Quantum Physics Lecture 3 Particles as waves and waves as particles Particle in a box -Quantisation of energy

Wave – Particle?? Wave groups and velocity

The Uncertainty Principle

Analysis in terms of waves

Particle in a Box “confinement of moving particle implies energy quantisation”

Particle in a box, making elastic collisions with rigid walls. Cannot go outside box Has a kinetic energy KE (& zero potential energy change in box)

Total energy (non-relativistic)

OK for particle, just bounces between walls.

What if it’s a wave? Means standing waves in box – nodes at walls

c.f. Guitar strings, Waves in cup, etc…

L

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E = KE = 12mov2 =

mov( )2mo

2

=hλ( )

2mo

2

=h2

2moλ2

Particle in a Box (cont.) λ1 = 2L/1 λ2 = 2L/2 λ3 = 2L/3 λ4 = 2L/4 Where n = 1, 2, 3, ….

En: “quantised” energy level. n is the “quantum number”

E ≠ 0 (since v=0 implies infinite λ)

So lowest energy level is E1 called “the ground state”

L

En =h2

2moλn2 =

h2

2mo2Ln( )

2 =n2h2

8moL2

λn =2L

n

E1

E3

E2

Conclude:

Confining a wave restricts possible wavelengths

⇒ only certain λ and hence Energies allowed!

Wider implication:Confining a wave (particle) restricts possible states

e.g. Diffraction – possible directions affected by spacing dThe atom (Lecture 6) and much else….

Waves of what? “everything in the future is a wave,

everything in the past is a particle”

Light: wave of E & M fields Matter: wave of “existence”

But: a particle is at a point, whereas a wave is extended…

So… “where” is the particle?

And… how does a point particle interact with more than one atom at the same time to give diffraction pattern?

Consider waves, groups and packets…

General fomula for Waves 1-D wave: Simple harmonic function of time t and distance xAt x = 0: (amplitude A, frequency f)

At general x? - travelling wave “speed” vwave travels a distance x in time t = x/vAmplitude at any x, at any time t, is amplitude at x = 0 but at the earlier time t - x/v

Also written Velocity v = f λ

Is v the same as the ‘particle velocity’ vp? E=hf p=mvp=h/λ

!?

Clearly not! ⇒ Wave ‘group’ Superposition of many waves

y = Acos2π ft = Acosωt

y = Acos ωt − kx( )

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ω = 2πf k = 2π λ

y = Acos2π f t − x v( ) = Acos2π ft − fx v( ) = Acos2π ft − x λ( )

v = fλ = hf λh=Ep=mc2

mvp=c2

vp

Wave Groups The wave y = A cos (ωt - kx) cannot reasonably be associated with a particle because of its (infinite) extent.Instead, consider a “wave group”

Simplest example of wave group is “beats”: 2 waves of slightly different frequencies:

y = y1 + y2 (using cos a + cos b = 2cos(a+b)/2.cos(a-b)/2)

For Δω and Δk small:

i.e. the basic wave ‘modulated’ by ‘beat’ frequency Δω/2 and wavenumber Δk/2

Velocity of group (beat) Group of many waves:

y1 = Acos ωt − kx( )

y2 = A cos ω + Δω( )t − k + Δk( )x[ ]

y = 2Acos 12 2ω + Δω( )t − 2k + Δk( )x[ ]cos 12 Δωt − Δkx[ ]

y = 2Acos ωt − kx[ ]cos Δω 2t −Δk2 x[ ]

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vg =Δω

2Δk2

=ΔωΔk

vg =dωdk

Group velocity and particle velocity

v g =dωdk

=dω

dvdk

dv

= vAre they the same? ??

So… ω =

E!=mc2

!=

m0c2

! 1− v2 c2

dωdv

=m0v

! 1− v2 c2( )32

k = p!=mv!=

m0v! 1− v2 c2

dkdv

=m0

! 1− v2 c2( )32

Now

And So…

⇒ v g =dω

dvdk

dv

= v

Group velocity equals particle velocity

Group velocity and “dispersion”

Can equate particle velocity (v) with group velocity (vg), not with phase velocity (vp)!

Wave group built from many individual waves; each wave has a

phase velocity (vp). If vp is independent of ω or λ (as for light in a vacuum),

then cannot represent a particle!

Corollary: it is a further requirement of de Broglie wave group that the phase velocity varies with wavelength: This “dispersion” has implications (later)

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vp =ωk

⎛

⎝ ⎜

⎞

⎠ ⎟

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vg =dωdk

= v

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vg =dωdk

=ωk

= vp

Particle versus photon General relations apply to both, some specifics do not:

photon particle mo = 0 mo ≠ 0

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E 2 = mo2c 4 + p2c 2

!ω( )2 = mo2c 4 + !k( )2c 2

!ω( )2 = !k( )2 c2

ω k = fλ = cω k ≠ c

The Uncertainty Principle Where is the particle?: somewhere within the length of thewave group? Most probably in the middle, where |ψ|2 is greatest. To be more precise, need narrower wave group!

Problem:the wavelength of a narrow wave group is poorly defined!(not enough oscillations to measure λ accurately). Therefore, using p = h/λ, momentum is poorly defined…..…need a wider wave group!

Problem:Now λ is well defined but the position of the particle is not!

The Uncertainty Principle Heisenberg (1927): “It is impossible to know the exact position and exact momentum of an object at the same time”

In general, there is an uncertainty in position (Δx) and in momentum (Δp);

Δx and Δp are “inversely” related:

reduce Δx (shorten the wave group), find Δp increases reduce Δp (lengthen the wave group), then Δx increases

Is this quantifiable?

Their product cannot be less than a certain minimum!

Uncertainty Principle analysed Plot ψ versus x, for four special cases (top curves)

“pulse” “wave group” “wave train” “Gaussian”

Bottom curves show corresponding Fourier transforms g(k)eg, “wave train” has a single value of k “wave group” has a narrow range of k “pulse” has a broad range of k

the half-widths Δx (top) and Δk (bottom) inversely related;

minimum value of the product Δx.Δk is the “Gaussian” case!

Uncertainty Principle analysed

minimum value of the product ΔxΔk is the “Gaussian” case!

Now So

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Gaussian :Δx.Δk = 12 ⇒ in general : Δx.Δk ≥ 12

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p =hλ

=hk2π

= !k

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Δp = !Δk

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∴ΔxΔk ≥ 12

⇒ ΔxΔp ≥ !2

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ΔxΔp ≥ h4π