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Quantum Harmonic Oscillator with Ladder Operators Kenneth Taliaferro January 29, 2010 In Quantum Mecahnics, the state of a particle ∫is given by a complex valued function Ψ of postion x and time t such that |Ψ(x, t)|2 dx = 1. I’m looking for a solution Ψ to the Schrödinger Equation ih̄ ∂Ψ p2 Ψ = +VΨ ∂t 2m where h̄, k, and m are constants and p = −ih̄ ∂ ∂x 1 V = kx2 2 (momentum operator) (potential energy operator) After separation of variables (Ψ(x, t) = ψ(x)ϕ(t)), this boils down to the eigenvalue problem of ﬁnding twice diﬀerentiable functions ψ in L2 (R) such that Hψ = Eψ where 1 2 mω 2 2 p + x 2m 2 √ k ω= m H= (right side of Schrödinger equation) 1 x is self adjoint because, for f, g ∈ D(x) ⊂ L2 (R), ∫ ⟨xf, g⟩ = xf (x)g(x)dx ∫ = f (x)xg(x)dx = ⟨f, xg⟩. By integration by parts, p is self-adjoint. Moreover, by the product rule, [x, p] := xp − px = ih̄. Deﬁne the ladder operators √ ) ( mω i a+ = x− p (raising operator) mω 2h̄ √ ( ) mω i a− = x+ p (lowering operator) mω 2h̄ Then a+ and a− are adjoints of each other. Using the fact that [x, p] = ih̄, it is straightforward to show that [a− , a+ ] = 1 h̄ω H= (2a+ a− + 1) 2 [H, a+ ] = h̄ωa+ (1) (2) Suppose that ψ is an eigenfunction of H with eigenvalue E. Then Ha+ ψ = (Ha+ − a+ H + a+ H)ψ = ([H, a+ ] + a+ H)ψ = (h̄ωa+ + a+ E)ψ = (h̄ω + E)a+ ψ i.e. a+ ψ (if nonzero) is an eigenfunction with eigenvalue E + h̄ω. Similarly, a− ψ (if nonzero) is an eigenfunction with eigenvalue E − h̄ω. This is why a+ and a− are called the raising and lowering operators. Lemma 1: The only possible eigenvalues of H are h̄ω(n+ 12 ), n = 0, 1, 2, 3, ... 2 Proof: Use the raising and lowering operators. See the last page of this document or [3], page 594. Let E0 be the smallest eigenvalue of H. Let ψ0 be a corresponding normalized eigenfunction. Then a− ψ0 = 0 (3) because otherwise a− ψ0 would be an eigenfunction with eigenvalue E − h̄ω < E0 . Thus, by (2), 0 = h̄ωa+ a− ψ0 ) ( h̄ω ψ0 = H− 2 ( ) h̄ω = E0 − ψ0 2 so E0 = h̄ω . An eigenfunction with eigenvalue En = (n + 12 )h̄ω is given by 2 n n a+ ψ0 (if a+ ψ0 ̸= 0). Let ψn be a corresponding normalized eigenfunction with eigenvalue En . Then, for some constant Cn+1 , ψn+1 = Cn+1 a+ ψn . But, 1 = ⟨ψn+1 , ψn+1 ⟩ = |Cn+1 |2 ⟨a+ ψn , a+ ψn ⟩ = |Cn+1 |2 ⟨ψn , a− a+ ψn ⟩ ) ( 1 H 2 = |Cn+1 | ⟨ψn , + ψn ⟩ h̄ω 2 ) ( (n + 12 )h̄ω 1 2 = |Cn+1 | ⟨ψn , + ψn ⟩ 2 h̄ω so Cn+1 = (n + 1)−1/2 . It follows that a+ ψn = √ Similarly, a− ψn = n + 1ψn+1 . √ 3 nψn−1 . (4) (5) Therefore 1 ψn = √ an+ ψ0 . (6) n! This is suﬃcient for computing expectation values of powers of x and p of the eigenfunctions because √ h̄ x= (a+ + a− ) √2mω h̄mω p=i (a+ − a− ) 2 For example, since eigenfunctions of a self-adjoint operator corresponding to diﬀerent eigenvalues are orthogonal and [a− , a+ ] = 1, the expectation value of x2 on the state ψn is given by ( ) h̄ 2 ⟨ψn , x ψn ⟩ = ⟨ψn , (a+ + a− )(a+ + a− )ψn ⟩ 2mω h̄ = ⟨ψn , (a2+ + a2− + a+ a− + a− a+ )ψn ⟩ 2mω h̄ = ⟨ψn , (a2+ + a2− + a+ a− + (a+ a− − a+ a− ) + a− a+ )ψn ⟩ 2mω h̄ = ⟨ψn , (a2+ + a2− + 2a+ a− + 1)ψn ⟩ 2mω h̄ ⟨ψn , (2n + 1)ψn ⟩ = 2mω h̄ = (2n + 1) 2mω √ ψ0 can be found explicitly by solving the separable ODE (3). Let α = mω/h̄. In terms of the variable y = αx, (3) says ) ( d y+ ψ0 (y) = 0 dy For any constant N0 , a solution is ψ0 (y) = N0 e−y i.e. ψ0 (x) = N0 e−α 4 2 /2 . 2 x2 /2 . By (6), ( ψn (x) = α √ n π2 n! )1/2 e−α 2 x2 /2 Hn (αx) where Hn is the Hermite polynomial of order n. By separation of variables, Ψn (x, t) = e(−iEn /h̄)t ψn (x). Since the position distribution |Ψn |2 doesn’t depend on time, these are called stationary states. References [1] Agnolet, G. Physics 412 Lecture Notes. [2] Bransden, B.H. and C.J. Joachain. Quantum Mechanics. [3] Kreyszig, E. Introductory Functional Analysis with Applications. 5 Proof of Lemma 1: As derived earlier, [a− , a+ ] = 1 h̄ω H= (2a+ a− + 1) 2 [H, a+ ] = h̄ωa+ (7) (8) Suppose ψ is an eigenfunction of H with eigenvalue E. Let Ẽ = E 1 − . h̄ω 2 (9) Then, by (8), a+ a− ψ = Ẽψ. Applying a− to each side, a− a+ (a− ψ) = Ẽ(a− ψ). By (7), a− a+ = a+ a− + 1, so a+ a− (a− ψ) = (Ẽ − 1)a− ψ. Repeating the last two steps j − 1 times yields a+ a− (aj− ψ) = (Ẽ − j)aj− ψ. (10) Now aj− ψ = 0 for suﬃciently large j; because, otherwise, by (10), j+1 j j ⟨aj+1 − ψ, a− ψ⟩ = ⟨a+ a− (a− ψ), a− ψ⟩ = (Ẽ − j)⟨aj− ψ, aj− ψ⟩ i.e. Ẽ − j = 2 ∥aj+1 − ψ∥ ≥0 ∥aj− ψ∥2 (11) for all j. Thus ∃n ∈ N such that an− ψ ̸= 0 but an+1 − ψ = 0. By (11), Ẽ −n = 0. By (9), ( ) 1 E = h̄ω n + for some n ∈ {0, 1, 2, 3, ...}. 2 6