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If the state |psi> is normalised, what is the value of |c|?

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data

    The observable A is represented by operator A-hat. The eigenvectors of A-hat are |phi1> and |phi2> which are orthonormal. The corresponding eigenvalues are +1 and -1 respectively. The system is prepared in a state |psi> given by

    |psi>=(4/10)|phi1>+c|phi2>

    a) If the state |psi> is normalised what is the value of |c|?
    b) Calculate the numerical value of the expectation value of A-hat in the state |psi>.
    c) With the system in the state |psi> a measurement of observable A is made. What is the probability of obtaining the eigenvalue -1?
    d) Immediately after the measurement is made, with the result -1, what is now the expectation value of A-hat? Explain your answer briefly.

    3. The attempt at a solution

    a) <psi| psi>=1

    integral from -infinity to infinity of (16/100)+(c^2)+(8/10)c=1

    [((16/100)+(c^2)+(8/10)c)x]=1

    (2*16/100)a+2(c^2)a+(16/10)ca=1
    (8/25)a+2(c^2)a+(16/10)ca=1
    (8/25)a+2(c^2)a+(8/5)ca=1
    a[(8/25)+2(c^2)+(8/5)c]=1
    (8/25)+(2(c^2))+(8/5)c=1/a
    2(c^2)+(8/5)c+(8/25)-(1/a)=0
    c=[(-8/5)+sqrt((8/5)^2 -4*2*((8/25)-(1/a))))]/4
    c=[(-8/5)-sqrt((8/5)^2 -4*2*((8/25)-(1/a))))]/4

    this seems way too complicated. Have I done something wrong?
     
  2. jcsd
  3. Oct 17, 2011 #2
    A is an observable, so Ahat is an Hermitian operator. This mean that its eigenfunctions are orthogonal. This means that <psi|psi> = 16/100 <phi1|phi1> + c^2 <phi2|phi2>. Now assume that |phi1> and |phi2> are also normalized, then we get 1 = 16/100 + c^2, so c=sqrt(84/100)
     
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