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

1. Oct 17, 2011

### blueyellow

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. Oct 17, 2011

### susskind_leon

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)