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Unformatted text preview: PHYS 3316 Solutions: Problem Set 5
1 Angular Momentum Operator
Consider a wavefunction ψ = ψ (θ) for θ ∈ (−π, π ], such that ψ (π ) = ψ (−π ) , and consider an operator ˆ L = −i ∂θ . (1) (2) (Henceforth, we will often use the notation used in lectures to denote the inner product O≡ dqψ (q )∗ Oψ (q ) ≡ (ψ, Oψ ) , (3) where q is a generalized coordinate and O some operator.) ˆ ˆ To show that L is real, we need only show that L ˆ L
∗ ∗ ˆ = L . That is ˆ = (ψ, Lψ )∗
−π dθ ∂θ ψ ∗ ψ ψ
2 π −π π (complex conjugate) (by parts) (by eq. (1)) (4) =i − dθψ ∗ ∂θ ψ
−π ˆ = (ψ, Lψ ) ˆ =L ˆ ˆ so the expectation value of L is real. We have shown here that the expectation value of L is real for all ˆ wavefunctions ψ . This is suﬃcient for L to be a Hermitian operator. (Somewhat extracurricular comment: ) To be more precise, for an operator O we deﬁne the Hermitian conjugate O† (often called the adjoint) of O as the operator such that (O† ψ, ψ ) ≡ (ψ, Oψ ) . (5) The operator O is a Hermitian operator (or a ‘self-adjoint’ operator) if O† = O. What we showed in eq. (4) was that (ψ, Lψ ) = (Lψ, ψ ), ∀ψ . It follows immediately from eq. (5) that Lψ = L† ψ , ∀ψ =⇒ L = L† . (6) So L is a Hermitian operator. Note that while L = L† , it is not true that L = L∗ . The diﬀerence between these two statements is exactly the same as for a matrix M : M † = M is not the same as M ∗ = M , where here the † is understood explicitly as a conjugate transpose. 1 2 Time Dependence of Expectation Values
(a) First, note that O is explicitly spatially independent, so d O /dt = ∂t O . Using our inner product notation d O = ∂t (ψ, Oψ ) dt = ∂t ψ, Oψ + ψ , O∂t ψ = i i∂t ψ, Oψ − i ψ , Oi∂t ψ i = H ψ, Oψ − ψ , OHψ i i ψ , H Oψ − ψ , OHψ H, O . (product rule; O is time independent) (TDSE i ∂t ψ = Hψ ) (H is Hermitian, so (Hψ, φ) = (ψ, Hφ)) (7) = = (b) For d x /dt, we consider the commutator
2 H, x ˆ = =− =− dx ψ ∗ −
2 2m 2 ∂x + U (x), x ψ ˆ 2m
2 2 2 dx ψ ∗ ∂x (xψ ) − ψ ∗ x∂x ψ (U (x) commutes with x) ˆ dx ψ ∗ 2∂x ψ 2m i =− dx ψ ∗ pψ ˆ m dx ˆ = dt p ˆ m . (8) Hence (9) 3 Time Dependent Solutions in the Square Well
(a) Consider the wavefunction deﬁned on x ∈ [0, L] by 1 πx ψ (x, t) = √ sin exp L L ≡ ψ1 (x, t) − ψ2 (x, t) , −i π2 t − sin 2mL2 2πx L exp −i 2 π2 t mL2 (10) and zero outside this domain. To check explictly that this is a particular solution of the time-dependent Schr¨dinger equation for the square well of length L, we have o i ∂t ψ1 = π ψ1 2mL2
2 22 Hψ1 = − ∂ 2 ψ1 2m x 22 π ψ1 . = 2mL2 (U (x) = 0 for x ∈ [0, L]) (11) Hence ψ1 satisﬁes the TDSE on x ∈ [0, L]. For ψ1 to be a solution of the TDSE for the square well potential, we must also check that ψ1 satsiﬁes the boundary conditions of the square well potential. That is ψ1 (0) = ψ1 (L) = 0. This is clearly satisﬁed by ψ1 . Hence ψ1 is a solution. By an identical calculation, one may show that ψ2 is a solution, and then by the linearity of the TDSE operators (i.e. ∂t and H ), it follows that ψ (x, t) = ψ1 (x, t) − ψ2 (x, t) , 2 (12) ...
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