Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wavefunction expansion coefficients

  1. May 18, 2005 #1


    User Avatar

    I'm working in Liboff, 4e, QM, page 114, problem 4.35.
    An electron in a 1-D box with walls at x= 0,a is in the state [tex]\psi(x) = A[/tex] for [tex]x\in (0,a/2)[/tex] and [tex]\psi(x) = -A[/tex] for [tex]x\in (a/2,a)[/tex]. What is the lowest possible energy that can be measured?

    From my understanding, the answer to this question will be the integer of first nonzero coefficient in the expansion [tex]\psi = \Sigma \limits_{n=1}^{\infty} a_n \phi_n[/tex], where [tex]\phi_n = \sqrt{\frac{2}{a}}\sin(\frac{n\pi x}{a})[/tex] are the basis functions given in eq (4.15) from the book (the eigenstates for the 1D box Hamiltonian). I do this and I get [tex]a_n = \frac{\sqrt{2}}{n\pi}(1+\cos(n\pi) - 2\cos(n\pi/2))[/tex]. Now correct me if I'm wrong, but is it not true that [tex]\psi(x)=A[/tex] for [tex]x\in (0,a)[/tex] represents the same state since only the square of the wavefunction is given significance? In that case, however, I get [tex]a_n=\frac{\sqrt{2}}{n\pi}(1-\cos(n\pi))[/tex]. It is my understanding that [tex]a_n^2[/tex] represents the probability of measuring the particle to be in the state [tex]\phi_n[/tex]. But in these two cases, we will get different [tex]a_n^2[/tex] indicating that the two states are physically different.

    Can anyone point out my mistake?
  2. jcsd
  3. May 18, 2005 #2


    User Avatar
    Science Advisor
    Homework Helper

    Advice:read more on Fourier series.U could see post #5 from here https://www.physicsforums.com/showthread.php?p=571801#post571801 which deals exactly with a function of type [itex] \psi [/itex] in your problem.

    Your problem should reduce to the question:does the fundamental state [itex] \phi_{1} [/itex]-which has the lowest possible energy-have a nonzero probability ...?

    U'll need to compute the Fourier coefficients of that wavefunction (whose sq.moduli give probabilities),so that's why i advised you to read.

  4. May 19, 2005 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No, that is not true. It is not because |psi1|^2 = |psi2|^2 that psi1 and psi2 represent the same physical states. In order for two descriptions A and B to represent the same physical state, you have to have that ALL POSSIBLE measurements cannot make any difference. |psi1| = |psi2| only means that all possible POSITION measurements cannot make a difference. But a momentum measurement can, for instance.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook