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Finding overlap of fuzzy energy states

  1. Apr 25, 2015 #1
    1. The problem statement, all variables and given/known data

    Recall the definition of the overlap of wave functions Φ and Ψ:

    [; (\Psi , \Phi ) = \int\limits_{-\infty}^\infty dx\: \Psi ^{*} (x)\Phi(x);]



    Let ψ1(x) and ψ2(x) be unit-normalised wavefunctions representing sharp-energy states with different energies (and hence zero overlap). Use real constants N1, N2 > 0 to normalise the wave functions

    [; \Phi (x) = N_1 \{ (\sqrt{3} -i) \psi_1(x) + (-2 + i) \psi_2(x)\} ;]
    [; \Psi (x) = N_2 \{ (1 +2i) \psi_1(x) - (1 -2i) \psi_2(x)\} ;]

    and find the overlap of the corresponding fuzzy-energy states.

    2. Relevant equations
    [; (\Psi , \Phi ) = \int\limits_{-\infty}^\infty dx \Psi ^{*} (x)\Phi(x);]


    3. The attempt at a solution
    I have normalised Φ and Ψ to get values for N_1 and N_2

    [; |N_1|= \frac{1}{3} and\: |N_2|=\frac{1}{\sqrt{5}};]

    and this is as far as I can get, as for finding the overlap of the corresponding fuzzy-energy states I have been through my lecture notes and cannot find where to go from here. Do I use the face that that in general

    [;N_1= \frac{1}{3} e^{i\theta} and \: N_2=\frac{1}{\sqrt{5}}e^{i\theta};]

    and then plug them into the above integral?

    Thanks for your help, I am completely lost on this one.
     
  2. jcsd
  3. Apr 25, 2015 #2

    BvU

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    Hi Woody, welcome to PF :smile:! Nice Avatar :wink:.

    It looks as if you are using delimiters that my computer does not recognize: ##[;## for open and ##;]## for close instead of ##\#\# ## for textstyle open and close and ## $$ ## for displaystyle open and close

    [;(\Psi , \Phi ) = \int\limits_{-\infty}^\infty dx\: \Psi ^{*} (x)\Phi(x);]

    First: I get something slightly different for ##N_2##

    Then to your real question: no, the ##\theta## does not have to be brought in. Simply write out the overlap definition in full and make good use of what you know from " ψ1(x) and ψ2(x) are unit-normalised", which means " ψ1(x) and ψ2(x) are orthonormal"
     
  4. Apr 25, 2015 #3
    Sorry, the second equation should be

    [; \Psi(x) =N_2\{2(1+2i)\psi_1(x)-(1-2i)\psi_2(x)\};]

    which (I hope ) gives [;|N_2|=\frac{1}{ \sqrt {15}};] (I am presuming you got [; N_2= \frac{1}{\sqrt{10}};] )

    Is there another term for overlap? I cannot find anything in my textbook regarding overlap and in my lecture notes we jump from the definition to an example where we just plug the following into the definition

    [; \psi_1(x)=\frac{1}{\sqrt{L}}e ^{ip_1/\hbar}\;,\;\psi_2(x)=\frac{1}{\sqrt{L}}e ^{ip_2/\hbar};]

    and solve using periodic boundary conditions [; -\frac{L}{2}\leq x\leq\frac{L}{2};]

    Thanks
     
  5. Apr 25, 2015 #4

    BvU

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    In that case I again get something else for ##N_2## than you :smile: .
    Overlap is defined in your problem statement. You have ##\Phi = a\psi_1 + b\psi_2## and ##\Psi = c\psi_1 + d\psi_2## ; carry out the multiplication of the complex conjugate of a sum of two terms times a sum of two terms .
     
  6. Apr 25, 2015 #5
    Damn:frown: and thanks

    here is my working
    [; \Psi(x) =N_2\{2(1+2i)\psi_1(x)-(1-2i)\psi_2(x)\};]
    [;1=|N_2|^{2}\{2(1+4)+1+4\};]
    where
    [;1=P_1+P_2=a^{2}+b^{2}+c^{2}+d^{2};]
    where a, b, c, d are read directly from c1 and c2

    And that is how I get
    [;N_2=\frac{1}{\sqrt{15}};]

    or do I ignore the 2 out front of Ψ1? and get [;N_2=\frac{1}{\sqrt{10}};]

    and back to the problem I am having real problem with :wink:

    So I just sub Ψ and Φ into the definition, do all the multiplication and a bunch of stuff will drop out leaving me with a much simpler integral?

    Thanks so much for your help.
     
  7. Apr 25, 2015 #6
    Damn:frown: and thanks

    here is my working
    ##\Psi(x) =N_2\{2(1+2i)\psi_1(x)-(1-2i)\psi_2(x)\}##

    ##1=|N_2|^{2}\{2(1+4)+1+4\}##
    where
    ##1=P_1+P_2=a^{2}+b^{2}+c^{2}+d^{2}##
    where a, b, c, d are read directly from c1 and c2

    And that is how I get
    ##N_2=\frac{1}{\sqrt{15}}##

    or do I ignore the 2 out front of Ψ1? and get ##N_2=\frac{1}{\sqrt{10}}##

    and back to the problem I am having real problem with :wink:

    So I just sub Ψ and Φ into the definition, do all the multiplication and a bunch of stuff will drop out leaving me with a much simpler integral?

    Thanks so much for your help.

    Fixed the layout. Thanks for that hint too!
     
  8. Apr 25, 2015 #7

    BvU

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    Looks a lot better. And I can simply right-click on the formatted TeX to get back the TeX input, copy it and try to colour the error: ##\
    1=|N_2|^{2} \{{\bf 2}(1+4)+1+4\}##

    (No I can't, makeup throws off LaTeX: ## 1=|N_2|^{2} (2(1+4)+1+4 )## ## so I boldfaced it ).

    You certainly don't ignore a factor in the coëfficiënt !

    And you confuse me mightily with ##1=P_1+P_2=a^{2}+b^{2}+c^{2}+d^{2}## (it may well be correct, but P1, P2, c1, c2 ? )
    when I expect you to interpret normalization as$$
    \int\Psi^*\Psi = 1 \\ \Leftrightarrow
    N_2^*N_2\ \int \bigl ( 2(1+2i)\psi_1 - (1-2i)\psi_2\bigr )^* \; \bigl ( 2(1+2i)\psi_1 - (1-2i)\psi_2\bigr ) = 1 \\
    \Leftrightarrow N_2^*N_2\ \int \bigl ( 2(1-2i) 2(1+2i) + (1+2i)(1-2i) \bigr ) = 1
    $$
    (where I used the orthonormality of ##\psi_1## and ##\psi_2## (*) )


    (*) -- check into that: I think you already know how it works for ##\Psi^*\Psi## ;
    same method comes in handy for ##\Psi^*\Phi## too :wink: :
    look ma, no integration needed to work out these integrals !

    and -- to check you 've mastered it:
    find out if (for the most general case) ## \Psi^*\Phi = (\Phi^*\Psi)^*##
    as one would expect.

    --​
     
  9. Apr 25, 2015 #8
    Thanks again, I have manged to get a better answer for N2 this time it is a nice number I I am guessing I did manage to get it right...N2 = 1/5

    Now I have done the easy bit, time to get back to the bit that I am struggling with...o_O
     
  10. Apr 26, 2015 #9

    BvU

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    Look carefully at what I did for ##\int\Psi^*\Psi## and do the same for the overlap as given in the problem statement (not the =1, though, because this time there will be a different answer :smile:). Write out all the intermediate steps (integral of sum = sum of integrals, bring constant coefficients in front of the ##\int## sign, etc.) and work towards being able to exploit the orthonormality of the ##\psi##.
     
  11. Apr 26, 2015 #10
    Thanks so much for your help on this but I think I might have to bail on this. I have a bunch of other work and I think it is a bit crap that our lecturer dumped this assignment on us on the last Friday of the break and it is due first day back, tomorrow, after being so vague on this in lectures. I think I know where it is going after a lot of work if I have time I will have another look tomorrow. Thanks again
     
  12. Apr 26, 2015 #11

    BvU

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    Pity. The point I was trying to make that for the overlap it's the same route and no more difficult than the normalization procedure. Not much work either. No integrals to be worked out, since you already know ##\int \psi_1^* \psi_1= 1, \ \ \int \psi_2^* \psi_2= 1 , \ \ \int \psi_1^* \psi_2= \int \psi_2^* \psi_1= 0##. So you only have to deal with the coefficients.
     
  13. Apr 26, 2015 #12
    I know I'm close thanks to you. I think it is of the form |c1|2+|c2|2

    Just need to work out the c's.

    Thanks again
     
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