Harmonic Oscillator with raising and lowering operators. We write the Schrödinger equation for the harmonic oscillator in one dimension as follows:

Transcription

1 We write the Schrödinger equation for the harmonic oscillator in one dimension as follows: H ˆ! = "!2 d 2! + 1 2µ dx 2 2 kx 2! = E! T ˆ = "! 2 2µ d 2 dx 2 V ˆ = 1 2 kx 2 H ˆ = ˆ T + ˆ V (1) where µ is the reduced mass and k is the force constant of the oscillator. We would rather use algebra than calculus to solve this problem. If we utilize operator algebra, we can find an alternative way to solve the problem that uses only algebra. First, let k = µ! 2 where µ = reduced mass ( µ = m 1 m 2 m 1 + m 2 ) and ω is the frequency of the oscillation. This is a standard substitution from classical physics. With this substitution, the Schrödinger equation becomes:!! 2 d 2 " = E! µ2 2µ dx 2 2 x2 Dividing both sides by!! 2 )" (2) ( 2µ yields: d 2! = µ 2 " 2 x 2 dx 2 2µE! 2! 2 )! (3) ( Now we let! = µ" 2µE and! = and substitute back into equation (3) so that the!! 2 Schrödinger equation becomes: d 2 dx 2! = " 2 x 2 ( )! (4) To simplify, we transform from the variable x, to a new variable ξ where! = "x or x =! " (5) d! = "dx or dx = d! " (6) Now we transform the Schrödinger equation into this new variable:

2 d! dx = d! d" d" dx = d! d" (via the chain rule) (7) and d 2! dx 2 = d dx d! d" d" ( = d dx dx ) d! ( = ) d d! ( = ) d d! d" ( = ) d 2! (8) d" dx d" d" d" dx d" 2 This makes it possible for us to change equation (4) to the variable ξ: d 2! = " d 2! = " 2 2 dx 2 d 2 ( " ) * +! = (" 2 x 2 )! d 2! = 2 ) d 2 "*! (9) If we substitute the values we defined for α and β, we find that! " = 2µE! 2 µ! = 2E! which we will call =! " (10) λ is a dimensionless number that compares two energies. We can think of! = as an " index that will be a measure of state and energy (as we will see). λ will start to seem like a quantum number of sorts. Now equation (9) becomes: d 2! d" 2 = " 2 ( )! or d 2! d" 2 " 2! =! (11) The left hand side of equation (11) is an operator equation, nearly the Hamiltonian, and right hand side looks like an eigenvalue. Thus, equation (11) is an eigenvalue problem waiting to be solved. It is easier if we regroup and multiply by negative one:! 2 " d 2 d! 2 () = *) (12)

3 The left side of equation (12) looks like it should be able to be factored if we wanted. That is what we often do with algebra so let us try remember that we are now dealing with operator algebra and operators do not necessarily commute. If the expression were not operators, we could just write down: a 2 x 2! b 2 y 2 = ( a! b) ( a + b) but if we try that we find:! " d d! (! + d d! () =! " d d! (!) + d) ( d! =! 2 ) "! d) d! + d ( d!!) ) " d 2 ) d! 2 =! 2 ) "! d) d! d! d) +) +! d! d! " d 2 ) d! 2 =! 2 ) "! d) d) +) +! d! d! " d 2 ) d! 2 =! 2 ) +) " d 2 ) d! 2 =! 2 +1" d 2 d! 2 () which is almost, but not quite what we wanted. If we do it in the opposite order, we get:! " d d! (! + d d! () =! 2 "1" d 2 d! 2 () (14) Note well that we did not get the same answer when we reversed the order. That is because we are dealing with operators and apparently ξ and d do not commute, that is, d! "!, d d! ( 0. This is not entirely unexpected as ξ is a modified position operator while d is a modified momentum operator. We know that position and momentum do not d! commute so it is not surprising that order should matter for the factorization. We also note that the factorizations did not work. We wanted equation (12) but we got:! 2 "1" d 2 d! 2 () and! 2 +1" d 2 d! 2 () (13)

4 However, we notice, that if we add these two expressions, we get almost what we wanted:! 2 "1" d 2 d! 2 ()! 2 +1" d 2 d! 2 () + 2! 2 " d 2 d! 2 () (15) Equation (15) is just equation (12) times 2 so we have to divide by 2 to get what we wanted. If we divide both factors by 2, we get: 1 2! 2 "1" d 2 d! 2 () + 1 2! 2 +1" d 2 d! 2 () =! 2 " d 2 d! 2 () = *) Now we go back to the factored forms of equations (13) and (14) and distribute the additional factors of 1/2 over the factored operators: 1 2! " d ( d! 1 2! + d ( d! 1 2! + d () = 1 d! 2! 2 "1" d 2 d! 2 () = (* "1)) 1 2! " d () = 1 d! 2! 2 +1" d 2 d! 2 () = (* +1)) Notice that the eigenvalue we got from each equation either adds one to or subtracts one from our index, λ. We will make a definition for the factors: 1 2! " d ( = 1 d! 2! + d ( = (16) d! Then, 1 2! " d ( d! 1 2! + d ( d! and 1 2! + d () = ˆ a ) = (* "1)) d! 1 2! " d () = ) = (* +1)) d! (17)

5 ! 2 " d 2 () = ( 2 +1)) d! 2 = ( 2 "1)) (18) We can rewrite equations (18) in terms of and ˆ a! = 1 2 (" 1)!! = 1 2 (" +1)! (19) The most amazing thing is that the expressions in equations (19), are still the original, unmultiplied, unapproximated Schrödinger equation that we set out to solve in equation (1). The only difference is that it is transformed into different variables, into operator form. Now we manipulate this operator form; combine the operator forms in equations (18) to get: ( 2 +1)! = (" 1)! ( 2 1)! = (" +1)! 2(! )" =!2" 20 or (! )" =!" ˆ a " = (!1)" 21 " = ( +1)" Careful perusal of equation 20 reveals that, while the separate terms do not present an eigenvalue problem for the wavefunction ψ, the combination! ˆ a ( ) does. In ( a! ˆ a ) =!1 or [, ] =!1 ; these operators do not commute (so we cannot addition, ˆ measure them simultaneously.)

6 So what exactly are and? We can find out by letting them operate on ψ. First let us define a new operator (it will become transparent why as we use it): N ˆ = (22) What the properties of N ˆ? We can look at the commutator with and [ N ˆ, ] = N ˆ! N ˆ = ˆ a! ( ) ˆ = ˆ a! +1 a = ˆ a!! =! [ N ˆ, ] =! Similarly, [ N ˆ, ] = N ˆ! N ˆ =! =! (!1) = ˆ [ N ˆ, ] = ˆ =! + ˆ a a a (23) (24) Let us take ψ n to be an eigenfunction of the operator ˆ N. We will assume that it is normalized to make our lives easier. Then: ˆ N = n (25) What happens if we allow or to operate on ψ n? If we do it with N ˆ we can get an idea: N ˆ = (" + N ˆ ) ( ) = " + n = ("1 + n ) = ( n "1) = ( n "1) (26)

7 From equations (26), it appears that has the property of turning ψ n into ψ n-1 because we know that the eigenvalue for N ˆ must be n. In order to get an eigenvalue of n-1, N ˆ must operate on ψ n-1. N ˆ = n N ˆ "1 = n "1"1 and N ˆ so ˆ a "1 = n "1 ˆ a (27) Therefore, we call the lowering operator. Similarly, is a raising operator, that is, it changes ψ n into ψ n+1 What happens if we allow ˆ a to operate on ψ 0, the lowest energy state? ˆ a! 0 = 0 (28) because you cannot go any lower than the lowest state. If we then try to raise the state we get:! 0 =! 0 = 0 = 0 What does this mean? We know from the equations in (19) that! = 1 2 (" 1)!! 0 = 1 2 (" 1)! 0 This means that when we raise the lowered lowest state, it must be ˆ a! 0 = 1 2 (" 1)! 0 = 0 (29)

8 Because we know that neither ψ 0 nor 1/2 are zero, then the only way to get zero is for (λ 1) to be equal to zero for the minimum level and thus the smallest value that λ can have is λ 0 =1. Remember, λ is not just an index, it is given in equation number (10) and is a ratio of energies! = " = 2E! and! 0 = 2E 0! = 1 (30) E 0 =! 2 which means that the lowest energy of the harmonic oscillator is not zero, it is a finite positive value. This energy is commonly known as the zero point energy. It follows that the energy levels of the harmonic oscillator are given by:! E n = n + 1! where n=0, 1, 2, 3, (31) " 2 Now we see that and make it possible for us to change from level to level in the harmonic oscillator potential well. For example:! 7 "! 6 and! 7 "! 8 However, we do not know the equality here, only the proportionality. Neither do we have a good physical feel for what and really are. Let us return to the definition of and from equation (16) and substitute back in for ξ = 1 " 2! + d d! = 1 " 2 = 1 " 2! * d d! = 1 " 2 ( x + 1 d ( dx = ( " 2 x + 1 d = µ) " x ˆ + ( dx 2! ( x * 1 d = ( " ( dx 2 x * 1 d = µ) " x ˆ * ( dx 2! i µ) ˆ p x i µ) ˆ p x (32) where p ˆ x =! i d dx or d dx = i! ˆ p x

9 Now we can define ˆ x and ˆ p x in terms of ˆ a and ˆ a : x ˆ = 1 a 2! ˆ + and ( ) (33) ˆ p x = µ! i 1 a 2" ˆ ˆ ( a ) (34) Now we can use equations (33) and (34) to regain the Schrödinger equation again: = 1 " 2! + d d! = = µ) " x 2! ˆ 2 (! µ) (!2 d 2 µ 2 ) 2 dx 2 and 1 " 2! ( d d! = µ) " x 2! ˆ 2 +! µ) (! 2 d 2 µ 2 ) 2 dx 2 1 " 2! ( d = 1 " d! 2! 2 (1( d 2 d! 3 1 " 2! + d d! = 1 " 2! 2 +1( d 2 d! 3 (35) As before, these are almost what we want but not quite because they have an extra term. However, if we combine them linearly, we get: + ˆ a = µ! 2! x ˆ 2 "! µ! "!2 d 2 µ 2! 2 dx 2 ( + µ! 2! x ˆ 2 +! µ! "!2 d 2 µ 2! 2 dx 2 ( = µ!! x ˆ 2 "!2 d 2 µ 2! 2 dx 2 ( Multiplying both sides by!! and dividing by 2 we get:!! + 2 ˆ H =!! 2 ( ) =!! 2 ( + ˆ a ) µ! x ˆ 2 "! 2 d 2! µ 2! 2 dx 2 ( = "!2 d 2 2µ dx µ!2 x ˆ 2 = H ˆ (36) which is the same thing that we started with for ˆ H in equation (1). If we return to our definitions and derivation in equation 21, then we see that

10 = ˆ a +1 ˆ H =!! 2 ( + ˆ a ) =!! 2 (( +1 ) + ) =!! 2 2ˆ a +1 ( ) =!! ˆ " a =!! " N ˆ so, we see that H ˆ =!" N ˆ H ˆ =!" N ˆ + 1 2! n =!" n = E n E n =!" n This means that N ˆ is an operator that is closely related to H ˆ ; in fact, N ˆ is essentially equivalent to H ˆ save a few constants. N ˆ tells us in which level of the harmonic oscillator the system is found. This may seem like a trivial result but it is the same as solving a lot of differential equations and yields insight into the system. We call N ˆ the number operator. The final thing that we need to do is to figure out exactly what the raising and lowering operators do to the system. We know (38) " 1 and " +1 or = c lower 1 and = c raise +1 What are c lower and c raise? Knowing that the solution to the harmonic oscillator must form an orthonormal set of eigenfunctions, we can write: 37 = 1 "1 "1 = = 1 and have an adjoint relation with each other, that is, = implies that (39) and ˆ a = ˆ a. This * = "1 c lower c lower "1 (40) because when the raising operator operates on the wavefunction to the left, it operates as its adjoint or as the lowering operator. but ˆ a ˆ a = ˆ N = n = n = n (41)

11 This suggests that * c lower c lower = n 2 If c lower is real, then c lower = n or c lower = n, and ˆ a = n "1 (42) In a similar fashion, ˆ a = n (43) Final note: While this derivation may seem tedious and incomprehensible, you will see when we do problems to predict infrared transitions or perturbation theory problems, that it makes solving the harmonic oscillator incredibly tractable.