Mean Value and RMS Deviation

The mean value in Quantum Mechanics is the average result obtained when a large number of measurements are made on identical systems, i.e. systems in the same state |Ψ(t)>.
For the harmonic oscillator
<X> = <Φn|X|Φn> = 0,  <P> = <Φn|P|Φn> = 0.
The mean value of X and P in an eigenstate of H is always zero.

The classical mean value can be defined as the average result obtained when a large number of measurements are made on the same system at different times, or when a large number of measurements are made at the same time on different systems, whose phases are randomly chosen.
xavg = (xM/T)∫0Tcos(ωt + φ)dt = 0,  T = 2π/ω = period,
or
xavg = (xM/2π)∫0cos(ωt + φ)dφ = 0.
pavg = -mv = -(mωxM/T)∫0Tsin(ωt + φ)dt = 0,  or  pavg = -(mωxM/2π)∫0sin(ωt + φ)dφ = 0.

For a harmonic oscillator of well-defined energy <X> = xavg = 0,  <P> = pavg = 0.
But we have to remember that the classical and quantum mechanical mean value are interpreted differently.

In a similar way the quantum mechanical and the classical root-mean-square deviation are interpreted differently.
In Quantum Mechanics we define
(ΔX)2 = <X2> - <X>2 = <X2>,  (ΔP)2 = <P2> - <P>2 = <P2>.
X2 = (ħ/(2mω))(a + a)(a + a) = (ħ/(2mω))(aa + aa + aa + aa),
P2 = -(mħω/2)(a - a)(a - a) = (mħω/2)(-aa + aa + aa - aa).
n|aan> = <aΦn|aΦn> ∝ <Φn-1n+1> = 0.
n|aa|Φn> = <aΦn|aΦn> ∝ <Φn+1n-1> = 0.
n|aa|Φn> = <aΦn|aΦn> = n.
n|aan> = <aΦn|aΦn> = n + 1.
<X2> = (ħ/(2mω))(2n + 1) = (ħ/(mω))(n + ).
<P2> = (mħω/2)(2n + 1) = mħω(n + ).
ΔXΔP = (n + )ħ.
ΔXΔP = ħ for n = 0, i.e. for the ground state.

For a classical harmonic oscillator with energy E = (n + )ħω we have
E = mω2xM2 = (n + )ħω,  xM2 = 2(n + )ħ/(mω) = 2(ΔX)2.
ΔX = xM/√2.
pM =  mωxM = √2mωΔX =  √2ΔP.
ΔP = PM/√2.
The quantum mechanical root mean square deviations ΔX and ΔP are proportional to the classical maximum amplitude and maximum momentum.
The classical root-mean-square deviation is given by
x2avg= (xM2/T)∫0Tcos2(ωt + φ)dt = (ΔX)2.
p2avg = (pM2/T)∫0Tsin2(ωt + φ)dt = (ΔP)2.
For a harmonic oscillator of well defined energy we obtain the same root-mean-square deviations.  But again we have to remember that the classical and quantum mechanical root-mean-square deviation are interpreted differently.


Coherent states
Ehrenfest’s theorem states d<X>/dt = <P>/m,  d<P>/dt = -<dU(x)/dt> = -mω2<X>.
In an eigenstate of the Hamiltonian <X>(t) and <P>(t) are zero at all times.  However if the state vector |Ψ(0)> is a linear superposition of energy eigenstates then <X>(t) and <P>(t) satisfy the classical equations of motion and behave like the classical position and momentum as a function of time.

Proof:
|Ψ(0)> = ∑ncnn>,  |Ψ(t)> = ∑nexp(-iEnt/ħ)cnn>.
<x(t)> =  ∑nn'cn*cnexp(-i(En - En)t/ħ)<Φn|X|Φn'>
n|X|Φn'> = (ħ/(2mω))1/2[(n'+1)1/2δn,n'+1 + (n+1)1/2δn+1,n'].
<x(t)> = (ħ/(2mω))1/2[ ∑n'cn'+1*cn(n'+1)1/2exp(iωt) + ∑ncn*cn+1(n+1)1/2exp(-iωt)]
=  (ħ/(2mω))1/2n(n+1)1/2(cn+1*cn exp(iωt) + cn*cn+1exp(-iωt))
=  (ħ/(2mω))1/2n(n+1)1/2(Re(cn+1*cn)2cos(ωt) - Im(cn*cn+1)2sin(ωt))
= A(cosφ cos(ωt) - sinφ sin(ωt)) = A cos(ωt + φ).
where Acosφ = (ħ/(2mω))1/2n√(n+1) 2Re(cncn+1*), 
Asinφ = (ħ/(2mω))1/2n√(n+1) 2Im(cncn+1*).

Similarly,  <P(t)> = -A mω sin(ωt + φ).

Conclusion:
If we want a quantum mechanical wave packet to behave like a classical particle we have to build it as a linear superposition of energy eigenstates.
Problem:

For a one-dimensional simple harmonic oscillator we may define raising and lowering operators
a = (mω/(2ħ))1/2(X + iP/(mω)),  a = (mω/(2ħ))1/2(X - iP/(mω)),
with properties
a|n> = √(n) |n - 1>,  n ≠ 0, a|n> = 0, b = 0, and
a|n> = √(n + 1) |n + 1>.
(a)  Show by direct calculation that the ground state of the oscillator satisfies
(ΔX)2(ΔP)2 = |<[x,p]>|2,
and hence is a minimum uncertainty state.
(b)  A coherent state of a one-dimensional oscillator, |b>, may be obtained by applying the finite displacement operator exp(-iPb/ħ) to the ground state eigenket |0>.   Here b is the displacement distance.  Using this definition of a coherent state, prove that the coherent state is also a minimum uncertainty state.