Quantum Mechanics (Fundamental Concepts) Solutions

1. FUNDAMENTAL CONCEPTS 25 1 Fundamental Concepts 1.1 Consider a ket space spanned by the eigenkets \(\{|\sigma^{\prime}\rangle\}\) of a Her- mitian operator A. There is no degeneracy. (a) Prove that \[\prod_{a^{\prime}}(A-a^{\prime})\] is a null operator. (b) What is the significance of \[\prod_{a^{2^{2}\ne a^{4}}\frac{(A-a^{1^{}})}{a^{\prime}-a^{y}};\] (c) Illustrate (a) and (b) using A set equal to \(S_{z}\) of a spin \(\frac{1}{7}\) system. (a) Assume that \(|\alpha\rangle\) is an arbitrary state ket. Then \[\prod_{a^{\prime}}(A-a^{\prime})|\alpha\rangle=\prod_{a^{\prime}}(A-a^{\prime})\sum_{a^{\prime\prime}}|a^{\prime\prime}\rangle\langle{a^{\prime\prime}|\alpha\rangle}{\epsilon_{a}\sigma^{\prime\prime}}=\sum_{a^{\prime\prime}}c_{a^{\prime\prime}}\prod_{a^{\prime}}(A-a^{\prime})|a^{\prime\prime}\rangle\] \[=\sum_{a^{\prime}}c_{a^{\mu}}\prod_{a^{\prime}}(a^{\mu}-a^{\prime})|a^{il}\rangle^{a^{\mu}\underline{E}^{\prime}\}}0.\] (1.1) (b) Again for an arbitrary state \(|\alpha\rangle\) we will have \[[\prod_{a^{\prime\prime}\ne a^{\prime}}\frac{(A-a^{\prime\prime})}{a^{\prime}-a^{\prime\prime}}]|\alpha\rangle=[\prod_{a^{\prime\prime}\ne a^{\prime}}\frac{(A-a^{\prime\prime})}{a^{\prime}-a^{\prime\prime}}]\sum_{a^{\prime\alpha}}|a^{\prime\prime}\rangle\langle a^{\prime\prime}|\alpha\rangle\] \[=\sum_{a^{\prime\prime}}\langle a^{\prime\prime\prime}|\alpha\rangle\prod_{a^{s}\ne a^{\prime}}\frac{(a^{\prime\prime}-a^{\prime\prime})}{a^{\prime}-a^{\prime\prime}}|a^{\prime\prime}\rangle=\] \[=\sum_{a^{m}}\langle a^{|m}|\alpha\rangle\delta_{a^{\prime\prime}a_{a^{\prime}}}|a^{\prime\prime}\rangle=\langle a^{\prime}|\alpha\rangle|a^{\prime}\rangle\Rightarrow\] \[[\prod_{a^{n\ne a^{\prime}}\frac{(A-a^{n})}{a^{\prime}-a^{n}}]=|a^{\prime}\rangle\langle a^{\prime}|\equiv\Lambda_{a^{\prime}}.\] (1.2) So it projects to the eigenket \(|a^{\prime}\rangle\) 26 (c) It is Sħ/2(1+)(+/--(-1). This operator has eigenkets |+) and |-) with eigenvalues h/2 and -1/2 respectively. So ΠΣ) = (-1) [(-1+1+1+1)-(-)--++1) = [(--++++---+++) = [---][+<+1]=---++1=0, (1.3) where we have used that +++-)(-1=1. Forah/2 we have Π (Sa) = Π 1 (S1) S +41 = 7/2/2+カ/2 [(--++++---++ = 1 = ++1 = 1+{+1. (1.4) Similarly for a' ħ/2 we have Π (Sa") a-a" = Π (S-1) = S-1 /2-1/2---1/2-1/2 [(--+++----++1)] = 1 (1.5) 1.2 A spin system is known to be in an eigenstate of 5-in with eigenvalue 7/2, where å is a unit vector lying in the zz-plane that makes an angle with the positive z-axis. (a) Suppose S. is measured. What is the probability of getting 1. FUNDAMENTAL CONCEPTS 35 (b) The operator exp() gives translation in momentum space. This can be justified by calculating the following operator [p,exp( ix Xi hbar )]=[p, sum n = 0 to ∞ 1/(n!) ( ix Xi hbar )^ n ]= sum n=0 ^ infty 1 n! ( i Xi hbar )^ n [p,x^ n ] = sum n=1 ^ infty 1 n! ( i Xi hbar )^ n sum k = 1 to n x ^ (n - k) * [p, x] * x ^ (k - 1) = sum n = 1 to ∞ 1/(n!) ( i Xi hbar )^ n sum k=1 ^ n (-i hbar)x^ n-1 = sum n=1 ^ infty 1 n! ( i Xi hbar )^ n n(-i hbar)x^ n-1 = sum n = 1 to ∞ 1/((n - 1)!) ( i Xi hbar )^ n-1 x^ n-1 (-i hbar)( i Xi hbar )= Xi sum n=0 ^ infty 1 n! ( ix Xi hbar )^ n = Xi*exp ( ix Xi hbar ) (1.47) So when this commutator acts on an eigenstate | p' rangle of the momentum oper-ator we will have [p.exp()]) = exp())]-[exp(4 dxad () Xi*exp ( ix Xi hbar )=p[exp( ix Xi hbar )|p^ prime rangle]-p^ prime [exp( ix Xi hbar )]|p^ prime rangle Rightarrow p[exp( ix Xi hbar )|p^ prime rangle]= (p' + Xi) [exp( ix Xi hbar )|p^ prime rangle] . dxa (1.48) Thus we have that exp()) = Alp/+3), (1.49) where A is a constant which due to normalization can be taken to be 1.