Griffith's Introduction To Quantum Mechanics Problem 3.30

Special Problem 2 Peter D Alison May 18, 2009

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Problem 3.30

Suppose A x 2 + a2

Ψ(x, 0) = for constants A and a.

(a) Determine A, by normalizing Ψ(x, 0). We determine A using the normalization condition Z ∞

|Ψ(x, 0)|2 dx = 1

−∞

Z ∞

(

−∞ x2

A )2 dx + a2

Using Mathematica we obtain 1 ax x ∞ A2 π + arctan( ))|−∞ = =1 = 3( 2 2a a + x2 a 2a3 s

A=a

3 2

2 π

(b) Find < x >, < x2 >, and σx (at time t = 0).

< x >=

Z ∞

x|Ψ(x, 0)|2 dx

−∞

1

< x >=

Z ∞

x(

−∞

=−

2(a2

x2

A )2 dx + a2

1 |∞ = 0 + x2 ) −∞

Anyways the function was odd, so this was quite clear. Z ∞

< x2 >=

x2 |Ψ(x, 0)|2 dx

−∞

Z ∞

A )2 dx + a2 −∞ arctan( xa ) ∞ x + |−∞ = a2 =− 2 2 2(a + x ) 2a √ σx = < x2 > − < x >2 √ σ x = a2 − 0 = a 2

< x >=

x2 (

x2

(c) Find the momentum space wave function Φ(x, 0), and check that it is normalized. 1 Z ∞ −ipx/¯h e Ψ(x, 0) dx 2π¯h −∞ 1 Z ∞ −ipx/¯h A Φ(p, 0) = e dx 2π¯ h −∞ x 2 + a2 Once again, using Mathematica to solve this nasty integral we obtain Φ(p, 0) = √

r

Φ(p, 0) =

a − ap e h¯ h ¯

This is a rather nice, pretty function, so notice how easy it will be to calculate expectation values of p with the momentum space wave function. (d) Use Φ(x, 0) to calculate < p >, < p2 >, and σp (at time t = 0). < p >=

Z ∞

p|Φ(x, 0)|2 dp

0

< p >=

Z ∞ 0

ap a p (e− h¯ )2 dp h ¯

2

This integral is −

e−

2ap h ¯

h ¯ (2ap + h ¯) ∞ |0 2 4a h ¯ < p >= 4a

< p2 >=

Z ∞

p2 |Φ(x, 0)|2 dp

0 2

< p >=

Z ∞ 0

ap a p2 (e− h¯ )2 dp h ¯

This integral is −

e−

2ap h ¯

h ¯ (2a2 p2 + 2ap¯h + h ¯) ∞ |0 3 4a h ¯2 < p2 >= 2 4a s

σp =

q

< p2 > − < p >2 =

√ h ¯ 2 ¯ h ¯2 3h −( ) = 2 4a 4a 4 a

(e) Check the Heisenberg uncertainty principle for this state. The uncertainty principle states σx σp ≥

h ¯ 2

√ √ 3h ¯ 3 h ¯ σx σp = (a)( )= h ¯≥ 4 a 4 2 The uncertainty principle is satisfied. Yay.

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