The Harmonic Oscillator, the Ladder Operator Solutions

University of Connecticut OpenCommons@UConn

Chemistry Education Materials Department of Chemistry

8-1-2006 The aH rmonic Oscillator, The Ladder Operator Solutions Carl W. David University of Connecticut, [email protected]

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Recommended Citation David, Carl W., "The aH rmonic Oscillator, The Ladder Operator Solutions" (2006). Chemistry Education Materials. 19. https://opencommons.uconn.edu/chem_educ/19 The Harmonic Oscillator, The Ladder Operator Solutions

C. W. David Department of Chemistry University of Connecticut Storrs, Connecticut 06269-3060 (Dated: August 1, 2006)

I. SYNOPSIS which means, of course,

The solution to the quantum mechanical harmonic os- [p, x] = px − xp = −ı¯h cillator using ladder operators is a classic, whose ideas permeate other problem’s treatments. For the Harmonic Oscillator, we form the two operators II. HAMILTONIANS AND COMMUTATORS

The Hamiltonian for the Harmonic Oscillator is a+ = p + ıµωx p2 k + x2 2µ 2 and where p is the momentum operator and x is the position − operator. We know that the Schrödinger prescription is a = p − ıµωx ∂ p → −ı¯h which differ solely by that intervening sign (Remember ∂x q k that ω = µ ). The entire derivation now hinges on while x → x, as usual. This means, as we well know, the properties of these two operators. We start with the that the commutator of x and p is non-zero. elementary question, what is the commutator of a+ and − ∂ ∂ a ? [p, x]f = (px − xp)f = −ı¯h xf − x(−ı¯h f) ∂x ∂x [a+, a−] = a+a− − a−a+ which leads to ∂x ∂ ∂ [p, x]f = (px−xp)f = −fı¯h −−x(ı¯h )fx(−ı¯h f) by definition; these terms expand to become ∂x ∂x ∂x

[a+, a−] = (p + ıµωx)(p − ıµωx) − (p − ıµωx)(p + ıµωx) which is

[a+, a−] = p2 − pıµωx + ıµωxp + µ2ω2x2 − (p2 + pıµωx − ıµωxp + µ2ω2x2)

[a+, a−] = −ıµω(px − xp + px − xp) = −2ıµω(px − xp) = −2ıµω(−ı¯h) = −2µω¯h

Now the product a+a− + a−a+ is i.e.,

2 2 2 2 2(p + µ ω x ) a+a− + a−a+ p2 + µ2 k x2 = µ = H 4µ 2µ so a+a− + a−a+ p2 + µ2ω2x2 We next need the commutator of these “a” operators = and the energy operator. We have 4 2

Typeset by REVTEX 2

a+a− + a−a+ a+a− + a−a+ [a+,H] = a+H − Ha+ = a+ − a+ 4µ 4µ

which is Now operate from the left with a+ and let’s see what happens. We have 1 [a+,H] = a+(a+a− + a−a+) − (a+a− + a−a+)a+ 4µ a+(Hψ) = (a+H)ψ = Ea+ψ

1 (operators ignore constants like E) which is, employing [a+,H] = a+a+a− + a+a−a+ − a+a−a+ − a−a+a+ + + + 4µ the commutator a H − Ha = −¯hωa (Equation 2.1) we have

1 (−¯hωa+ + Ha+)ψ = E(a+ψ) [a+,H] = a+(a+a− − a−a+) + (a+a− − a−a+)a+ 4µ

(−¯hωa+)ψ + (Ha+)ψ = E(a+ψ) −1 [a+,H] = a+(2µ¯hω) + (2µ¯hω)a+ (2.1) 4µ H(a+ψ) = E(a+ψ) + ¯hω(a+ψ) [a+,H] = −¯hωa+ + + This is a crucial result, since it contains the seed of H(a ψ) = (E +hω ¯ )(a ψ) what a ladder operator does. What does this last statement say? It says that a+ψ is also an eigenfunction of H if ψ was. The only thing is III. COMMUTING THE LADDER OPERATOR that the energy is higher than the energy associated with WITH THE HAMILTONIAN ψ (byhω ¯ ). What we have shown is that a+ is an operator, a lad- Consider if we had in hand an eigenfunction ψ of H. der operator, which takes an existing eigenfunction, and That would mean that transforms it into the next highest eigenfunction, sepa- rated in energy by ¯hω. Wonderful. Hψ = Eψ Now we need to see what happens with a−. We have

a−a+ + a+a− a−a+ + a+a− [a−,H] = a−H + Ha− = a− − a− 4µ 4µ

which is [a−,H] = ¯hωa− 1 [a−,H] = a−(a−a+ + a+a−) − (a−a+ + a+a−)a− Again, given 4µ Hψ = Eψ 1 [a−,H] = a−a−a+ + a−a+a− − a−a+a− − a+a−a− Now operate from the left with a− and let’s see what 4µ happens. We have

1 (a−)Hψ = Ea−ψ [a−,H] = a−(a−a+ − a+a−) − (a−a+ − a+a−)a− 4µ (operators ignore constants like E) which is, employing a−H − Ha− = +¯hωa−, we have − 1 − − [a ,H] = a (2µ¯hω) + (2µ¯hω)a − − − 4µ (+¯hωa + Ha )ψ = E(a ψ) 3

(+¯hωa−)ψ + (Ha−)ψ = E(a−ψ) or √ kµ 2 0 − x C xe h¯ 2 → ψ H(a−ψ) = E(a−ψ) − ¯hω(a−ψ) 1 Continuing, we have − − √ √ H(a ψ) = (E − ¯hω)(a ψ) kµ 2 kµ 2 + 0 − x ∂ 0 − x a C xe h¯ 2 → −ı¯h + ıµωx C xe h¯ 2 What does this last statement say? It says that a−ψ is ∂x also an eigenfunction of H if ψ was. The only thing is that which is, carrying out the diﬀerentiation the energy is lower than the energy that was associated √ with ψ byhω ¯ . kµ x2 √ 0 − kµ 2 − ∂C xe h¯ 2 0 − x What we have shown is that a is an operator, a ladder = −ı¯h + ıµωxC xe h¯ 2 operator, which ladders us down! ∂x

√ √ 2 2 kµ 2 0 kµ 0 x 0 x − x IV. FINDING THE LOWEST EIGENVALUE = −ı¯hC + 2ı¯h C + ıµωC e h¯ 2 AND ITS ASSOCIATED EIGENVECTOR ¯h 2 2

√ The last part of this wonderful derivation follows. 2 kµ 2 0 p 2 p x − x What happens when we continuously “down ladder” from ıC −¯h + kµx + kµ e h¯ 2 2 any eigenfunction? Clearly, at some point, we must run out, since the energy can not be lowered indeﬁnitely. Said √ kµ 2 another way, there must be a lowest eigenfunction, call 0 p 2 − x ıC −¯h + kµx e h¯ 2 it ψ0, which is destroyed if we ladder down from it.

a−ψ → 0 0 √ √ kµ 2 0 kµ 2 − x ıC ¯h x − 1 e h¯ 2 → ψ Then, we have ¯h 2

(p − ıµωx)ψ0 = 0 You will recognize, from this result, that life would have been simpler had we used a dimensionless coordinate for which means the distance, rather than x. We keep bringing square 0 ∂ψ roots, and ¯h s, which could have been eliminated had we −ı¯h 0 = ıµωxψ ∂x 0 used a dimensionless “x”. which is immediately integrable. We obtain √ V. MATRIX ELEMENTS USING LADDER k √ 2 µ 2 kµ 2 − µω x − µ x − x OPERATORS ψ0 = Ce h¯ 2 = Ce h¯ 2 = Ce h¯ 2

If this is the ground state, then laddering up with the Be that as it may, we see that we can now evaluate up-ladder operator, a+ should give the ﬁrst excited state, common matrix elements in a simple manner now. Con- sider < n|x|m > which is a transition matrix element √ √ kµ 2 kµ 2 + − x − x associated with selection rules. Since a Ce h¯ 2 = (p + ıµωx)Ce h¯ 2 a+ − a− = p + ıµωx − (p − ıµωx) which is √ We have kµ x2 √ − kµ 2 ∂Ce h¯ 2 − x −ı¯h + ıµωxCe h¯ 2 a+ − a− ∂x = x 2ıµω i.e., so √ √ √ kµ 2 kµ 2 kµ − x − x + − (−ı¯h C − 2x e h¯ 2 + ıµωxCe h¯ 2 a − a ¯h < n|x|m >=< n| |m > 2ıµω which equals and we know what the eﬀect of the ladder operators are √ s ! √ on the bra’s and ket’s involved. once we’ve expanded kµ 2 2 kµ k − x ı¯h 2x + ıµ x Ce h¯ 2 this maximally, and used the ladder operator properties ¯h µ all one need do is preserve the < i||i > terms, since the 4

< i||j > terms vanish (orthogonality) when i is diﬀerent Similar schemes can be used for powers of x or the mo- from j! mentum, or mixed expressions. But the central idea is Consider < n|px|m > which is a transition matrix ele- that when these ladder operators operate on the right ment associated with selection rules. Since hand ket, they generate, aside from constants, other kets, which may or may not be orthogonal to the bra’s in the a+ + a− = p + ıµωx + (p − ıµωx) expression. This leads to obvious simpliﬁcations. We have a+ + a− = p 2 x so a+ + a− < n|p |m >=< n| |m > x 2