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Apply Equation 3.71 to the following special cases: (a)Q=1; (b)Q=H; (c)Q=x; (d)Q=p. In each case, comment on the result, with particular reference to Equations 1.27,1.33,1.38, and conservation of energy (comments following Equation 2.39).

Short Answer

Expert verified

a) When Q^=1thenddtΨΨ=0 .

b) When Q^=H^then dHdt=0.

c) When Q^=x^then ddtx^=p^m.

d) When Q^=p^then ddtp^=-∂V∂x.

Step by step solution

01

Expectation value of an operator Q

The rate of change of the expectation value of an operator Qis:

ddt<Q^>=ddt<ΨQ^Ψ>=ih<H^,Q^>+<∂Q^∂t>

Where,localid="1656314946232" H^is the Hamiltonian of the system.

02

Apply Equation 3.71 to the cases whenQ^=1.

(a)

Any operator commutes with a constant, so Q^=1commutes with everything, that isH^,Q^=0, substitute into (1):

dQ^dtddtΨΨ=0this is the conservation of normalization. which is proved in equation 1.27.

03

Apply Equation 3.71 to the cases when

(b)

Any operator commutes with itself, soQ^=H^commutes with H^, that is role="math" localid="1656315468447" H^,H^=0, substitute into (l):

dHdt=∂H∂t

If the energy has no explicit time dependence, thendHdt=0this result expresses the conservation of energy.

04

Apply Equation 3.71 to the cases when Q^=x^

(c)

For Q^=x^,find the commutator of [H, x], as:

x,H=x,p22m+V=12mx,p2+x,V

Where:

x,p2=xp2-p2x=xp2-pxp+pxp-p2x=x,pp+px,p

Butx,p2=ih,so:x,p2=2ihandx,V=0hence,x,H=12m2ihp=ihpm

Substitute into (1), to get:

ddtx^=ihx,H+∂x^∂t=p^m+∂x^∂t

But and are independent variables, then ∂x/∂t=0, substitute to get:

ddtx^=p^m

This is the quantum equivalent of the classical equation p=mv.

05

Apply Equation 3.71 to the cases when Q^=P^

(d)

For Q^=p^,find the commutator of [H, p] as:

H,pg=h22m∂2∂x2+V,hi∂∂xg=hiV∂∂x-∂∂xVg=-hi∂V∂xg

The above equation will be,

H^,p^=hi∂V∂x

Substitute into , to get:

ddtp^=ihH^,p^+∂p^∂t=-∂∂x+∂p^∂t

But pand tare independent variables, then ∂p^/∂t=0, substitute to get:

ddtp^=-∂V∂x

This is Ehrenfest's theorem.

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