130_notes.dvi

Letscheck the normalization.

∫∞

−∞

|f(x)|^2 dx=

√

1

2 πα

∫∞

−∞

e−

x 22 α

dx=

√

1

2 πα

√

2 απ= 1

Given a normalizedA(k), we get a normalizedf(x).

The RMS deviation, or standard deviation of a Gaussian can be read from the distribution.

P(x) =

1

√

2 πσ^2

e−

(x−X)^2

2 σ^2

Squaringf(x), we get

P(x) =

√

1

2 πα

e−

x 2 α^2

Reading from either the coefficient or the exponential we see that

σx=

√

α

For the width in k-space,

P(k) =

√

2 α

π

e−^2 α(k−k^0 )

2

Reading from the coefficient of the exponential, we get

σk=

1

√

4 α

We can see that as we vary the width in k-space, the width in x-spacevaries to keep the product
constant.

σxσk=

1

2

Translating this into momentum, we get the limit of theHeisenberg Uncertainty Principle.

σxσp=

̄h

2

In fact the Uncertainty Principle states that

σxσp≥

̄h

2

so the Gaussian wave packets seem to saturate the bound!

5.6.5 Time Dependence of a Gaussian Wave Packet*.

Assume we start with our Gaussian (minimum uncertainty) wavepacketA(k) =e−α(k−k^0 )

2
att= 0.
We are not interested in careful normalization here so we will drop constants.

ψ(x,t) =

∫∞

−∞

A(k)ei(kx−ω(k)t)dk