Class 5: Quantum harmonic oscillator – Ladder operators Ladder operators The time independent Schrödinger equation for the quantum harmonic oscillator can be written as 1 ()pmx2+ 222ω ψ = E ψ , (5.1) 2m where the momentum operator p is d p= − i ℏ . (5.2) dx If p were a number, we could factorize p2+ m 222ω x =−+( ipmxipmx ω)( + ω ). (5.3) However, we need to remember that p is an operator. In fact, because d d ()()−+ipmxipmxω + ωψ =−+ℏ mx ω ℏ + mx ωψ dx dx d d ψ =−+ℏmxω ℏ + mx ω ψ dx dx d dψ d ψ =−ℏ ℏ +mxωψ + mx ω ℏ + mx2 ωψ 2 2 dxdx dx (5.4) d2ψ d d ψ =−ℏ2 − ℏ()mxωψ + mx ω ℏ + mx2 ωψ 2 2 dx2 dx dx d 2ψ =−ℏ2 −mωψ ℏ + m 222 ω x ψ dx 2 =p2ψ + mx 222 ωψ − m ωψℏ we see that there is an extra term, such that p2+ m 222ω x =−+( ipmxipmx ω)( + ωω) + m ℏ (5.5) Even so it is useful to consider the two dimensionless operators 1 a+ =() − ipmx + ω , (5.6) 2ℏmω and 1 1 a− =() + ipmx + ω . (5.7) 2ℏmω The reason for labeling the operators with subscripts + and – will become clear later. Since the order of operation is important, we have 1 1 aa− + =() ++ ipmxω() −+ ipmx ω 2ℏmω 2 ℏ m ω (5.8) 1 2 222 =p + mω x − im ω () xp − px , 2ℏmω and 1 1 aa+ − =() −+ ipmxω() ++ ipmx ω 2ℏmω 2 ℏ m ω (5.9) 1 2 222 =p + mω x + im ω () xp − px . 2ℏmω Each expression contains the combination xp− px which is called the commutator of x and p. The standard notation for the commutator of two operators A and B is [ A, B] = AB − BA . (5.10) Since dψ d []xp,ψ=− ixℏ + i ℏ() x ψ = i ℏ ψ , (5.11) dx dx we can write [x, p] = i ℏ . (5.12) Also, because the Hamiltonian is p2 mω 2 x 2 1 H=+ =() pmx2 + 222ω , (5.13) 2m 2 2 m we have 1 1 H 1 a a= H +ℏω =+ . (5.14) − + ℏω2 ℏ ω 2 Similarly 2 1 1 H 1 a a= H +ℏω =− . (5.15) + − ℏω2 ℏ ω 2 These last two equations can be combined to show that 1 ℏ H=ω () aaaa−+ + +− . (5.16) 2 We also see that [aa−+,] = aa −+ − aa +− = 1. (5.17) The Schrödinger equation can be written as 1 E a− a + − ψ = ψ , (5.18) 2 ℏω or as 1 E a+ a − + ψ = ψ . (5.19) 2 ℏω Suppose we have a solution ψ of the Schrödinger equation corresponding to energy E. Then a+ψ is also a solution to the Schrödinger equation but for energy E + ℏω. To see why, consider ℏ 1 Ha()+ψ= ω aa +− + () a + ψ 2 ℏ 1 =ωaaa+−+ ψ + a + ψ (5.20) 2 ℏ 1 =ωa+ a − a + ψ + ψ . 2 Now using equation (5.14), this gives ℏ1 1 1 ℏ Ha()+ψω= aH + ψψψ ++= aE+ () ψωψ + ℏω 2 2 (5.21) ℏ =()E + ω a + ψ , which proves our contention. Similarly, if ψ is a solution of the Schrödinger equation corresponding to energy E, then a−ψ is also a ℏ solution to the Schrödinger equation but for energy E − ω. The operators a+ and a− are called ladder 3 ℏ operators , because the raising operator a+ moves up the energy ladder by a step of ω and the lowering ℏ operator a− moves down the energy ladder by a step of ω. Since the minimum value of the potential energy is zero and occurs at a single value of x, the lowest energy for the QHO must be greater than zero. Let the wave function for the minimum energy be ψ 0 ( x). Since there is no energy level below this minimum value, we must have a−ψ 0 = 0. (5.22) When combined with the normalization condition, this equation allows us to find ψ 0 ( x), since it gives (+ip + mω x ) ψ 0 = 0. (5.23) The normalized solution is found to be 1 4 mω x 2 − mω 2ℏ ψ 0 (x )= e . (5.24) π ℏ The energy of this state is easily found by applying the Hamiltonian operator given by equation (5.15) ℏ1 1 ℏ Hψω0= a+ a − + ψ 0 = ωψ 0 . (5.25) 2 2 We see that the lowest energy is 1 E = ℏω, (5.26) 0 2 and by applying the raising operator, the energy of the nth state is 1 ℏ En = n + ω, (5.27) 2 Normalization of the wave functions Although the ladder operators can be used to create a new wave function from a given normalized wave function, the new wave function is not normalized. To determine the normalization constant, we need to explore some more properties of the ladder operators. First consider ∞ ∗ ∫ f() ag+ dx , (5.28) −∞ where f and g are functions of x. Putting in the explicit form of the operator, we have 4 ∞ ∞ ∗ 1 ℏ ∗dg ∗ fagdx()+ = − f + mxfgdxω ∫ℏ ∫ −∞2 mω −∞ dx ∗ 1 ∞ d() f g df ∗ = −ℏ ++ ℏ g mxfgdxω ∗ (5.29) ℏ ∫ dx dx 2 mω −∞ ∞ ∗ 1∗ ∞ 1 df ∗ =−+ℏf g ℏ g + mω xf g dx . ℏ −∞ ℏ ∫ 2mω 2 m ω −∞ dx Provided f∗ g → 0 as x → ∞ , the first term on the last line is zero. Also the remaining integral is 1 ∞ df ∗ ∞ ℏg+ mxfgdxω ∗ = gaf ∗ dx . (5.30) ℏ ∫ ∫ ()− 2 mω −∞ dx −∞ Hence we have shown that ∞ ∞ ∗ ∗ ∫f() agdx+= ∫ gaf() − dx . (5.31) −∞ −∞ In a similar manner, we can show that ∞ ∞ ∗ ∗ ∫f() agdx−= ∫ gaf() + dx . (5.32) −∞ −∞ (The operators a− and a+ are said to be Hermitian conjugates .) Suppose ψ n represents a normalized wave functions for any value of n. Let g =ψ n and f= a +ψ n in equation (5.31). Then ∞ ∞ ∞ ∗ ∗ ∗ (5.33) ∫()()()a++ψψnn a dx= ∫ ψ n aa −+ ψ n dx = ∫ ψψ nn() aa −+ dx . −∞ −∞ −∞ Using equation (5.14) and equation (5.27), we have * * a− a + ψn=( n + 1) ψ n . (5.34) On using this in equation (5.33), we get ∞ ∞ ∗ ∗ ∫()()()a+ψψnn a + dx= ∫ ψ n n + 1 ψ n dx . (5.35) −∞ −∞ Since the wave functions are assumed normalized, we see that 5 a+ψn= n + 1 ψ n + 1 . (5.36) In a similar way, by starting with equation (5.32), we find a−ψn= n ψ n − 1. (5.37) Since 1 12 1 3 ⋯ ψψn==a+− n1() a +− ψ n 2 =() a +− ψ n 3 = , n nn()−1 nnn()() − 1 − 2 we see that 1 n ψn = ()a+ ψ 0. (5.38) n! Problem 2.12 Find x,, p x2 , p 2 , and T for the nth stationary state of the harmonic oscillator. Check that the Uncertainty Principle is satisfied. From equations (5.6) and (5.7), we find 2mω a+ a = x , (5.39) − + ℏ and 2 aa− = ip . (5.40) − + ℏmω Hence ∞ℏ ∞ x=∫ψψnn xdx = ∫ ψ n() aadx− += + ψ n 0, −∞2mω −∞ where use has been made of the orthogonality of the wave functions, and the lowering and raising properties of the ladder operators. Similarly we see that p = 0. Also 6 ∞ℏ ∞ ℏ ∞ 2 2 2 2 2 x=ψψn xdx n = ψψ n() aa−+ += n dx ψ n () aaaaaa −−++−+ +++ ψ n dx ∫2mω ∫ 2 m ω ∫ −∞ −∞ −∞ E 1 ℏ =n =n + , mω2 2 m ω where we have used equation (5.16). Similarly ∞ℏ ∞ ℏ ∞ 2 2 mω2 m ω 2 2 p=ψψn pdx n =− ψψ n() aa− − + n dx =− ψ n () aaaaaa− −−+ −+ +_ + ψ n dx ∫2 ∫ 2 ∫ −∞ −∞ −∞ ℏmω∞ ℏ m ω 2 E 1 =ψaaaa + ψ dx =n =+ nℏ m ω . ∫ n()− + + _ n ℏ 2−∞ 2ω 2 Also, 2 p 1 1 ℏ 1 T= =+ n ω = E n . 2m 22 2 To check whether the Uncertainty Principle holds, consider ℏ 2 2 1 1ℏ 1 ℏ σx σ p =xp =+ n ⋅+ nmnω =+ . 2mω 2 2 We see that the principle holds, with equality for n = 0. Orthonormality of the wave functions The stationary state wave functions for the harmonic oscillator are orthonormal, i.e. ∞ ∫ ψmn ψdx = δ mn . (5.41) −∞ The orthogonality of the wave function can be shown by considering ∞ ∞ ∞ ∫ψψmnHdx= ∫ ψψ mnn EdxE = nmn ∫ ψψ dx . (5.42) −∞ −∞ −∞ Now ∞ ∞ ℏ22 ℏ 2∞ 2 ∞ d d ψ n ∫∫ψψψmnmHdx=−+2 Vdx ψ n =− ∫∫ ψ m2 dx + ψψ mn Vdx (5.43) −∞ −∞ 2mdx 2 m−∞ dx −∞ By integration by parts, we have 7 ∞ 2 ∞ ∞ dψn d ψ n dd ψψ mn ∫ψm2 dx= ψ m − ∫ dx −∞ dx dx −∞ −∞ dxdx (5.44) ∞ ∞ ∞ 2 dψn d ψ m d ψ m =ψm − ψ n + ∫ 2 ψ n dx . dx −∞ dx −∞ −∞ dx Applying the boundary conditions, the first two terms on the right hand side are both equal to zero. Putting the result in equation (5.43), we have ∞ℏ2 ∞ 2 ∞∞ ∞ d ψ m ∫ψψmnHdx=− ∫∫∫ ψ n2 dx + ψψ nm Vdx = ψψ nmmnm HdxE = ∫ ψψ dx . (5.45) −∞2m −∞ dx −∞ −∞ −∞ Hence ∞ ∞ Em∫ψψ nm dxE= n ∫ ψψ nm dx . (5.46) −∞ −∞ Because the energy values are not equal if m≠ n , we see that the wave functions must be orthogonal. This is a quite general result, which does not depend on the particular potential energy function. Two wave functions that have different energies are orthogonal. 8 .