130_notes.dvi

6.6.2 Verify Energy Operator

E(op)

1

√

2 π ̄h

ei(p^0 x−E^0 t)/ ̄h=

1

√

2 π ̄h

i ̄h

−iE 0

̄h

ei(p^0 x−E^0 t)/ ̄h

=E 0

1

√

2 π ̄h

ei(p^0 x−E^0 t)/ ̄h

6.7 Examples

6.7.1 Expectation Value of Momentum in a Given State

A particle is in the stateψ(x) =

( 1

2 πα

) 1 / 4

eik^0 xe−

x 4 α^2

. What is the expectation value ofp?

We will use the momentum operator to get this result.

〈p〉ψ = 〈ψ|p|ψ〉=

∫∞

−∞

ψ∗(x)p(op)ψ(x)dx

=

∫∞

−∞

(

1

2 πα

) 1 / 4

e−ik^0 xe−

x 4 α^2 ̄h

i

∂

∂x

(

1

2 πα

) 1 / 4

eik^0 xe−

x 4 α^2

dx

=

(

1

2 πα

) 1 / 2

̄h

i

∫∞

−∞

e−ik^0 xe−

x 4 α^2 ∂

∂x

eik^0 xe−

x 4 α^2

dx

=

(

1

2 πα

) 1 / 2

̄h

i

∫∞

−∞

e−ik^0 xe−

x 4 α^2

(

ik 0 eik^0 xe−

x 4 α^2

−

2 x

4 α

eik^0 xe−

x 42 α

)

dx

=

(

1

2 πα

) 1 / 2

̄h

i

∫∞

−∞

(

ik 0 e−

x 22 α

−

2 x

4 α

e−

x 2 α^2

)

dx

The second term gives zero because the integral is odd aboutx= 0.

〈ψ|p|ψ〉 =

(

1

2 πα

) 1 / 2

̄h

i

∫∞

−∞

(

ik 0 e−

x 22 α)

dx

〈ψ|p|ψ〉 =

(

1

2 πα

) 1 / 2

̄hk 0

√

2 πα= ̄hk 0

Excellent.

6.7.2 Commutator ofEandt

Again use the crutch of keeping a wave function on the right to avoidmistakes.

[E,t]ψ(x,t) =

(

i ̄h

∂

∂t

t−ti ̄h

∂

∂t

)

ψ(x,t)