5.61 Fall 2007 Lecture Summary 12-15, Supplement page 1
Ehrenfest’s Theroem
In the lecture notes for the harmonic oscillator we derived the
expressions for x̂ (t) and p̂x (t) using standard approaches – integrals
involving Hermite polynomials (see pages 17 and 18, Lecture Summary 12-15).
The calculations are algebraically intensive, but showed that x̂ (t) and
p̂x (t) oscillate at the vibrational frequency. The results were as follows:
12
⎛ ! ⎞
x (t) = ( 2α ) cos (ω vibt ) = ⎜ cos (ω vibt )
−1 2
⎝ 2 µω ⎟⎠
and
1 ⎡ ⎛α⎞ ⎤ ⎛ !µω ⎞
12 12
p (t) = ⎢i! ⎜ ⎟
2 ⎣⎢ ⎝ 2 ⎠
(e iω vib t
−e −iω vib t
) ⎥ = −⎜
⎝ 2 ⎟⎠
sin (ω vibt )
⎦⎥
The issue considered here is an approach to calculate x (t) and p (t) in a
more straightforward manner.
Classically, (we use m instead of µ since we are dealing with a free
particle)
dx
p = mv = m
dt
So, quantum mechanically we might expect
d x (t)
p (t) = m .
dt
But, is this expression valid ? We can show that in fact it is with the
following argument.
d x (t)
For our original expression was …
dt
d x (t) d ⎧ ∞ * ⎫ ∞ dψ * ∞
dψ
= ⎨ ∫ ψ x̂ ψ dx ⎬ = ∫ x̂ ψ dx + ∫ ψ * x̂ dx
dt dt ⎪⎩ −∞ ⎪⎭ −∞ dt −∞
dt
Recall the time dependent Schrödinger equation is
∂ψ ∂ψ 1
i! = Hψ or = Hψ
∂t ∂t i!
Inserting these results into the expression above yields
Ehrenfest’s Theroem
In the lecture notes for the harmonic oscillator we derived the
expressions for x̂ (t) and p̂x (t) using standard approaches – integrals
involving Hermite polynomials (see pages 17 and 18, Lecture Summary 12-15).
The calculations are algebraically intensive, but showed that x̂ (t) and
p̂x (t) oscillate at the vibrational frequency. The results were as follows:
12
⎛ ! ⎞
x (t) = ( 2α ) cos (ω vibt ) = ⎜ cos (ω vibt )
−1 2
⎝ 2 µω ⎟⎠
and
1 ⎡ ⎛α⎞ ⎤ ⎛ !µω ⎞
12 12
p (t) = ⎢i! ⎜ ⎟
2 ⎣⎢ ⎝ 2 ⎠
(e iω vib t
−e −iω vib t
) ⎥ = −⎜
⎝ 2 ⎟⎠
sin (ω vibt )
⎦⎥
The issue considered here is an approach to calculate x (t) and p (t) in a
more straightforward manner.
Classically, (we use m instead of µ since we are dealing with a free
particle)
dx
p = mv = m
dt
So, quantum mechanically we might expect
d x (t)
p (t) = m .
dt
But, is this expression valid ? We can show that in fact it is with the
following argument.
d x (t)
For our original expression was …
dt
d x (t) d ⎧ ∞ * ⎫ ∞ dψ * ∞
dψ
= ⎨ ∫ ψ x̂ ψ dx ⎬ = ∫ x̂ ψ dx + ∫ ψ * x̂ dx
dt dt ⎪⎩ −∞ ⎪⎭ −∞ dt −∞
dt
Recall the time dependent Schrödinger equation is
∂ψ ∂ψ 1
i! = Hψ or = Hψ
∂t ∂t i!
Inserting these results into the expression above yields
