that these eigenvectors are orthonormal as well as orthogonal (that is, they are taken to be unit vectors), then 3U1+ 4U2 has length
    [tex]\sqrt{3^2+ 4^2}= \sqrt{25}= 5[/tex]
    In order to make k(3U1+ 4U2) a unit vector, k must equal 1/5.

    In that case k(3U1+ 4U2)= 1/5(3U1+ 4U2)= (3/5)U1+ (4/5)U2+ (0/5)U3. The "amplitudes" for U1, U2, U3 are 3/5, 4/5, and 0 respectively. Their probabilities are the squares of those: 9/25, 16/25, and 0.
     
  4. Sep 4, 2005 #3
    ok, what you said makes sense, except, why is it a vector? am i being utterly stupid?
     
  5. Sep 4, 2005 #4
    (a) In general a wavefunction (stationary state) looks like this:
    [tex] \Psi = K( c_{1} u_{1} + c_{2} u_{2} + c_{3} u_{3} + ...)[/tex]
    where K is the normalization constant.

    The numbers [tex]c_{1},c_{2},c_{3},c_{4}....[/tex] are the coefficients.

    You can calculate K by [tex]K=\sqrt{|c_{1}|^2+|c_{2}|^2+|c_{3}|^2 +...}[/tex],
    so what you do is you take the absolute square of each coefficient, sum them up and take the squareroot of this sum =>
    that's K

    Your wavefunction is:
    [tex] \Psi = K(3 u_{1} + 4 u_{2}) [/tex]
    [tex] \Psi = K(c_{1} u_{1} + c_{2} u_{2})[/tex]

    Your coefficients are [tex]c_{1}=3[/itex] and [itex]c_{2}=4[/tex].
    Plugging this into the formula for K you get:
    [tex]K= \sqrt{|3|^2+|4|^2} = \sqrt{25} = 5[/tex]

    Thus your wavefunction is:
    [tex]\Psi = \frac{1}{5}(3 u_{1} + 4 u_{2})[/tex]
    [tex]= \frac{3}{5} u_{1} + \frac{4}{5} u_{2}[/tex]

    Now you have the new, normalized coefficients:
    [tex]\Psi = b_{1} u_{1} + b_{2} u_{2} [/tex]
    [tex]= \frac{3}{5} u_{1} + \frac{4}{5} u_{2}[/tex]

    The new normalized coefficients are:
    [tex] b_{1} = \frac{3}{5} [/tex]
    [tex] b_{2} = \frac{4}{5} [/tex]


    (b) You can get [tex]p_{1}[/itex] by the formula [tex] p_{1} = |b_{1}|^2[/tex],
    that is you just have to pick out the "normalized" coeficient [tex]b_{1}[/tex]
    in front of the eigenfunction [tex]u_{1}[/tex] and take the absolute square of it.
    Look above, the coefficient in front of
    [itex] u_{1}[/itex] is [itex] b_{1}= \frac{3}{5}[/itex]. Thus, using the formula [itex] p_{1} = |b_{1}|^2[/itex] you obtain 9/25.
    I repeat: Use the normalized coefficient! (not the [tex]c_{1}[/tex])


    (c) You do the same as in (b), pick out the coefficient in front of [itex]u_{2}[/itex], which
    is [itex]b_{1} = \frac{4}{5}[/itex] (see above). You obviously get [itex]p_{2} = \frac{9}{25}[/itex]


    (d) Just take again the formula [tex]p_{3} = |c_{3}|^2[/tex].
    But then the question is where do you get [tex]|c_{3}|^2[/tex] from? It's the coefficient in front of [tex]u_{3}[/tex], but since [tex] u_{3}[/tex] doesn't appear in the wavefunction [tex] \Psi = \frac{3}{5} u_{1} + \frac{4}{5} u_{2}[/tex], the coefficient is zero.

    To better understand this you could also write:
    [tex] \Psi = \frac{3}{5} u_{1} + \fract{4}{5} u_{2}[/tex]
    [tex] = \frac{3}{5} u_{1} + \fract{4}{5} u_{2} + 0 u_{3}[/tex], right? (I just added a zero there)
    Then you see the coefficient is zero in front of [tex] u_{3}[/tex].


    (e) The average energy is defined as:
    Expectation value for energy [itex]= \sum_{i} p_{i} E_{i} [/itex]
    (just as you know it from school).

    Just plug in the values you obtained before:
    Expectation value for energy [itex] = \frac{9}{25} E_{1} + \frac{16}{25} E_{2} = \frac{1}{25} \left (9E_{1} + 16E_{2} \right )[/itex]

    (f) The expectation value is defined as [tex] \langle \hat{H} \rangle = \langle \Psi| \hat{H} | \Psi \rangle [/tex].
    (i) First calculate [tex]\hat{H} | \Psi \rangle [/tex],
    (ii) then "multiply" from the left side [tex] \langle \Psi |[/tex].

    Hint for (i): [tex] \hat{H} |u_{1}\rangle = E_{1} |u_{1} \rangle [/tex],
    and [tex] \hat{H} |u_{2}\rangle = E_{2} |u_{2}\rangle[/tex]
    Hint for (ii): [tex]\langle u_{1} | u_{1} \rangle = 1[/tex] and [tex]\langle u_{2} | u_{2} \rangle = 1[/tex], and: [tex] \langle u_{1} | u_{2} \rangle =0, [/tex] and [itex]\langle u_{2} | u_{1} \rangle = 0[/tex].
     
    Last edited: Sep 4, 2005
  6. Sep 4, 2005 #5
    oh thats all so easy, thanks edgardo!
     
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