Quantum Mechanics ¯h = 1.0545716 10 kg m sec

with
A = − (p − p 0) 2
4σ 2 − ī ( p 2 )
t
h 2m − px ( it
= −
2m¯h + 1 ) (
4σ 2 p 2 p0
+
2σ 2 + ix¯h
)
p − p 0 2
4σ 2
+ 2itσ2
= −¯hm
(p
4σ 2¯hm − m(p 0¯h + 2ixσ 2 ) 2
¯hm + 2itσ 2 + B ,
B = −mσ2 x 2 + i¯hmp 0 x − i¯hp 0 2 t/2
¯h(¯hm + 2itσ 2 )
and the integral over p leaves as a result
ψ(x, t) ∝ e B
2. By taking the real parto of B, since
|ψ(x, t)| 2 = ψ(x, t)ψ(x, t) ∗ = e B+B∗
B + B ∗ = −mσ2 x 2 + i¯hmp 0 x − i¯hp 0 2 t/2
¯h(¯hm + 2itσ 2 )
+ −mσ2 x 2 − i¯hmp 0 x + i¯hp 0 2 t/2
¯h(¯hm − 2itσ 2 )
= −2σ 2 (mx − p ot) 2
m 2¯h 2 + 4σ 4 t 2
The value of N is obtained by applying Eq.(366) iwth
c = 2σ 2 (mx − p ot) 2
m 2¯h 2 + 4σ 2 t 2
3. The peak of the lump is that combination of x and t for which |ψ| 2 is
maximal, i.e. c vanishes. This happens when mx = p 0 t, in other words
when
x = t p 0
m
4. |ψ(x, t)| 2 is a Gaussian distribution in terms of x, so that the expected
value of x coincides with the peak value (the Gaussian is symmetrical
around its peak). From the last remark made in section 12 we see that
σ(x) 2 must be given by
σ(x) 2 = 1 2c = ¯h2
4σ 2 + σ2 t 2
m 2
141