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Chapter 5 Harmonic Oscillator and Coherent States

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Chapter 5 Harmonic Oscillator and Coherent States Chapter 5 Harmonic Oscillator and Coherent States 5.1 Harmonic Oscillator In this chapter we will study the features of one of the most important potentials in physics, it's the harmonic oscillator potential which is included now in the Hamiltonian m!2 V (x) = x2 : (5.1) 2 There are two possible ways to solve the corresponding time independent Schr¨odinger equation, the algebraic method, which will lead us to new important concepts, and the analytic method, which is the straightforward solving of a differential equation. 5.1.1 Algebraic Method We start again by using the time independent Schr¨odingerequation, into which we insert the Hamiltonian containing the harmonic oscillator potential (5.1) 2 d2 m!2 H = − ~ + x2 = E : (5.2) 2m dx2 2 We rewrite Eq. (5.2) by defining the new operatorx ¯ := m!x " # 1 d 2 1 H = ~ + (m!x)2 = p2 +x ¯2 = E : (5.3) 2m i dx 2m We will now try to express this equation as the square of some (yet unknown) operator p2 +x ¯2 ! (¯x + ip)(¯x − ip) = p2 +x ¯2 + i(px¯ − xp¯ ) ; (5.4) but since x and p do not commute (remember Theorem 2.3), we only will succeed by taking the x − p commutator into account. Eq. (5.4) suggests to factorize our Hamiltonian by defining new operators a and ay as: 95 96 CHAPTER 5. HARMONIC OSCILLATOR AND COHERENT STATES Definition 5.1 a := p 1 (m!x + ip) annihilation operator 2m!~ ay := p 1 (m!x − ip) creation operator 2m!~ These operators each create/annihilate a quantum of energy E = ~!, a property which gives them their respective names and which we will formalize and prove later on. For now we note that position and momentum operators are expressed by a's and ay's like r r m! x = ~ a + ay p = −i ~ a − ay : (5.5) 2m! 2 Let's next calculate the commutator of the creation and annihilation operators. It's quite obvious that they commute with themselves [ a ; a ] = ay ; ay = 0 : (5.6) To find the commutator of a with ay we first calculate aay as well as aya 1 a ay = (m!x)2 + im! [ p ; x ] + p2 (5.7) 2m!~ 1 aya = (m!x)2 − im! [ p ; x ] + p2 : (5.8) 2m!~ Since we know [ p ; x ] = −i~ we easily get a ; ay = 1 : (5.9) Considering again the Hamiltonian from Eq. (5.3) we use expressions (5.7) and (5.8) to rewrite it as 1 ! H = p2 + (m!x)2 = ~ aya + a ay : (5.10) 2m 2 Using the commutator (5.9) we can further simplify the Hamiltonian a ; ay = a ay − aya = 1 ) a ay = aya + 1 ; (5.11) y 1 H = ~!(a a + 2) : (5.12) For the Schr¨odingerenergy eigenvalue equation we then get y 1 ~!(a a + 2 ) = E ; (5.13) 5.1. HARMONIC OSCILLATOR 97 which we rewrite as an eigenvalue equation for the operator aya E 1 aya = − ; (5.14) ~! 2 having the following interpretation: Definition 5.2 N := aya occupation (or particle) number operator and which satisfies the commutation relations N ; ay = ay [ N ; a ] = −a : (5.15) Next we are looking for the eigenvalues ν and eigenfunctions ν of the occupation number operator N, i.e. we are seeking the solutions of equation N ν = ν ν : (5.16) To proceed we form the scalar product with ν on both sides of Eq. (5.16), use the positive definiteness of the scalar product (Eq. (2.32)) and the definition of the adjoint operator (Definition 2.5) y ν h ν j ν i = h ν j N ν i = h ν a a ν = h a ν j a ν i ≥ 0 : (5.17) | {z } 6= 0 The above inequality vanishes only if the corresponding vector a is equal to zero. Since we have for the eigenvalues ν ≥ 0 , the lowest possible eigenstate 0 corresponds to the eigenvalue ν = 0 ν = 0 ! a 0 = 0 : (5.18) Inserting the definition of the annihilation operator (Definition 5.1) into condition (5.18), i.e. that the ground state is annihilated by the operator a, yields a differential equation for the ground state of the harmonic oscillator 1 ~ d a 0 = p (m!x + i ) 0 = 0 2m!~ i dx m! d ) x + 0 = 0 : (5.19) ~ dx We can solve this equation by separation of variables Z Z d 0 m! m! 2 = − dx x ) ln 0 = − x + ln N ; (5.20) 0 ~ 2~ where we have written the integration constant as ln N , which we will fix by the normal- ization condition 98 CHAPTER 5. HARMONIC OSCILLATOR AND COHERENT STATES 0 m! 2 m! 2 ln = − x ) 0(x) = N exp − x : (5.21) N 2~ 2~ We see that the ground state of the harmonic oscillator is a Gaussian distribution. The 1 R 2 normalization dx j 0(x)j = 1 together with formula (2.119) for Gaussian functions −∞ determines the normalization constant 1 rm! m! N 2 = )N = 4 : (5.22) π~ π~ We will now give a description of the whole set of eigenfunctions ν of the operator N based on the action of the creation operator by using the following lemma: Lemma 5.1 If ν is an eigenfunction of N with eigenvalue ν, then y a ν also is an eigenfunction of N with eigenvalue (ν + 1). Proof: y Eq: (5:15) y y y N a ν = (a N + a ) ν = a (N + 1) ν y y = a (ν + 1) ν = (ν + 1) a ν : q.e.d. (5.23) y The state a ν is not yet normalized, which means it is only proportional to ν + 1 , to find the proportionality constant we use once more the normalization condition y y y a ν a ν = h ν j a a ν i = h ν j (N + 1) j ν i = (ν + 1) h ν j ν i (5.24) |{z} | {z } aya + 1 1 p y ) a ν = ν + 1 ν + 1 : (5.25) This means that we can get any excited state ν of the harmonic oscillator by succes- sively applying creation operators. In total analogy to Lemma 5.1 we can formulate the following lemma: Lemma 5.2 If ν is an eigenfunction of N with eigenvalue ν, then a ν also is an eigenfunction of N with eigenvalue (ν − 1). 5.1. HARMONIC OSCILLATOR 99 Proof: Eq: (5:15) N a ν = (a N − a) ν = a (N − 1) ν = a (ν − 1) ν = (ν − 1) a ν : q.e.d. (5.26) We again get the proportionality constant from the normalization y h a ν j a ν i = h ν j a a ν i = h ν j N j ν i = ν h ν j ν i (5.27) |{z} N | {z } | {z } ν j ν i 1 p ) a ν = ν ν − 1 : (5.28) Summary: Eigenvalue equation for the harmonic oscillator The time independent Schr¨odingerequation, the energy eigenvalue equation H n = En n ; (5.29) 1 y with the Hamiltonian H = ~! (N + 2 ) and the occupation number operator N = a a provides the energy eigenvalues 1 En = ~! (n + 2) : (5.30) The corresponding state vectors, the energy eigenfunctions, are given by 1 1 y n 1 m! 4 y n m! 2 n(x) = p (a ) 0(x) = p (a ) exp(− x ) ; (5.31) n! n! π~ 2~ with the ground state 1 m! 4 m! 2 0(x) = exp(− x ) : (5.32) π~ 2~ The explicit form of the excited state wave functions will be calculated later on but we can for now reveal that they are proportional to a product of the ground state and a family of functions, the so-called Hermite polynomials Hn. The wave functions thus form a ladder of alternating even and odd energy states, see Fig. 5.1, which are each separated by a quantum of energy ~!, i.e. equally spaced. The creation and annihilation operators then "climb" or "descend" this energy ladder step by step, which is why they are also called ladder operators. 100 CHAPTER 5. HARMONIC OSCILLATOR AND COHERENT STATES Figure 5.1: Harmonic oscillator: The possible energy states of the harmonic oscillator potential V form a ladder of even and odd wave functions with energy differences of q ~ ~!. The ground state is a Gaussian distribution with width x0 = m! ; picture from http://en.wikipedia.org/wiki/Quantum mechanical harmonic oscillator 5.1.2 Zero Point Energy We already learned that the lowest possible energy level of the harmonic oscillator is not, 1 as classically expected, zero but E0 = 2 ~!. It can be understood in the following way. The ground state is an eigenfunction of the Hamiltonian, containing both kinetic and potential energy contributions, therefore the particle has some kinetic energy in the vicinity of x = 0 , where the potential en- ergy V (x ! 0) ! 0 . But this implies according to Heisenberg's uncertainty relation (Eq. (2.80)) that the momentum uncertainty increases pushing the total energy up again until it stabilizes ∆p / ~ % for ∆x ! 0 ) E 6= 0 : (5.33) ∆x We will now illustrate the harmonic oscillator states, especially the ground state and the zero point energy in the light of the uncertainty principle. We start by calculating the position and momentum uncertainties using Definition 2.10 Eq: (5:5) y h x i = h n j x j n i / h n j a + a j n i / h n j n−1 i + h n j n+1 i = 0 | {z } | {z } 0 0 (5.34) x2 = h j x2 j i = ~ h j a2 + aay + aya + (ay)2 j i : (5.35) n n 2m! n n Because the the eigenstates n for different n are orthogonal, the expectation values of a2 and (ay)2 vanish identically and we proceed by using Eq.
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