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Normalizing a wave function.

  1. Mar 28, 2012 #1
    1. The problem statement, all variables and given/known data

    Okay, so i have a wave function from a particle in an infinite square well that has an initiate wave function with an even mixture of the first two stationary states.

    ψ(x,0) = A[ψ1(x) + ψ2(x)]

    a. Normalize ψ(x,0)

    b Find ψ(x,t) and |ψ(x,t)|2 (use Euler's formula eitheta = cos(theta) + isin(theta) to elimiate all exponentials from the equation.) let ω= pi2hbar/2ma2 and we know en = hbarn2ω

    (there are K parts to the actually problem, but after these two i should be able to solve the rest myself)

    2. Relevant equations

    Eulers equation as stated in the question, as well as the solved first two stationary states of a particle in an infite square well (SQRT(2/L)*Sin(npi/L*X) which can be substituted in for ψn.

    3. The attempt at a solution

    Okay so i know to normalize an equation i need to set it equal to one, square it and integrate from negative to positive infinity. However, i can also integrate from 0 to L in this case.

    So i square it (use X for the first state, Y for the second state)

    1 = A2 ∫2/L(X2+xy+y2) from 0 to L

    So i dnt know how to do that integral, but Wolfram alpha does and spits out "2" which im kind of dubious on, anyone know if im going in the right direction?
     
  2. jcsd
  3. Mar 28, 2012 #2

    fzero

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    Yes, the integral gives "2." One way to see this is just to use the orthonormality of the wavefunctions. However, you should figure out how to do the integrals by hand, since you're just going to be faced with more difficult ones later on in your course. In this case, you can use the trig identities for half-angles to simplify the expressions.
     
  4. Mar 28, 2012 #3
    (wrote this before i saw the above response)
    Maybe i cant evaluate it from 0 to L?

    Should I be evaluating it from negative infinity to positive infinity (i could then double the integral, and evaluate from 0 to infinity because Sin^2 is a even function)

    The bad part is we went over this part in class and there was a trick to it but i cant find my notes :/

    EDIT:

    Thank you, i have been taught how to evaluate sin^2 as an integral but i had forgotten and needed a refresher. now that i know to trust the answer though i can chug my way through it.
     
  5. Mar 28, 2012 #4
    Okay, so now knowing that A=1/4 i return to my equations...

    ψ(x,0) = A[ψ1(x) + ψ2(x)]

    and im looking to find Find ψ(x,t).

    In order to do this i need to know change my equations for ψ so that they are related to t, time. I am given ω (angular frequency) which i know is related to t. The frequency of the equation (2/L) ½*Sin(n∏/L*X) varies clearly with x and the quantum number n of the wave function it describes. In order to change it to vary with time, i would need to equate X with T...

    Or perhaps im just overthinking it and need to just multiply it be En, since the wave function ψ(x,0) = A[ψ1(x) + ψ2(x)] just describes the dependence with respect to x, and 0. So in order to describe its relation with respect to x,t i would need to just multiply by hbarn2ω, which has a time dependence.
     
  6. Mar 28, 2012 #5
    You can see im quite new (and perhaps not the best) at this? so i am reading and figuring this out as i go and i found that


    ψ(x,t) = ψ(x)*e-iEt/hbar

    Since i was given En, i can substitute that in for E in the above equation.

    yielding:


    ψ(x,0) = 1/4[ψ1(x)e-iE1t/hbar + ψ2(x)e-iE2t/hbar]

    Is this the equation i should be simplifying?
     
  7. Mar 28, 2012 #6

    fzero

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    In general, wavefunctions must be defined on all of space, so your integrals would run from negative infinity to positive infinity. However, in this case, the potential is infinite outside the interval from 0 to L, so the wavefunction is only nonzero from 0 to L. So it's OK to just integrate over this region.

    You should double-check that. Your formula for [itex]\psi_n[/itex] looked like it was normalized already, so you should find [itex]A=1/\sqrt{2}[/itex].

    Yes, but I have a feeling that the grader might be looking for you to give the expressions for [itex]E_1[/itex] and [itex]E_2[/itex], so you might want to include them. Also recheck the normalization coefficient.
     
  8. Mar 28, 2012 #7
    I rechecked the normalization coeffecient, and substituted in my expressions for E1
    and E2, (Hbarω and 4hbarω respectively) t get an expression for ψ(x,t).

    Now the directions said to use eulers formula to simplify and elminate exponentials... soo....

    ψ(x,t) = SQRT(1/2)*[ψ1(x)e-iE1t/hbar + ψ2(x)e-iE2t/hbar]

    Where ψ(x) = (SQRT(2/L)*Sin(npi/L*X)

    In order to simplify my simplifications and spare you a giant block eyesore, i will simplify some things first.

    i will merge the SQRT(1/2) that is distributed to both functions of ψ, thereby canceling the SQRT of 2 in the SQRT(2/L) funtion preceeding the trignometric function.

    I am left with ψ(x,t) = [SQRT(1/L)*Sin(npi/L*X)e-iE1t/hbar + SQRT(1/L)*Sin(npi/L*X)e-iE2t/hbar]

    Making the relevant substitutions for E, and then setting the common variables to a notation V in the exponent of e, as well as letting everything inside the sin arguement equal U....

    ψ(x,t) = SQRT(1/L)*[Sin(u)e-iv + Sin(2u)e-i4v]

    much more concise.

    applying Eulers formula (e-iv=cos(v)-isin(v))

    (dropping the SQRT(1/L) for now)

    = Sin(u)(cos(v)-isin(v)) + Sin(2u)(cos(4v)-isin(4v))

    = Sin(u)cos(v)-isin(u)sin(v) + sin(2u)cos(4v)-isin(2u)sin(4v)

    using the trig identity sin(2u)=2sin(u)cos(u)

    = Sin(u)cos(v)-isin(u)sin(v) + 2sin(u)cos(u)cos(4v)-i2sin(u)cos(u)sin(4v)
    Factoring
    = sin(u)(cos(v)-isin(v)+2cos(u)cos(4v)-i2cos(u)sin(4v))

    I could apply the same identity to the cos(4v) but i feel as though it would clutter my expression more.

    Im afriad moving into |ψ(x,t)|2 would be extremely cluttered and if im being honest i don't know exactly how to tackle it.
     
  9. Mar 29, 2012 #8

    fzero

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    You could proceed like that, but as you say it's cluttered. Luckily there are some tricks that can be used to simplify the expression along the way so that the end result isn't as bad looking.

    One trick that will help is to note that

    [tex]|\psi(x,t)|^2 = \frac{1}{L} \left| \sin u ~ e^{-iv} + \sin (2u) ~ e^{-4iv} \right|^2
    = \frac{1}{L} \left|e^{-iv} \right|^2 \left| \sin u + \sin (2u) ~ e^{-3iv} \right|^2, [/tex]

    then use [itex]|e^{-iv} |=1[/itex]. After using Euler's formula, we have

    [tex]|\psi(x,t)|^2
    = \frac{1}{L} \left| \left( \sin u + \sin (2u) \cos (3v) \right) - i \sin (2u) \sin (3v) \right|^2.[/tex]

    Here we've grouped the real and imaginary parts together, which makes it easier to compute using

    [tex] | x + i y |^2 = x^2 + y^2.[/tex]

    You can actually clean it up a lot by using the double angle formula that you figured out already. You'll have a few terms to deal with in the integral, but it shouldn't be too bad.
     
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