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Energy Eigenstates

  1. Nov 13, 2005 #1
    I have a particle of mass m in a box of length L. The energy eigenstates of this particle have wave functions
    [tex]\phi_{n}(x)=\sqrt{2/L}sin(n \pi x/L)[/tex]
    and energies
    [tex]E_n = n^{2}\pi^{2}\hbar^{2}/2mL^{2}[/tex]
    where n=1, 2, 3,... At time t=0, the particle is in a state described as follows.
    [tex]\Psi(t=0)=\frac{1}{\sqrt{14}}[\phi_1 + 2\phi_2 + 3\phi_3][/tex]
    To find the energy for state [tex]\Psi[/tex] I did the following.
    [tex]\sum_{1, 2, 3} E_n = (1^2 + 2^2 +3^2) \frac{\pi^2\hbar^2}{2mL^2}=14\frac{\pi^2\hbar^2}{2mL^2}= 14E_1 [/tex]
    where [tex]E_1=\frac{\pi^2\hbar^2}{2mL^2}[/tex]
    I have made a mistake somewhere because the actual answer is [tex]9 E_1[/tex]. Does anyone know where my error is?
    Last edited: Nov 13, 2005
  2. jcsd
  3. Nov 13, 2005 #2
    Well just apply the Hamiltonian onto the state psi. this state is the sum of three phi-terms. So each phi-state yields :
    [tex]H \phi_1 = E_1 \phi_1[/tex]
    [tex]2H \phi_2 = 2E_2 \phi_2[/tex]
    [tex]3H \phi_3 = 3E_3 \phi_3[/tex]

    Just add up everything and you get :

    [tex]H \Psi =\frac {1}{\sqrt14} (E_1 \phi_1 + 2E_2 \phi_2 + 3E_3 \phi_3)[/tex]

    The clue is to write down each energy term as a function of [tex]E_1[/tex]. You have a formula given to do that. Keep in mind that the coefficients of the psi-wave-function denote the possible energy values for the system

    Good Luck

    Last edited: Nov 13, 2005
  4. Nov 13, 2005 #3
    Ps, as an additional question : can you give me the probability that the psi-system has energy value 9E_1 ?

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