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Ben Gurion University of the Negev www.bgu.ac.il/atomchip, www.bgu.ac.il/nanocenter Physics 3 for Electrical Engineering Lecturers: Daniel Rohrlich, Ron Folman Teaching Assistants: Daniel Ariad, Barukh Dolgin Week 8. Quantum mechanics – raising and lowering operators, 1D harmonic oscillator • harmonic oscillator eigenvalues and eigenfunctions • matrix representation • motion of a minimumuncertainty wave packet • 3D harmonic oscillator • classical limit Sources: Merzbacher (2nd edition) Chap. 5 Sects. 1-4; Merzbacher (3rd edition) Chap. 5 Sects. 1, 3 and Chap. 10 Sect. 6; Tipler and Llewellyn, Chap. 6 Sect. 5. We already stated that Schrödinger’s wave equation, with its continuous solutions, and Heisenberg’s matrix algebra, with its quantum jumps, are equivalent. Let’s see and compare how these two different methods apply to the quantum harmonic oscillator. The 1D harmonic oscillator x 0 The 1D harmonic oscillator t1 t2 t3 t4 t5 t6 x 0 x 0 x 0 x 0 x 0 x 0 The 1D harmonic oscillator x t1 0 x t2 0 F(t2) x t3 0 F(t3) x t4 0 F(t4) x t5 0 x t6 F(t6) 0 The 1D harmonic oscillator x 0 d F ( x) kx V ( x) dx 1 2 V ( x) kx 2 This system is a model for many systems, e.g. molecules made of two atoms. 2 d 2 k 2 Eψ( x) x ψ( x ) 2 2 2m dx The 1D harmonic oscillator x 0 d F ( x) kx V ( x) dx 1 2 V ( x) kx 2 Any system with a potential minimum at some x = x0 may behave like a harmonic oscillator at low energies: 2 d 2 k 2 ( x) 0)2 V′′(x 0) + …. x ψ( x ) V(x) = V(x0)E+ψ(x–x 2 2 2m dx x0 The 1D harmonic oscillator x 0 d F ( x) kx V ( x) dx 1 2 V ( x) kx 2 Also, a mode of the electromagnetic field of frequency ν behaves like a 1D harmonic oscillator of frequency ν. Its energy levels nhν correspond 2 tod n2 photons k of frequency ν. 2 Eψ( x) x ψ( x ) 2 2 2m dx The 1D harmonic oscillator x 0 d F ( x) kx V ( x) dx 1 2 V ( x) kx 2 Schrödinger’s equation: 2 d 2 k 2 Eψ( x) x ψ( x ) 2 2 2m dx Solving Schrödinger’s equation Schrödinger’s way: m Define a new variable x , where k / m. In terms of ξ , Schrödinger’s equation is d2 2E 2 . 2 ψ 0 , where d Solutions: Try ψ e 2 / 2 H ( ) , where H(ξ) is a polynomial. ψ is a solution when H(ξ) is one of the Hermite polynomials: H0(ξ) = 1, H1(ξ) = 2ξ, H2(ξ) = 4ξ2–2, H3(ξ) = 8ξ3–12ξ, …. Solving Schrödinger’s equation Heisenberg’s way: Define new variables aˆ m pˆ xˆ i , aˆ 2 m m pˆ xˆ i , 2 m where k / m. Let’s prove that 1 ˆ H aˆ aˆ , 2 aˆ , aˆ 1 , Hˆ , aˆ aˆ ˆ , Bˆ , Cˆ ) that Try also to prove (for any three operators A Aˆ Bˆ, Cˆ Aˆ Bˆ, Cˆ Aˆ, Cˆ Bˆ . . Raising and lowering operators To prove: ˆ ˆ 1 ˆ H a a , 2 aˆ , aˆ 1 , Hˆ , aˆ aˆ m pˆ pˆ 1 ˆ ˆ 1 a a xˆ i xˆ i 2 m m 2 2 pˆ 2 m 2 2 ˆ ˆ ˆ x i x , p 2m 2 2 2 pˆ 2 m 2 2 xˆ Hˆ . 2m 2 . Raising and lowering operators To prove: ˆ ˆ 1 ˆ H a a , 2 aˆ , a ˆ aˆ , aˆ 1 , Hˆ , aˆ aˆ m 2 pˆ xˆ i m i 2 pˆ , xˆ 2 i 2 i 1 . 2 pˆ , xˆ i m . Raising and lowering operators To prove: ˆ ˆ 1 ˆ H a a , 2 aˆ , aˆ 1 , Hˆ , aˆ aˆ . 1 ˆ ˆ ˆ ˆ H , a a a , aˆ 2 aˆ aˆ (aˆ aˆ ) aˆ aˆ aˆ (aˆ aˆ 1) aˆ aˆ aˆ aˆ . Suppose ψ is an eigenstate of Ĥ with eigenvalue E. Then Hˆ aˆ ψ aˆ Hˆ aˆ ψ E aˆ ψ , using Hˆ , aˆ aˆ . Therefore, aˆ ψ is an eigenstate of Ĥ with eigenvalue E . We call â a raising operator. Homework: Show that â is a lowering operator, i.e. Hˆ aˆ ψ aˆHˆ aˆ ψ E aˆ ψ . Suppose ψ is an eigenstate of Ĥ with eigenvalue E. Then Hˆ aˆ ψ aˆ Hˆ aˆ ψ E aˆ ψ , using Hˆ , aˆ aˆ . Therefore, aˆ ψ is an eigenstate of Ĥ with eigenvalue E . We call â a raising operator. The harmonic oscillator must have a ground state – call it ψ0(x) or 0 – with minimum energy. For 0 we have aˆ 0 0 which means pˆ d 0 xˆ i ψ 0 ( x) x ψ 0 ( x) , m m dx and therefore m x 2 / 2 ψ0 ( x) A0 e . 1 Since Hˆ aˆ aˆ and aˆ 0 0, the ground-state energy 2 is E0 , and the energy of the state n , defined by 2 1 1 / 2 ˆ n n (n!) a 0 , is En n . 2 1 1 ˆ ˆ ˆ From a a n H n En n n n 2 2 we learn that aˆ aˆ n n n . Thus we call aˆ aˆ the number operator. Since n aˆ aˆ n n we can write aˆ n n n 1 . Likewise, since n aˆ aˆ n n aˆ aˆ 1 n n 1 , we can write aˆ n n 1 n 1 . 1 Since Hˆ aˆ aˆ and aˆ 0 0, the ground-state energy 2 is E0 , and the energy of the state n , defined by 2 1 1 / 2 ˆ n n (n!) a 0 , is En n . 2 Normalization: 1 | ψ 0 ( x) |2 dx | A0 | | A0 | 2 2 m x 2 / e m dx . So the ground state normalization is | A 0 | all n, n n 1. (Prove it!) 4 m . Then for Harmonic oscillator eigenvalues and eigenfunctions What are the lowest eigenstates of the harmonic oscillator? m m x 2 / 2 ψ 0 ( x) e 1/ 4 m ˆ ψ1 ( x) a ψ 0 ( x) 1/4 1/2 2 3/ 4 m x 2 / 2 xe 1 2 2m 2 m x 2 / 2 1/2 m ψ 2 ( x) aˆ ψ 0 ( x) 2 x 1e 2 1/ 4 1 3 1 m ψ 3 ( x) aˆ ψ 0 ( x) 1/4 6 3 3/ 4 2m 3 m x 2 / 2 x 3 x e Note that the eigenfunctions ψn(x) are even or odd in x. Why? Harmonic oscillator eigenvalues and eigenfunctions This Figure is taken from here. Harmonic oscillator eigenvalues and eigenfunctions This Figure is taken from here. If the harmonic oscillator represents a mode of the electromagnetic field, an energy level En (n ½ ) represents n photons each having energy , plus additional “zero-point energy” of 2 per mode. Matrix representation In the basis of harmonic-oscillator eigenvectors, we can represent the operators Hˆ , aˆ, aˆ , etc., and even xˆ and pˆ , as matrices. Since m Hˆ n (n ½ ) if m = n and vanishes otherwise, we can represent Ĥ as an infinite matrix: 1 0 Hˆ 2 0 0 ... 0 0 3 0 0 5 0 0 ... ... 0 ... 0 ... 0 ... 7 ... ... ... Matrix representation In the basis of harmonic-oscillator eigenvectors, we can represent the operators Hˆ , aˆ, aˆ , etc., and even xˆ and pˆ , as matrices. Since m aˆ n n if m = n–1 and vanishes otherwise, we can represent â as an infinite matrix: 0 0 â 0 0 ... 1 0 0 0 2 0 0 ... 0 ... ... 0 ... 3 ... 0 ... ... ... 0 Matrix representation In the basis of harmonic-oscillator eigenvectors, we can represent the operators Hˆ , aˆ, aˆ , etc., and even xˆ and pˆ , as matrices. Since m aˆ n n 1 if m = n+1 and vanishes otherwise, we can represent â as an infinite matrix: 0 1 â 0 0 ... 0 0 2 0 ... 0 ... 0 0 ... 0 0 ... 3 0 ... ... ... ... 0 The normalized harmonic-oscillator eigenvectors themselves are the basis vectors: 1 0 0 0 , 0 ... 0 1 1 0 , 0 ... 0 0 2 1 , 0 ... 0 0 3 0 , ... . 1 ... The transpose of a matrix M̂ is written M̂ T and defined by Mˆ T ij i Mˆ T j j Mˆ i Mˆ ji : a b ... a c ... T ˆ ˆ M c d ... , M b d ... . ... ... ... ... ... ... The adjoint of a matrix M̂ is written M̂ and defined by * * ˆ ˆ ˆ ˆ M ij i M j j M i M ji : a b ... a * c * ... ˆ ˆ M c d ... , M b * d * ... . ... ... ... ... ... ... The adjoint of a matrix M̂ is written M̂ and defined by * * ˆ ˆ ˆ ˆ M ij i M j j M i M ji : a b ... a * c * ... ˆ ˆ M c d ... , M b * d * ... . ... ... ... ... ... ... Any observable  is self-adjoint, i.e. Aˆ Aˆ . “ is Hermitian” and “ is self-adjoint” mean the same thing. The adjoint of a matrix M̂ is written M̂ and defined by * * ˆ ˆ ˆ ˆ M ij i M j j M i M ji : a b ... a * c * ... ˆ ˆ M c d ... , M b * d * ... . ... ... ... ... ... ... Any observable  is self-adjoint, i.e. Aˆ Aˆ . “ is Hermitian” and “ is self-adjoint” mean the same thing. The raising and lowering operators â and â are not selfadjoint, but are adjoints of each other: aˆ ˆ a , a ˆ aˆ . Hˆ , xˆ and pˆ are manifestly self-adjoint: 1 0 0 0 3 0 ˆ H 2 0 0 5 ... ... ... ... ... , ... ... Hˆ , xˆ and pˆ are manifestly self-adjoint: xˆ ˆa aˆ 2m 0 1 2m 0 ... 0 m 1 m aˆ aˆ i pˆ i 2 0 2 ... 1 0 2 ... ... 2 ... , 0 ... ... ... 0 0 1 0 2 0 2 ... ... ... ... . ... ... Vectors, too, have adjoints. For any ψ we have ψ ψ , and, for example, 1 0 0 ... 1 0 . 0 ... Try to prove, for any two operators Aˆ , Bˆ , the rule ( Aˆ Bˆ ) Bˆ Aˆ . Motion of a minimum-uncertainty wave packet The ground state wave function ψ0(x) is a minimum-uncertainty wave function: we can calculate m , x p . 2 2 , p 2m x In general, the time evolution of any initial wave function Ψ(x,0) can be obtained from the expansion of Ψ(x,0) in the basis of energy eigenstates. If the initial wave function is ( 0) cn n , n 0 for given cn, then the wave function at time t is (t ) n 0 cn e iEnt / n e it / 2 n 0 cn e int n . Motion of a minimum-uncertainty wave packet (t ) cn e iEnt / n e it / 2 n 0 cn e int n . n 0 The period of a classical harmonic oscillator having angular frequency ω is T = 2π /ω . If we add T to t in Ψ(x,t), the wave function does not change, up to an overall phase factor –1: (t T ) (t ) . Schrödinger even found a solution Ψ(x,0) that moves between x = R and x = –R while the probability distribution |Ψ(x,0)|2 keeps its (minimum uncertainty) shape: 1/ 4 m 2 | ( x, t ) | e m ( x R cos t ) 2 / 2 . Motion of a minimum-uncertainty wave packet Schrödinger thought his solution might be typical, but it is not. Usually the probability distribution spreads over time. Prove that a free 1D wave packet spreads, that if it is initially ( x,0) (2 ) 1 / 4 ( x) 1 / 2 i p x / x 2 / 4( x)2 e e , then at time t the wave packet is ( x, t ) e i p 2 t / 2 m i p x / ( x p t / m) 2 /[ 4( x ) 2 2it / m] e ( 2 )1 / 4 x it 2 m x 1 / 2 Show that Δp = 2x is constant in time, but x(t ) x 2 p 2 t 2 / m 2 . . 3D harmonic oscillator The Schrödinger equation for a general 3D harmonic oscillator, 2 2 kx 2 k y 2 kz 2 Eψ( x, y, z ) x y z ψ , 2 2 2 2m allows for a harmonic potential with different strengths along the three axes. Show that the eigenfunctions are products of 1D harmonic oscillator eigenfunctions. What are the lowest energies and their degeneracies as a function of kx, ky and kz? Classical limit A classical particle in a square well has equal probability to be at any point inside. How about a quantum particle in its ground state? a quantum particle in a highly excited state? The probability P(x) for classical harmonically oscillating particle to be at any point x is inversely proportional to its speed at that point: P(x) ~ (2E/m – ω2x2)–1/2, where E is total energy. n=0 . n = 10