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> = <Φ_{n}|X|Φ_{n}> = 0, <P> = <Φ_{n}|P|Φ_{n}>
= 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.

x_{avg}
= (x_{M}/T)∫_{0}^{T}cos(ωt + φ)dt = 0, T =
2π/ω = period,

or

x_{avg}
= (x_{M}/2π)∫_{0}^{2π}cos(ωt + φ)dφ = 0.

p_{avg}
= -mv = -(mωx_{M}/T)∫_{0}^{T}sin(ωt + φ)dt = 0,
or p_{avg}
= -(mωx_{M}/2π)∫_{0}^{2π}sin(ωt + φ)dφ = 0.

For a harmonic oscillator of **well-defined energy** <X> =
x_{avg}
= 0, <P> =
p_{avg} = 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} = <X^{2}> - <X>^{2} = <X^{2}>,
(ΔP)^{2} = <P^{2}> - <P>^{2} = <P^{2}>.

X^{2} = (ħ/(2mω))(a^{†} + a)(a^{†} + a) =
(ħ/(2mω))(aa + aa^{†} + a^{†}a + a^{†}a^{†}),

P^{2} = -(mħω/2)(a^{†} - a)(a^{†} - a) = (mħω/2)(-aa + aa^{†} + a^{†}a - a^{†}a^{†}).

<Φ_{n}|a^{†}a^{†}|Φ_{n}> =
<aΦ_{n}|a^{†}Φ_{n}> ∝
<Φ_{n-1}|Φ_{n+1}> = 0.

<Φ_{n}|aa|Φ_{n}> =
<a^{†}Φ_{n}|aΦ_{n}> ∝
<Φ_{n+1}|Φ_{n-1}> = 0.

<Φ_{n}|a^{†}a|Φ_{n}> =
<aΦ_{n}|aΦ_{n}> = n.

<Φ_{n}|aa^{†}|Φ_{n}> =
<a^{†}Φ_{n}|a^{†}Φ_{n}> = n +
1.

<X^{2}> = (ħ/(2mω))(2n + 1) = (ħ/(mω))(n + ½).

<P^{2}> = (mħω/2)(2n + 1) = mħω(n + ½).

ΔXΔP = (n + ½)ħ.

ΔXΔP = ½ħ for n = 0, i.e. for the ground
state.

For a classical harmonic oscillator with energy E = (n + ½)ħω we have

E = ½mω^{2}x_{M}^{2} = (n + ½)ħω, x_{M}^{2}
= 2(n + ½)ħ/(mω) = 2(ΔX)^{2}.

ΔX = x_{M}/√2.

p_{M} = mωx_{M} = √2mωΔX = √2ΔP.

ΔP = P_{M}/√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

x^{2}_{avg}= (x_{M}^{2}/T)∫_{0}^{T}cos^{2}(ωt + φ)dt
=
(ΔX)^{2}.

p^{2}_{avg}
= (p_{M}^{2}/T)∫_{0}^{T}sin^{2}(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.

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)> = ∑_{n}c_{n}|Φ_{n}>, |Ψ(t)> = ∑_{n}exp(-iE_{n}t/ħ)c_{n}|Φ_{n}>.

<x(t)> = ∑_{n}∑_{n'}c_{n}*c_{n}exp(-i(E_{n}
- E_{n})t/ħ)<Φ_{n}|X|Φ_{n'}>

<Φ_{n}|X|Φ_{n'}> = (ħ/(2mω))^{1/2}[(n'+1)^{1/2}δ_{n,n'+1}
+ (n+1)^{1/2}δ_{n+1,n'}].

<x(t)> = (ħ/(2mω))^{1/2}[ ∑_{n'}c_{n'+1}*c_{n}(n'+1)^{1/2}exp(iωt)
+ ∑_{n}c_{n}*c_{n+1}(n+1)^{1/2}exp(-iωt)]

= (ħ/(2mω))^{1/2} ∑_{n}(n+1)^{1/2}(c_{n+1}*c_{n}
exp(iωt) + c_{n}*c_{n+1}exp(-iωt))

= (ħ/(2mω))^{1/2} ∑_{n}(n+1)^{1/2}(Re(c_{n+1}*c_{n})2cos(ωt)
- Im(c_{n}*c_{n+1})2sin(ωt))

= A(cosφ cos(ωt) - sinφ sin(ωt)) = A cos(ωt + φ).

where Acosφ = (ħ/(2mω))^{1/2}∑_{n}√(n+1) 2Re(c_{n}c_{n+1}*),

Asinφ = (ħ/(2mω))^{1/2}∑_{n}√(n+1) 2Im(c_{n}c_{n+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.

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} = ¼|<[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.

- Solution:

(a) Let X_{s}= (mω/ħ)^{1/2}X, P_{s}= (mωħ)^{-1/2}P.

a = (2)^{-1/2}(X_{s}+ iP_{s}), a^{†}= (2)^{-1/2}(X_{s}- iP_{s}).

X_{s}= (2)^{-1/2}(a^{†}+ a), P_{S}= i(2)^{-1/2}(a^{†}- a), {X_{s},P_{s}] = i.

(ΔX)^{2}= <X^{2}> - <X>^{2}, (ΔP)^{2}= <P^{2}> - <P>^{2}.

For the harmonic oscillator in its ground state <X> = <P> = 0.

<X_{s}^{2}> = ½<0|aa + aa^{†}+ a^{†}a + a^{†}a^{†}|0> = ½<0|aa^{†}|0> = ½.

<P_{s}^{2}> = -½<0|aa - aa^{†}- a^{†}a + a^{†}a^{†}|0> = ½<0|aa^{†}|0> = ½.

<X_{s}^{2}><P_{s}^{2}> = ¼. <X^{2}><P^{2}> = ¼ħ^{2}. [X,P] = iħ. |<[X,P]>|^{2}= ħ^{2}.

Therefore (ΔX)^{2}(ΔP)^{2}= ¼|<[X,P]>|^{2},

the ground state is a minimum uncertainty state.

(b) Consider two operators A and B.

If Each of these operators commutes with the commutatot [A,B], then exp(A + B) = exp(A) exp(B) exp(-½(A,B]). (Proof)

The operator exp(-iPb/ħ) = exp(c(a^{†}- a)). [a^{†},-a] = I. The commutator is the identity operator and commutes with a and a^{†},

Therefore exp(ca^{†}) exp(-ca) exp(-c^{2}/2).

Here c = -b(mω/(2ħ))^{1/2}.

|b> = exp(c(a^{†}- a))|0> = exp(-|c|^{2}/2)∑_{n=0}^{∞}(c^{n}/√(n!))a^{†}|0>,

since |n> = (n!)^{-1/2}(a^{†})^{n}|0>.

|b> is an eigenstate of the lowering operator a.

a|b> = exp(-|c|^{2}/2)∑_{0}^{∞}(c^{n}/(√n!))a|n> = c exp(-|c|^{2}/2)∑_{1}^{∞}(c^{n}(√n)/(√n!))|n-1>

= c exp(-|c|^{2}/2)∑_{1}^{∞}(c^{n-1}/√((n-1)!))|n-1> = c exp(-|c|^{2}/2)∑_{0}^{∞}(c^{n}/(√n!))|n> = c|b)>.

The eigenvalue of a is c.

|Ψ(t)> = U(t,0)|b> = exp(-iHt/ħ)|b> = exp(-|c|^{2}/2)∑_{0}^{∞}(c^{n}/(√n!))exp(-i(n + ½)ωt)|n>

= exp(-iωt/2) exp(-|c|^{2}/2)∑_{0}^{∞}(c^{n}/(√n!))exp(-inωt)|n> = exp(-iωt/2) exp(-|c'|^{2}/2)∑_{0}^{∞}(c'^{n}/(√n!))|n>,

with c' = c exp(-iωt), |c'| = |c|.

|ψ(t)> is an eigenstate of a with eigenvalue c’.

<x(t)> = (ħ/(2mω))^{1/2}<a^{†}+ a>(t) = (ħ/(2mω))^{1/2}(<aψ(t)|ψ(t)> + <ψ(t)|aψ(t)>)

= (ħ/(2mω))^{1/2}(c'* + c').

<X(t)>^{2}= (ħ/(2mω))(c'* + c')^{2}.

<X^{2}(t)> = (ħ/(2mω))<a^{†2}+ a^{2}+ 2a^{†}a + 1>(t) = (ħ/(2mω))((c'* + c')^{2}+ 1).

(ΔX)^{2}= (ħ/(2mω)).

<P(t)> = i(mωħ/2)^{1/2}<a^{†}- a>(t) = i(mωħ/2)^{1/2}(c'* - c').

<P(t)>^{2}= -(mωħ/2)(c'* - c')^{2}.^{ }<P^{2}(t)> = -(mωħ/2)<a^{†2}+ a^{2}- 2a^{†}a - 1>(t) = -(mωħ/2)((c'* - c')^{2}- 1).

(ΔP)^{2}= (mωħ/2).

ΔXΔP = ħ/2.

The coherent state is also a minimum uncertainty state.