Solutions to Tutorial Sheet 1: Mainly revision 1. Given the expansion ...

The state |Ψ, t〉 is said to be normalised if 〈Ψ, t|Ψ, t〉 = 1. Show that this implies that∑|c i (t)| 2 = 1iHint: use the expansion |Ψ, t〉 = ∑ ic i (t)|u i 〉 and the corresponding conjugate expansion〈Ψ, t| = ∑ jc ∗ j(t)〈u j |.Substituting for 〈Ψ, t| and |Ψ, t〉 in 〈Ψ, t|Ψ, t〉 we find〈Ψ, t|Ψ, t〉 = ∑ ∑c ∗ j(t)c i (t)〈u j |u i 〉 = ∑ ∑c ∗ j(t)c i (t)δ ji = ∑j ij ii|c i (t)| 2where we have used the orthonormality of the eigenbasis 〈u j |u i 〉 = δ ji and the siftingproperty of the Kronecker delta. Thus we have the result quoted in Lecture 1:∑|c i (t)| 2 = 1iIf the expectation value 〈Â〉 t = 〈Ψ, t|Â|Ψ, t〉, show by making use of the same expansionsthat〈Â〉 t = ∑ |〈u i |Ψ, t〉| 2 A iiand give the physical interpretation of this result.The suggested expansion of the state vector is, in Dirac notation,|Ψ, t〉 = ∑ ic i (t)|u i 〉 where c i (t) = 〈u i |Ψ, t〉and, correspondingly,〈Ψ, t| = ∑ jc ∗ j(t)〈u j |Substituting for |Ψ, t〉 and 〈Ψ, t| in the expression for the expectation value gives〈Â〉 t = ∑ j∑ic ∗ j(t)c i (t)〈u j |Â|u i〉 = ∑ jwhere we have used the eigenvalue equation for Â:Â|u i 〉 = A i |u i 〉∑ic ∗ j(t)c i (t)A i 〈u j |u i 〉We again use the orthonormality of the eigenbasis 〈u j |u i 〉 = δ ji to write〈Â〉 t = ∑ j∑ic ∗ j(t)c i (t)A i δ ji = ∑ i|c i (t)| 2 A i = ∑ i|〈u i |Ψ, t〉| 2 A iwhich is the desired result. As discussed in lectures, the interpretation is that |c i (t)| 2is the probability of getting the result A i in a measurement of the observable A, andthe mean value of a set of repeated measurements of A is just a sum over the possiblevalues weighted by the probabilities of obtaining them.2