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Homework Help: Challenging QM problem on coupling two 2-level systems with dipole moments

  1. May 11, 2006 #1

    My teacher gave me a very challenging problem on QM.
    I've only had one introductiary course, but he said he wanted me to figure this problem out by using whatever resources, so now I turn to you guys ;-). The problem is the following:

    London attraction forces can be thought of as a pure QM effect that can be calculated with the usage of 2 level systems that can feel each other due to transition dipole moments (this is also called exciton interaction).

    If we couple two 2-level systems (in NMR experiments this is called spin coupling) then four different energy levels will be possible. (These levels were present before coupling, but then they were two by two identical). The four levels can be found by determining the eigenvalues of the following matrix:

    http://img91.imageshack.us/img91/585/matrix2ej.gif [Broken]

    where E is the energy of the excited state of the original 2-level system (the ground state energy is 0) and U is given by:

    http://img168.imageshack.us/img168/734/u8yw.gif [Broken]

    with u1 and u2 two transition dipole moments and r = the vector that connects u1 and u2.

    a) Determine the eigenvalues of the matrix. Hint: think of it as two easier-to-calculate 2x2 matices that you can diagonalize much easier. Sketch the position of the energy levels in a energy diagram.

    I thought using the first and fourth row in one 2x2 matrix and the others in another one, but where to go from here?

    b)What is the new ground state energy?

    I really don't know how I'd solve this problem..

    c) The two dipole moments are both pointing in the x- direction and the first one is located in the origin. The other one is in the position (0,0,2). Distances are measured in angstroms. Now sketch the ground state energy as function of the position of the second dipole moment along the z-axis.

    I really don't know how I'd solve this problem too..

    d) What is the force that both systems exert on each other if their distance is z?

    Perhaps I need to take the - derivative to the position of the energy??? (but what would the energy function be....?)

    e) Show that the force, in a good approximation, is proportional to U2 and thus proportional to 1/distance6, just like the London force. Hint: Use the Taylor series.

    Again: I really don't know how I'd solve this problem..

    Could anybody of you PLEASE PLEASE PLEASE help me with this problem?!?! I'd REALLY appreciate it very very much!!!!
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. May 11, 2006 #2


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    You are correct. Then you have two different 2 by two matrices, right? Just find the eigenvalues of each. Do you know how to find the eigenvalues of a matrix (you probably have learned that in linear algebra)?
    Once you have found the four eigenvalues, the new ground state energy will simply be the smallest of these.

    The question is confusing since they say that z is fixed at 2 Angstroms and then they say to vary z:uhh:
    I guess they want z to be varied so you should treat it as a variable.

    It is quite simple, really, you just have to go through each term of U and simplify as much as possible. For example, what is the vector r? If the first dipole is at the origin and the second is at (0,0,z), then the vector r is simpply the vector going from the origin to (0,0,z). What is this vector? Now, use the fact the the two dipole vectors are pointing along x and you can explicitly calculate the dot products (the expression will simplify quite a bit).

    The force would be *minus* the derivative of the expression you have just found with respect to z.
    You vae to find U explicitly in terms of z and then it will be easy, so start with that.
    Last edited by a moderator: May 2, 2017
  4. May 12, 2006 #3
    So I have





    (I'll use L for lambda)

    Eigenvalues of A are found by using: det(A-LI)=0

    [tex]\left(\begin{array}{cc}-L&U\\U&2E-L\end{array}\right) = 0[/tex]

    L2 - 2EL -U2 =0

    Using the abc formula with
    a=1, b= -2E and c= -U2 yields:

    [tex]\frac{2E \pm sqrt{4E^2 + 4U^2}}{2} [/tex] =

    [tex] E \pm sqrt{E^2+U^2} [/tex]

    Eigenvalues of B are found by using: det(B-LI)=0

    [tex]\left(\begin{array}{cc}E-L&U\\U&E-L\end{array}\right) = 0[/tex]

    (E-L)2 -U2 =0
    E2- 2EL + L2 - U2 =0
    L2 -2E - (U2-E2) =0

    Using the abc formula with
    a=1, b= -2E and c= - (U2-E2) yields:

    [tex]\frac{2E \pm sqrt{4E^2 + 4 (U^2-E^2)}}{2} [/tex] =

    [tex] E \pm sqrt{U^2} = E \pm U [/tex]

    That would be

    [tex] E - sqrt{E^2+U^2} [/tex]


    Yeah I guess that's what they mean!

    This would give me:

    [tex] U= \frac{1}{4 \pi \epsilon_0 z^3} \mu_1 \mu_2 [/tex]

    I dont know if u1=u2?

    Substitution would give me:

    [tex] E - sqrt{E^2+ (\frac{1}{4 \pi \epsilon_0 z^3} \mu_1 \mu_2)^2} [/tex]

    which is quite an equation.... how on earth would I sketch it?

    Same as above: the equation is quite difficult to take the derivative to z, first I'd like to know if I'm correct up till now, then I'll try to move on....

    Thanks a lot btw for helping me on this problem :) I hope I'll finish it all the way :D
  5. May 13, 2006 #4
    Oh and as for:

    The *minus* derivative of:
    [tex] E - sqrt{E^2+ (\frac{1}{4 \pi \epsilon_0 z^3} \mu_1 \mu_2)^2} [/tex]

    would be

    [tex]\frac{-3 \mu_1^2 \mu_2^2}{8 \pi^2 \epsilon_0^2 z^7} \frac{1}{2 sqrt{E^2 + \frac{\mu_1^2 \mu_2^2}{16 \pi^2 \epsilon_0^2 z^6}}} [/tex]

    assuming E is independent of z (is that correct?) and u1 and u2 are NOT the same (or are they?) This seems like a huge equation to me too.... is it correct?
  6. May 13, 2006 #5


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    evrything you have done so far seems right.

    Asa for plotting the ground state energy, notice that as z->0 it goes to minus infinity. As z goes to infinity, what does it tend toward? And you have actually already calculated the derivative with respect to z and you can see that it is positive everywhere (for all positive z) so the slope is always positive. From this fact and the two asymptotes at z->0 and z->infinity, you can get a rough sketch easily. It is not a very exciting graph :wink:

    Your force equation seems correct.

    However, I am puzzled by the last question...maybe you could double check with your prof if there is not a mistake in the question. If you treat the two dipole moments (and btw, I think you should keep different symbols, i.e. don't assume they are equal unless told otherwise) as being very small parameters then you can treat U as being much smaller than E (this makes sense only as z is not too small because no matter how small the mu's are, if z is small enough, the U term will of course get quite large at some point).

    In any case, if U<<E then you can easily do a Taylor expansion of the potential and *that* will be proportional to U^2 and hence to 1/z^6. And, if I recall correctly, it's the London *potential* which goes like 1/distance^6. So what I am saying is that I *think* that in the last question the word "force" should be replaced everywhere by "potential". Otherwise the question does not make sense to me (I might be misisng something and someone more knowledgeable will come along and correct me but I would suggest that you double check with your prof)
  7. May 14, 2006 #6
    I've checked my notes and the only thing my prof told me about this, is that he mentioned the Lennard Jones potential. It shows that at a short distance the potential is proportional to 1/r12 (molecules are close to one another and repulse each other), though at large distances it goes with 1/r6, just like the London/ van der waals potential (and indeed, just like you said, when z is big enough to make U << E).
    Because it's the potential they are talking about one might expect the force to be proportional to 1/r7

    I've checked Google and found
    http://www.ocms.ox.ac.uk/mirrored/xplor/manual/htmlman/node117.html" [Broken] that says:


    And http://www.psiainc.com/menu037_001.asp" [Broken] confirms that:

    But I guess these people just mixed things up, because I found a more professional looking explanation in the abstract section at
    http://www.citebase.org/cgi-bin/fulltext?format=application/pdf&identifier=oai:arXiv.org:physics/0007084" [Broken]

    I guess the force we are looking for is called the ''non-retarded force'' in this case....

    Anyways... I'll consult with my teacher tomorrow and hope to be able to clear things up.

    Btw: my teacher told me that in order to solve this problem I need to have good understanding of quantum mechanics, but to be honest, I really don't know when I had to use some theorem from QM in this problem in order to get to a correct answer....
    Last edited by a moderator: May 2, 2017
  8. May 14, 2006 #7


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    Good work...It does sound like people go from "potential" to "force" without making much of a distinction!! But I am pretty convinced that the question was aimed at the potential energy.

    You used that at the very beginning when you found the energies by calculating the eigenvalues of the Hamiltonian. You see, there are two ways to present Hamiltonians in quantum mechanics: as differential operators or as matrices. In both cases, one must solve for the eigenvalues and eigenvectors (or "eigenstates"). When the Hamiltonian is written as differential operator, solving for the eigenvalues and eigenstates corresponds to solving a differential equation. When the operator is given as a matrix, one solves for the eigenvalues the way you did it (and one could also solve for the eigenstates but that was not needed for you).

    If you have never worked with a Hamiltonian written as a matrix, then this exercise was a way to get you to do this.


    Last edited by a moderator: May 2, 2017
  9. May 15, 2006 #8
    Hi Patrick!

    Today in college I've checked it with my professor. He immediately confessed the word force should be replaced by potential as you suggested. So I've done a Taylor expansion on

    [tex]L= E - sqrt{E^2 + U ^2}[/tex]

    Using U << E the Taylor series are:

    [tex]L(U)= L(0) + \frac{1}{1!} L'(U)|_U_=_0 U + + \frac{1}{2!} L''(U)|_U_=_0 U ^2 + ... [/tex]

    L(0)= [tex]L= E - sqrt{E^2 + 0 ^2}[/tex]= E- E= 0

    L'(U)= [tex]\frac{-2U}{2 sqrt{E^2 + U^2}} [/tex]

    L'(0)= 0

    L''(U)= [tex]\frac{-1}{sqrt{E^2 + U^2}} - 4 U^2 ( -E^2 + U^2)[/tex]

    L''(0)= [tex]\frac{-1}{E}[/tex]

    L(U)= 0 + 0 U +[tex] \frac{1}{2} \frac{-1}{E} U^2[/tex] = [tex] \frac{-U^2}{2E} [/tex]

    L(U)= [tex] \frac{-1}{2E} U^2[/tex] so L(U) ~ U2 and therefore L(U) ~ 1/distance6

    .... but now I used the second term of the Taylor series.... I don't know if that's a problem, because up till now we were explicitely told to only use the first one (cause in the second one the small variable is raised to the power 2, which gives an even smaller outcome)...

    Have I done it the right way?

    As for the Hamiltonian, thanks for explaining that to me! Now I know where the term eigenstates comes from (I never knew the Hamiltonian could be writtin in matrix form). I always wondered, because eigen means ''own'' in Dutch (and German, where it actually comes from) and I just could not figure out why an English term would contain a word in Dutch :D
  10. May 15, 2006 #9


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    I agree with this. The way I do it is to remember that for small epsilon, [itex] {\sqrt{1+ \epsilon}} \approx 1 + {\epsilon \over 2}[/itex].
    So [itex] E- {\sqrt{E^2 +U^2}} = E - E {\sqrt{1+ U^2/E^2}} \approx E - E (1 + {U^2 \over 2 E^2} ) = -{U^2 \over 2 E} [/itex]. But you proved the whole thing from scratch so that's even better.
    The *real* rule is to keep the first nonvanishing term. Here, there is no term of order U^1, so you have to keep the term of order U^2. (and if the term of order U^2 would have happened to be zero, you would have needed to keep the term or der U^3 and so on). So that's fine.

    Great. If you had never worked with a Hamiltonian in matrix form, then this is the main lesson of this calculation! (Note: you may not have realized this but Schrodinger's equation becomes a matrix equation when the Hamiltonian is given as a matrix. Basically, solving for the energies corresponds to finding the eigenvalues of the Hamiltonian matrix. So when you solved for the eigenvalues, you were actually solving Schrodinger's equation!!)

    You might had encountered "eigenvectors" of matrices in linear algebra. This is indeed why the solutions of the Schrodinger equation written as a differential operator are also called "eigenstates" of the Hamiltonian.

    You did all the work! Good job!
    I am glad I could guide you a bit.

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