Today: Review
Exam topics: All of Chapters 6 & 7 with some emphasis on:
Exam: Tomorrow 7:30 - 9:00pm, DUAN G1B30 Topics: TZD: 6&7 Allowed materials: One letter paper-sized, handwritten cheat-sheet (both sides). Calculator. (No textbooks, cell phones, laptops, radios, etc.) Helproom: Make use of the Helproom hours! Today: 1-5pm Tomorrow: 1-5pm
Exam is on concepts (not on memorizing examples and their solutions!) Why? Solutions can be found in any textbook. (No value in that!) But:
if you can apply your knowledge to new situations, then you learned something valuable. This requires understanding of concepts!
• Wave-functions: deBroglie and ψ(x), Ψ(x,t) • Schrödinger eqn. Why & how? • How to solve Schrödinger eqn. (properties of results, superposition principlet2
Crests (and zero crossings) travel to the right
Crests (& zero crossings) are stationary
x
x
What about ψ(x) and Ψ(x,t)?
What about ψ(x) and Ψ(x,t)?
In class 30 we found the wave h 2 ∂ 2ψ ( x) − = Eψ ( x) function for an electron in free 2 m ∂ x 2 space (V=0):
Remember: ψ(x) is the solution to the time independent Schrödinger equation: h 2 ∂ 2ψ ( x) + V ( x)ψ ( x) = Eψ ( x) − 2m ∂x 2 Ψ(x,t) = ψ(x)⋅e-iEt/ħ is the solution to the Schrödinger equation for time-independent h 2 ∂ 2 Ψ ( x, t ) ∂Ψ ( x, t ) − + V ( x ) Ψ ( x, t ) = ih potentials V(x): ∂t 2m ∂x 2
With the solution:ψ ( x ) and:
A) B) C) D) E)
ψ(x) is traveling, Ψ(x,t) is stationary wave ψ(x) is stationary, Ψ(x,t) is traveling wave ψ(x) and Ψ(x,t) are both stationary waves ψ(x) and Ψ(x,t) are both traveling waves Depends. ψ(x) and/or Ψ(x,t) could be either.
h2
= A cos kx , k2 = E 2m (or: Asinkx)
Ψ ( x, t ) = ψ ( x)e − iEt / h
A bit of algebra, and identity eiθ = cosθ + isinθ gives:
Ψ( x, t) = A cos( kx − ω t) + Ai sin( kx − ω t) Ψ(x,t) is a traveling wave, traveling along the x-axes. ψ(x) is not going anywhere (no time dependence!).
In class 32 we found the wave function for an electron infinite potential well: −
h 2 ∂ 2ψ ( x) = Eψ ( x) 2m ∂x 2
The solution: ψ ( x )
With BC: ψ(0)=ψ(L) =0
= A sin k n x,
(or: Acoskn x)
and: Ψ ( x, t ) = ψ ( x)φ (t ) =
with: k n
=
nπ L
2 nπx −iEnt / h sin( )e L L
What brought us to the Schrödinger eqn? n 5 4 3 2 1
Ψ(x,t) is a standing wave (zeros are stationary @x=0,L) ψ(x) is still not going anywhere (no time dependence).
What brought us to the Schröd. equ? EM-waves
Wave equation:
(1D In free space)
∂2E 1 ∂2E = ∂x 2 c 2 ∂t 2
EM wave function:
Probability to find photon:
E(x,t)
|E|2⋅ dx
E = hf
Energy
Wave equation: −
n=3
h 2 ∂ 2 Ψ ( x, t ) ∂Ψ( x, t ) + V ( x, t ) Ψ ( x, t ) = ih 2m ∂x 2 ∂t
1D In free space:
∂ Ψ ( x, t ) ∂Ψ ( x, t ) = const. ∂x 2 ∂t 2
n=2 n=1
Wave function:
Probability to find particle:
|Ψ|2⋅ dx
Ψ(x,t)
EM waves vs. matter waves Matter Waves (electrons, etc.): Amplitude ψ = “wave function” |Ψ|2 tells us probability of finding particle. Schrödinger Equation:
E ( x, t ) = A sin(kx − ωt ) E ( x, t ) = A cos(kx − ωt )
v = speed of wave Solutions: y(x,t)
Solutions: E(x,t)
Magnitude is spatial: = Vertical displacement of String
Want to know how light or a radio signal propagates? Want to know how light diffracts or reflects etc.? Solve the Wave equation (Maxwell)!
∂2E 1 ∂2E = ∂x 2 c 2 ∂t 2
Solutions are complex sine/cosine waves:
Ψ ( x, t ) = Aei ( kx −ωt ) =
E(x,t)
A(cos(kx − ωt ) + i sin(kx − ωt ))
Want to know where the particle is and how it propagates? Solve the Schrödinger equation! h ∂ Ψ ( x, t ) ∂Ψ ( x, t ) + V ( x , t ) Ψ ( x , t ) = ih 2 ∂t 2m ∂x 2
Ψ(x,t)
Energy
−
c = speed of light
∂Ψ h2 ∂ 2Ψ = ih − 2 2m ∂x ∂t
Solutions are sin/cosine waves:
2
∂2 y 1 ∂2 y = ∂x 2 v 2 ∂t 2
(for V=0)
∂2E 1 ∂2E = ∂x 2 c 2 ∂t 2
x
∂2E 1 ∂2E = ∂x 2 c 2 ∂t 2
Magnitude is non-spatial: = Strength of Electric field
x
EM Waves (light/photons): Amplitude E = electric field E2 tells us probability of finding photon. Maxwell’s Equations:
Vibrations on a string: y
‘Matter is a wave’
λ=h/p
V(x) ψ(x)
Electromagnetic waves: E x
De Broglie: ‘Nature is symmetric’
Matter waves:
Classical wave Equations
V(x) ψ(x) n=3 n=2 n=1
x
How to solve Schrödinger eqn: −
h 2 ∂ 2 Ψ ( x, t ) ∂Ψ ( x, t ) + V ( x , t ) Ψ ( x , t ) = ih 2 ∂t 2m ∂x
1. Figure out what V(x,t) is for the given physical situation. (If V is time-independent, switch to time-independent equ.) 2. Guess or look up functional form of solution Ψ(x,t) (or ψ(x)). 3. Plug in to check if Ψ’s (or ψ’s), and all x’s drop out, leaving equation involving only bunch of constants; showing that trial solution is correct functional form. 4. Figure out what boundary conditions make sense physically. (5. If you solved time-independent equation: multiply ψ(x) by time dependence Φ(t)=e-iEt/ħ to have full solution: Ψ(x,t) has time dependance, even if dV/dt=0.) ∞ 6. Figure out values of constants to meet |Ψ(x)|2dx =1 boundary conditions and normalization -∞
Time-independent eqn.
Time Independent Schrodinger Equation
General, time-dependent form in 1 dimension:
h 2 ∂ 2ψ ( x) − + V ( x)ψ ( x) = Eψ ( x) 2m ∂x 2
h 2 ∂ 2 Ψ ( x, t ) ∂Ψ ( x, t ) − + V ( x , t ) Ψ ( x , t ) = ih 2 ∂t 2m ∂x Often times V(x) is time-independent! That helps a lot! Can use math trick: Solve this one instead:
KE + PE = E (Basically just conservation of Energy) If Ψ=Aei(kx-ωt) (plane wave): dΨ/dx = ikAei(kx-ωt) d2Ψ/dx2 = -k2Aei(kx-ωt) = -k2Ψ(x)
h 2 ∂ 2ψ ( x) − + V ( x)ψ ( x) = Eψ ( x) 2m ∂x 2 Once you found ψ(x) simply multiply by it e-iEt/ħ:
Ψ(x,t)= ψ(x)e-iEt/ħ
de Broglie: ħk = p = mv first term: ½mv2 = KE
This is a solution to the first equation above.
Normalizing wave function (last step in our solution recipe) The probability of finding the particle anywhere in space (i.e. between –∞ ∞ and +∞ ∞) must be 100%! ∞
|Ψ(x,t)|2dx
≡ 1, ∀t
-∞
Example: simplest case, free space V(x) = const.
−
h 2 ∂ 2ψ ( x) + V ( x)ψ ( x) = Eψ ( x) 2m ∂x 2 Smart choice: constant V(x) V(x) ≡ 0!
−
h 2 ∂ 2ψ ( x) = Eψ ( x) 2m ∂x 2
If your solution Ψ(x,t) doesn’t fulfill this: must ‘normalize it!’
∞ |Ψ(x,t)|2dx = C -∞ 2 ∞ Ψ(x,t) Then: dx ≡ 1 C -∞
Ψ(x,t) C
Ψ(x,t)
2
h 2 k = E, 2m
p = hk
k, and therefore E, can take on any value. So almost have solution, but remember still have to include time dependence:
Ψ ( x, t ) = ψ ( x)φ (t )
ψ ( x) = A cos kx, or: ψ ( x) = B sin kx with:
So we found the solution for the time-independent Schrödinger equation for the special case with V(x) = 0:
ψ ( x) = A cos kx ,
Solution:
φ (t ) = e
h 2k 2 =E 2m
No boundary conditions not quantized!
Superposition principle (Similar to what you already know about EM waves!) • If ψ1(x,t) and ψ2(x,t) are both solutions to wave equation, so is Aψ1(x,t) + Bψ2 (x,t). → Superposition principle • E.g. homework (HW8, Q7b) – superposition of waves traveling left and right create standing wave:
E ( x, t ) = A sin(kx − ωt ) − A sin(kx + ωt ) = −2 A sin(kx ) sin(ωt )
− iEt / h
A bit of algebra, and identity eiθ = cosθ + isinθ gives:
Ψ( x, t) = A cos( kx − ω t) + Ai sin( kx − ω t) (Solution of time-independent Schrödinger eqn. with V(x)=0)
• We can also make a “wave packet” by combining plane waves of different energies:
ψ (x,t) = ΣnAnexp(i(knx-ωnt))
Plane Waves vs. Wave Packets
Superposition
Plane Wave: Ψ(x,t) = Aexp(i(kx-ωt)) – Wavelength, momentum, energy well-defined. – Position not well-defined: Amplitude is equal everywhere, so particle could be anywhere! Wave Packet: Ψ(x,t) = ΣnAnexp(i(knx-ωnt))
25
Heisenberg Uncertainty Principle • In math: ∆x∆k ≥ ½ With hk =p: ∆x∆p ≥ h/2 (or better: ∆x∆px ≥ h/2) In time domain: ∆t∆ω ≥ ½ ⇔ ∆t∆E ≥ h/2 • In words: Position and momentum cannot both be determined completely precisely. The more precisely one is determined, the less precisely the other is determined.
– λ, p, E not well-defined: made up of a bunch of different waves, each with a different λ,p,E – x much better defined: amplitude only non-zero in small region of space, so particle can only be found there.
Heisenberg Uncertainty Principle ∆x small ∆p – only one wavelength
∆x medium ∆p – wave packet made of several waves
• This is weird if you think about particles, not very weird if you think about waves. ∆x large ∆p – wave packet made of lots of waves
A slightly different scenario: Plane-wave propagating in x-direction. ∆y: very large ∆py: very small y
Restriction in y direct.: Small ∆y large ∆py wave spreads out strongly in y direction!
x
Weak restriction in y: medium ∆y medium ∆py wave spreads out weakly in y direction!
∆y∆py ≥ h/2
Example: The infinite square-well potential (aka. ‘The rigid box’)
So, a clever physicist (or mathematician) just has to solve (for 0
