math4104: quantum nonlinear dynamics. lecture three. review … · 2009. 9. 12. · lecture three:...
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![Page 1: MATH4104: Quantum nonlinear dynamics. Lecture Three. Review … · 2009. 9. 12. · Lecture three: A simple introduction to quantum theory. Quantum mechanics with particles. A free](https://reader036.vdocuments.us/reader036/viewer/2022071106/5fe0938b667c74582e2b9359/html5/thumbnails/1.jpg)
Lecture three: A simple introduction to quantum theory.
MATH4104: Quantum nonlinear dynamics.
Lecture Three.
Review of quantum theory.
G J Milburn
The University of Queensland
S2, 2009
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Lecture three: A simple introduction to quantum theory.
Quantum mechanics with particles.
A free particle is one that is not acted on by a force. In onedimension Newtonian physics gives the position as a function oftime as
x(t) = x0 + p0t/m
where x0, p0 are initial position and momentum.
The kinetic energy, depends only on momentum
E =p20
2m
is a constant of the motion
Note: ±p0 have the same energy.
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Lecture three: A simple introduction to quantum theory.
Quantum mechanics with particles.
A phase-space picture for free particles of definite energy, E .A distribution of states, all with the same energy,
p
x
p0
-p0
Note, the position distribution is uniform.
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Lecture three: A simple introduction to quantum theory.
Quantum mechanics with particles.
How to describe position measurements?
P(x)
x=0 x=k a
discrete ‘bins’ of width a
x=-k a
histogram of meas. outcomes
Plot a histograms of number of measurement results that lie ink-th bin.
Can make bin size — a — arbitrarily small.
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Lecture three: A simple introduction to quantum theory.
Classical position distribution for states of definite energy.
p
x
p0
-p0
P(x)
x=0 x=k ax=-k a
The distribution is uniform (within experimental sampling error).
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
p
x
p0
-p0
P(x)
x=0 x=k ax=-k a
classical phase-space picture quantum position distribution
The distribution is oscillatory.There are some bins where the particle is never seen.
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
Explanation: there are two indistinguishable ways to find a particlewith definite energy at the k−th bin:
• in k and travelling with positive momentum and
• in k and travelling with negative momentum.
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
Explanation: there are two indistinguishable ways to find a particlewith definite energy at the k−th bin:
• in k and travelling with positive momentum and
• in k and travelling with negative momentum.
A(k) = A(k : +p0) + A(k : −p0)
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
Explanation: there are two indistinguishable ways to find a particlewith definite energy at the k−th bin:
• in k and travelling with positive momentum and
• in k and travelling with negative momentum.
A(k) = A(k : +p0) + A(k : −p0)
P(k) = |A(k : +p0) + A(k : −p0)|2
= |A(k : +p0)|2 + |A(k : −p0)|2 + 2Re [A(k : +p0)A(k : −p0)∗]
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
Explanation: there are two indistinguishable ways to find a particlewith definite energy at the k−th bin:
• in k and travelling with positive momentum and
• in k and travelling with negative momentum.
A(k) = A(k : +p0) + A(k : −p0)
P(k) = |A(k : +p0) + A(k : −p0)|2
= |A(k : +p0)|2 + |A(k : −p0)|2 + 2Re [A(k : +p0)A(k : −p0)∗]
The last term is not always positive, so can cancel the first twoterms... interference.
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
How to assign the amplitudes — Schrodinger!
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
How to assign the amplitudes — Schrodinger!
Need uniform amplitude but varying phase
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
How to assign the amplitudes — Schrodinger!
Need uniform amplitude but varying phase
A(n : +p0) =1√N
e−2πinap0/h
N is total number of bins and h is Planck’s constant.
Note: ap0 has units of action.
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite energy.
How to assign the amplitudes — Schrodinger!
Need uniform amplitude but varying phase
A(n : +p0) =1√N
e−2πinap0/h
N is total number of bins and h is Planck’s constant.
Note: ap0 has units of action.
In continuum limit: na → x
A(x : +p0) =1√N
e−ixp0/~
~ = h/2π
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite
momentum
Schrodinger’s method for finding the position probability amplitudefor states of definite momentum.
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite
momentum
Schrodinger’s method for finding the position probability amplitudefor states of definite momentum.
Take the classical phase space function, f (x , p), representing thephysical quantity of interest, in this case f (x , p) = p.
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite
momentum
Schrodinger’s method for finding the position probability amplitudefor states of definite momentum.
Take the classical phase space function, f (x , p), representing thephysical quantity of interest, in this case f (x , p) = p.
Replace p by a differential operator
p → p = −i~d
dx
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite
momentum
Schrodinger’s method for finding the position probability amplitudefor states of definite momentum.
Take the classical phase space function, f (x , p), representing thephysical quantity of interest, in this case f (x , p) = p.
Replace p by a differential operator
p → p = −i~d
dx
The solve the eigenvalue problem:(
−i~d
dx
)
ψ(x) = pψ(x)
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Lecture three: A simple introduction to quantum theory.
Quantum position distribution for states of definite
momentum
Schrodinger’s method for finding the position probability amplitudefor states of definite momentum.
Take the classical phase space function, f (x , p), representing thephysical quantity of interest, in this case f (x , p) = p.
Replace p by a differential operator
p → p = −i~d
dx
The solve the eigenvalue problem:(
−i~d
dx
)
ψ(x) = pψ(x)
In this case: ψ(x) ∝ e−ixp/~
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Lecture three: A simple introduction to quantum theory.
Example: particle in a box.
An oscillator, but period depends on energy and motion is notharmonic.
L
x
p
p = 2mE0
L/2-L/2
p = 2mE0
-
The period,
T (E ) =2mL
p= L
√
2m
E
The area is given by 2L√
2mE ,
En =n2h2
8mL2
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Lecture three: A simple introduction to quantum theory.
Example: particle in a box.
x
x=L
x = 0
x = 0 x = L
Probability to find particle outside the box is zero.
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Lecture three: A simple introduction to quantum theory.
Example: particle in a box.
x
x=L
x = 0
x = 0 x = L
Probability to find particle outside the box is zero.
Probability to find particle at x = 0,L is zero.
P(x = 0) = P(x = L) = 0
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Lecture three: A simple introduction to quantum theory.
Example: particle in a box.
x
x=L
x = 0
x = 0 x = L
Probability to find particle outside the box is zero.
Probability to find particle at x = 0,L is zero.
P(x = 0) = P(x = L) = 0
The smallest value of p0 is not zero, (p0)min = h2L
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Lecture three: A simple introduction to quantum theory.
Example: particle in a box.
x
x=L
x = 0
x = 0 x = L
Probability to find particle outside the box is zero.
Probability to find particle at x = 0,L is zero.
P(x = 0) = P(x = L) = 0
The smallest value of p0 is not zero, (p0)min = h2L
The minimum allowed energy is Emin = 12m
(p0)2min = h2
8mL2
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Lecture three: A simple introduction to quantum theory.
Example: particle in a box.
In general, the allowed momenta are: ±nh2L
.
and allowed energies
En =n2h2
8mL2=
n2~
2π2
2mL2
and corresponding position probability amplitudes are
un(x) =
√
2
Lsin
(nπx
L
)
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Lecture three: A simple introduction to quantum theory.
Example: simple harmonic oscillator.
Simple harmonic oscillator: period is independent of energy.
x(t) = x0 cos(2πt/T )
where T is the period of the motion.
Surfaces of constant energy:
E =p2
2m+
k
2x2 (1)
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Lecture three: A simple introduction to quantum theory.
Example: simple harmonic oscillator.
Classical position prob. distribution for a state of definite energy.
-3 -2 -1 1 2 3
2
4
6
8
V(x)= kx 2_
21
E
turning points
xx0-x0
-3 -2 -1 1 2 3
turning points
x0-x0
P(x)
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Lecture three: A simple introduction to quantum theory.
Example: simple harmonic oscillator.
Quantum SHO.Fix energy, then
p(x) =
√
2m
(
E − mω2x2
2
)
The momentum is not fixed.
Schorodinger equation is required.
− ~2
2m
d2
dx2ψ(x) +
mω2
2x2ψ(x) = Eψ(x)
In the form Hψ(x) = Eψ(x) where formally replace p inHamiltonian by
p → p = −i~d
dx
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Lecture three: A simple introduction to quantum theory.
Example: simple harmonic oscillator.
Solution to Schrodinger equation requires restriction on the energy,
En = ~ω(n +1
2) n = 0, 1, 2 . . .
the position measurement probability amplitudes are then
ψn(x) = (2π∆)−1/4 (2nn!)−1/2Hn
(
x√2∆
)
e−x2
4∆
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Lecture three: A simple introduction to quantum theory.
Quantisation of Sommerfeld-Epstein.
Area of the orbit is E .T
Sommerfeld’s rule: only those orbits are allowed for which theaction is an integer multiple of Planck’s constant. The allowedenergies are given by
En = nhf
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Lecture three: A simple introduction to quantum theory.
Example: simple harmonic oscillator.
Check some averages
Pn(x) = |ψn(x)|2
•∫
∞
−∞dx Pn(x) = 1
•∫
∞
−∞dx xPn(x) = 0
•∫
∞
−∞dx x2Pn(x) = ∆(2n + 1)
Prove these results!
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Lecture three: A simple introduction to quantum theory.
Probability for momentum measurements.
The position prob. amp. state of definite momentum is
up(x) ∝ e−ip0x/~
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Lecture three: A simple introduction to quantum theory.
Probability for momentum measurements.
The position prob. amp. state of definite momentum is
up(x) ∝ e−ip0x/~
Consider a general state, with position prob. amp. ψ(x)
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Lecture three: A simple introduction to quantum theory.
Probability for momentum measurements.
The position prob. amp. state of definite momentum is
up(x) ∝ e−ip0x/~
Consider a general state, with position prob. amp. ψ(x)
If find a particle between x and x + dx , we have no knowledge ofits momentum
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Lecture three: A simple introduction to quantum theory.
Probability for momentum measurements.
The position prob. amp. state of definite momentum is
up(x) ∝ e−ip0x/~
Consider a general state, with position prob. amp. ψ(x)
If find a particle between x and x + dx , we have no knowledge ofits momentum
By Feynman’s rule, we need to sum over all states of definitemomentum which correspond to finding the particle between x andx + dx :
ψ(x) =1√2π~
∫
∞
−∞
dp φ(p)e−ixp/~
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Lecture three: A simple introduction to quantum theory.
Probability for momentum measurements.
ψ(x) =1√2π~
∫
∞
−∞
dp φ(p)e−ixp/~
is just a Fourier transform
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Lecture three: A simple introduction to quantum theory.
Probability for momentum measurements.
ψ(x) =1√2π~
∫
∞
−∞
dp φ(p)e−ixp/~
is just a Fourier transform
Invert to get the probability amplitudes for momentum
φ(p) =1√2π~
∫
∞
−∞
dp ψ(x)e ixp/~
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Lecture three: A simple introduction to quantum theory.
Probability for momentum measurements.
ψ(x) =1√2π~
∫
∞
−∞
dp φ(p)e−ixp/~
is just a Fourier transform
Invert to get the probability amplitudes for momentum
φ(p) =1√2π~
∫
∞
−∞
dp ψ(x)e ixp/~
Momentum probability density is: P(p) = |φ(p)|2.