130_notes.dvi

Our Fourier Transform can now be read to say that weadd up states of definite momentum
to getψ(x)

ψ(x) =

∫∞

−∞

φ(p)up(x)dp

and we add up states of definite position to getφ(p).

φ(p) =

∫∞

−∞

ψ(x)vx(p)dx

There is a more abstract way to write these states. Using the notation of Dirac, the state with
definite momentump 0 ,up 0 (x) =√ 21 π ̄heip^0 x/ ̄hmight be written as

|p 0 〉

and the state with definite positionx 1 ,vx 1 (p) =√ 21 πh ̄e−ipx^1 / ̄hmight be written

|x 1 〉.

The arbitrary state represented by eitherψ(x) orφ(p), might be written simple as

|ψ〉.

The actual wave functionψ(x) would be written as

ψ(x) =〈x|ψ〉.

This gives us the amplitude to be atxfor any value ofx.

We will find that there are other ways to represent Quantum states. This was a preview. We will
spend more time on Dirac Bra-ket notation (See section 6.4) later.

5.5 Time Development of a Gaussian Wave Packet*

So far, we have performed our Fourier Transforms att= 0 and looked at the result only att= 0.
We will now put time back into the wave function and look at the wave packet at later times. We
will see that the behavior of photons and non-relativistic electronsis quite different.

Assume westart with our Gaussian(minimum uncertainty) wavepacketA(k) =e−α(k−k^0 )
2
at
t= 0. We can do the Fourier Transform to position space, including thetime dependence.

ψ(x,t) =

∫∞

−∞

A(k)ei(kx−ω(k)t)dk

We write explicitly thatωdepends onk. For our free particle, this just means that the energy
depends on the momentum. For a photon,E =pc, so ̄hω = ̄hkc, and henceω=kc. For an

non-relativistic electron,E=p

2

2 m, so ̄hω=

̄h^2 k^2

2 m, and henceω=

̄hk^2

2 m.