Title: Physics 114 Lecture 9
1But why must I treat the measuring device
classically? What will happen to me if I
dont?? --Eugene Wigner
When I hear of Schrödingers cat, I reach for my
gun. --Stephen W. Hawking
There is obviously no such limitation I can
measure the energy and look at my watch then I
know both energy and time! --L. D. Landau, on
the time-energy uncertainty principle
2Lab 3 Comments
-Lab 3 meets this week. (So does Discussion, and
there is a quiz, so dont skip...)-If you are
normally assigned to 132 Loomis, go to 257 Loomis
instead.-You will need your Active Directory
Login www.ad.uiuc.edu-You can save a lot of
time by reading the lab ahead of time its a
tutorial on how to draw wavefunctions
- Quantum Information
- One of the most modern applications of QM
- quantum computing, quantum communication
cryptography, teleportation, quantum metrology - Prof. Kwiat will give an optional 214-level
lecture on this topic - Sunday, March 1
- 3 pm, 151 Loomis
3Lecture 9 Superposition Time-Dependent Quantum
States
y (x,t0)2
U?
U?
x
0
x
L
period T 1/f 2t0 with f (E2-E1)/h
Snapshot at two timesProbability density
oscillates backand forth between left and right
sides
4Overview
- Superposition of states and particle motion
- Packet States in a Box
- Measurement in quantum physics
- Schrödingers Cat
- Time-Energy Uncertainty Principle
5Time-independent SEQ
- Up to now, we have considered quantum particles
in stationary states, and have ignored their
time dependence
Remember that these special states were
associated with a single energy (from solution to
the SEQ)
Functions that fit (l 2L/n)
Doesnt fit
6Review Complex Numbers
The equation, eiq cosq isinq, might be new to
you. It is a convenient way to represent complex
numbers. It also (once you are used to it) makes
trigonometry simpler. a) Draw an Argand diagram
of eiq. b) Suppose that q varies with time, q
wt. How does the Argand diagram behave?
Solution
a) b)
a) The Argand diagram of a complex number, A,
puts Re(A) on the x-axis and Im(A) on the y-axis.
Notice the trig relation between the x and y
components. q is the angle of A from the real
axis. In an Argand diagram, eiq looks like a
vector of length 1, and components (cosq,
sinq). b) At t 0, q 0, so A 1 (no
imaginary component). As time progresses, A
rotates counterclockwise with angular frequency w.
ceiq (c and q both real), is a complex number of
magnitude, c. The magnitude of a complex
number, A, is A ?(AA), where A is the
complex conjugate of A Im(A) -Im(A).
7Lecture 9, Act i
We know that 1. What is (-i)i? a. i b.
-1 c. 1 2. What is 1/i? a. 1 b. -i c.
i 3. What is ei?2 ? a. 0 b. e2i? c. 1
8Lecture 9, Act i
We know that 1. What is (-i)i? a. i b.
-1 c. 1 2. What is 1/i? a. 1 b. -i c.
i 3. What is ei?2 ? a. 0 b. e2i? c. 1
(-i)i -i2 -(-1) 1
9Time-dependent SEQ
- To explore how particle wavefunctions evolve with
time, which is useful for a number of
applications as we shall see, we need to consider
the time-dependent SEQ
i2 -1
This equation describes the full time- and space
dependence of a quantum particle in a potential
U(x), replacing the classical particle dynamics
law, Fma
- Important feature Superposition Principle
- The time-dependent SEQ is linear in Y (a constant
times Y is also a solution), and so the
Superposition Principle applies - If Y1 and Y2 are solutions to the time-dependent
SEQ, then so is any linear combination of Y1 and
Y2 (example Y 0.6 Y1 0.8iY2)
10Time-dependence of Energy Eigenstates
- Example 1 Time-evolution of an eigenstate
- An eigenstate ? is described by a single E, so
we can write
- This equation has the solution
- This wavefunction has a complex time-dependence,
containing i.
- But, we are mostly interested in what we
measure, often Y(x,t)2
- As previously stated, the probability density
Y(x,t)2 associated with eigenstates of the SEQ
doesnt change with time.
- Thus the name for states with well-defined
energies Stationary States
11Time-dependence of Superpositions with different
Es
- It is possible that a particle can be in a
superposition of eigenstates with different
energies. - Such superpositions are also solutions of the
time-dependent SEQ! - What does it mean that a particle is in two
states. What is its E?
Lets see how these superpositions evolve with
time.
- Consider a simple example using our trusty
particle in an infinite square well system - A particle is described by a wavefunction
involving a superposition of the two lowest
infinite square well states (n1 and 2) - See nice animations at http//www.falstad.com/qm1d
/
12Particle Motion in a Box
- The probability density is given by Y(x,t)2
You can prove this using Y on the previous page.
Because the cos term oscillates between 1,
Y(x,t)2 oscillates between
Probability
The frequency of oscillation between these two
extremes is w (E2-E1)/h, or f (E2-E1)/h.
This is precisely the frequency of a photon that
would make a transition between the two states.
13Particle Motion in a Box Example
y (x,t0)2
- Consider the numerical example
U?
U?
An electron in the infinite square well potential
is initially (at t0) confined to the left side
of the well, and is described by the following
wavefunction
0
x
L
y (x,t0)2
U?
U?
If the well width is L 0.5 nm, determine the
time to it takes for the particle to move to
the right side of the well.
0
x
L
14Particle Motion in a Box Example
- Consider the numerical example
An electron in the infinite square well potential
is initially (at t0) confined to the left side
of the well, and is described by the following
wavefunction
y (x,t0)2
U?
U?
If the well width is L 0.5 nm, determine the
time to it takes for the particle to move to
the right side of the well.
0
x
L
period T 1/f 2t0 with f (E2-E1)/h
15Interference beats
- Note that all the motion comes from changing
interference terms between components with
different Es. - The frequencies of those changes are the
frequencies at which the interference terms
change the beat frequencies between the
different components - beat frequency frequency difference e.g.,
f2-f1 - So the important energies determining how things
move are energy differences hf1-hf2, etc. - E1-E2 etc.
- Just like in Newtons physics absolute energies
arent important (you can pick where to call U0
by convenience), its energy differences that are
important. - http//www.falstad.com/qm1d/
16Normalizing Superpositions
- Its a mathematical fact that any two eigenstates
with different eigenvalues (of any measurable,
including energy) are ORTHOGONAL - Meaning
To normalize a superposition of normalized energy
eigenstates, make the sum of the absolute
squares of their coefficients equal 1.
a2 is the probability that the particle would
be found in state 1 b2 is the probability
that the particle would be found in state
2 a2 b2 1 a2 and b2 dont change in
time because y1 and y2 are energy eigenstates!
17Lecture 9, Act 2
Consider a particle in an infinite potential
well, which at t 0 is in the
state with ?2(x) and ?4(x) both normalized. 1.
What is A2? a. 0.5 b. 0.707 c. 0.866 2.
At some later time t, what is the probability
density at the exact center of the well? a.
0 b. 1 c. It depends on the precise time t.
18Lecture 9, Act 2
Consider a particle in an infinite potential
well, which at t 0 is in the
state with ?2(x) and ?4(x) both normalized. 1.
What is A2? a. 0.5 b. 0.707 c. 0.866 2.
At some later time t, what is the probability
density at the exact center of the well? a.
0 b. 1 c. It depends on the precise time t.
As stated, the question is not well posed, since
A2 could be complex. However, lets assume that
A2 is real (or that we were asked for A2). We
are told that ?2(x) and ?4(x) are both
normalized. Therefore 0.52 A22 1, or A2
sqrt(1 0.25) sqrt(0.75) 0.866
19Lecture 9, Act 2
Consider a particle in an infinite potential
well, which at t 0 is in the
state with ?2(x) and ?4(x) both normalized. 1.
What is A2? a. 0.5 b. 0.707 c. 0.866 2.
At some later time t, what is the probability
density at the exact center of the well? a.
0 b. 1 c. It depends on the precise time t.
As stated, the question is not well posed, since
A2 could be complex. However, lets assume that
A2 is real (or that we were asked for A2). We
are told that ?2(x) and ?4(x) are both
normalized. Therefore 0.52 A22 1, or A2
sqrt(1 0.25) sqrt(0.75) 0.866
In general, the probability distribution of a
superposition of energy eigenstates does depend
on time. However, both of these solutions always
have a node at L/2. Therefore, every possible
superposition of them also has a node at L/2.
20Measurements of E or x
- The important new result concerning
superpositions of energy eigenstates is that
these superpositions represent quantum particles
that are moving. Consider
- But what happens if we try to measure E on a
wavefunction which involves more than one energy?
- We can still only measure one of the allowed
energies, i.e., one of the eigenstate energies
(e.g., only E1 or E2 in ?(x,t) above)!
If Y(x,t) is normalized, A12 and A22 give us
the probabilities that energies E1 and E2,
respectively, will be measured in an experiment!
- What about measurements of Y(x,t)2 ?
If we make a large of measurements at time t,
the result should look like the probability
function Y(x,t)2 at that time.
21Lecture 9, Act 3
Consider a particle in an infinite potential
well, which at t 0 is in the
state with ?2(x) and ?4(x) both normalized. 1.
We now measure the energy of the particle. What
value is observed? a. E2 b. E4 c. 0.25 E2
0.75 E4 d. It depends on when we measure the
energy. 2. If E2 is observed, what is the
state of the particle after the
measurement?
22Lecture 9, Act 3
Consider a particle in an infinite potential
well, which at t 0 is in the
state with ?2(x) and ?4(x) both normalized. 1.
We now measure the energy of the particle. What
value is observed? a. E2 b. E4 c. 0.25 E2
0.75 E4 d. It depends on when we measure the
energy. 2. If E2 is observed, what is the
state of the particle after the measurement?
Since we are measuring energy, we can only get
one of the eigen-energies, E2 or E4. The
probability of measuring E2 is 25. The
probability of measuring E4 is 75. The average
energy (if we were to measure a large ensemble of
similar particles) is the weighted sum of the
energies 0.25 E2 0.75 E4
23Lecture 9, Act 3
Consider a particle in an infinite potential
well, which at t 0 is in the
state with ?2(x) and ?4(x) both normalized. 1.
We now measure the energy of the particle. What
value is observed? a. E2 b. E4 c. 0.25 E2
0.75 E4 d. It depends on when we measure the
energy. 2. If E2 is observed, what is the
state of the particle after the measurement?
Since we are measuring energy, we can only get
one of the eigen-energies, E2 or E4. The
probability of measuring E2 is 25. The
probability of measuring E4 is 75. The average
energy (if we were to measure a large ensemble of
similar particles) is the weighted sum of the
energies 0.25 E2 0.75 E4
Our measurement has collapsed the original
wavefunction. It is now simply in the eigenstate
associated with the measurement result E2 ?2(x)
24Schrödingers CatHow far do we take
superpositions?
- We now know that we can put a quantum object into
a superposition of states. - We also know that we can measure it. But if
quantum mechanics is completely correct,
shouldnt our measurement apparatus end up in a
superposition state too? - This result is best exemplified by the famous
Schrödingers cat paradox
- A radioactive nucleus can decay, emitting an
alpha particle. - The alpha particle is detected with a geiger
counter, whose firing releases a hammer, which
breaks a bottle, which releases cyanide, which
kills a cat.
25Schrödingers Cat
- Strictly according to QM, because we cant know
without measuring when the decay happened, until
we look inside the box, the cat is in a
superposition of being both alive and dead!
http//www.physics.uiuc.edu/Research/QI/Photonics/
movies/cat.swf
- And in fact, strictly according to QM, we then
are put into a quantum superposition of having
seen a live and a dead cat!! - Where does it end?!?
- it doesnt end (wavefunction of the universe)
- there is some change in physics (quantum ?
classical) - many-worlds interpretation
- In any event, the correlations to the rest of the
system cause decoherence and the appearance of
collapse.
More correctly, the atom and the cat and us
become entangled.
26Motion of a Free Particle
- Wavefunction of a free particle with single K.E.
- A free particle moves without applied forces so
we set U(x) 0.
Wavefunction of free particle
Traveling wave solution
Check it. Take the derivatives
From DeBroglie, p hk h/l. Now
we see that E hw hf
27Wavepackets
- The plane-wave wavefunction for a free particle
is problematic
- It describes a particle with well defined
momentum, p hk, but completely uncertain
position. Its equally likely to be anywhere!
- By adding together (superposing) waves with a
range of wave vectors Dk, we can produce a
localized wave packet. We can imagine such a
packet in space
- We saw in Lecture 6 that the required spread in
k-vectors (and by p hk, momentum states) is
determined by the Heisenberg Uncertainty
Principle DpDx h
Well now talk about states with some k, but
really we mean packets witha narrow range of
ks. They do have an (approximate) position so it
means something to say where are.
28Free particle motion
- Consider a free particle that is moving in space
- At any time it is located in some region of space
Dx - The particle must have a spread in momenta as
required by the uncertainty principle - Dp gt
h/Dx - The different components (different ps,
different Es) of the wave packet move at
different velocities they change phase at
different rates (fE/h). - At a point where the components had been in
phase, they wont be any more. - At a point where they had been out of phase, they
will become in phase.
Dx
Position of maximum probability moves and the
width of probability distribution spreads out
NOTE This only happens for massive particles,
not photons ?
29FYI Time-Energy Uncertainty Principle
- Now that we are considering time-dependent
problems, it is a good time to introduce another
application of the Heisenberg Uncertainty
Principle, based on measurements of energy and
time. We start from our previous result - Sometimes this is further transformed as follows
- The last line is a standard result from Fourier
wave analysis this should not surprise us the
Uncertainty Principle arises simply because
particles behave as waves!
30FYIDE Dt Uncertainty Principle Example
A particular optical fiber transmits light over
the range 1300-1600 nm (corresponding to a
frequency range of 2.3x1014 Hz to 1.9x1014 Hz).
How long (approximately) is the shortest pulse
that can propagate down this fiber?
This problem obviously does not require quantum
mechanics per se. However, due to the
Correspondence Principle, the quantum constraints
on single photons also apply at the
classical-pulse level.
31FYIDE Dt Uncertainty Principle Example
A particular optical fiber transmits light over
the range 1300-1600 nm (corresponding to a
frequency range of 2.3x1014 Hz to 1.9x1014 Hz).
How long (approximately) is the shortest pulse
that can propagate down this fiber?
Note This means the upper limit to data
transmission is 1/(4fs) 2.5x1014 bits/second
250 Gb/s
This problem obviously does not require quantum
mechanics per se. However, due to the
Correspondence Principle, the quantum constraints
on single photons also apply at the
classical-pulse level.
32Supplement Quantum Information References
- Quantum computing
- employs superpositions of quantum states -
astoundingly good for certain parallel
computation, may use entangled states for error
checking - www.newscientist.com/nsplus/insight/quantum/48.htm
l - http//www.cs.caltech.edu/westside/quantum-intro.
html
- Quantum cryptography
- employs single photons or entangled pairs of
photons to generate a secret key. Can determine
if there has been eavesdropping in information
transfer - cam.qubit.org/articles/crypto/quantum.php
- library.lanl.gov/cgi-bin/getfile?00783355.pdf
- Quantum teleportation
- employs an entangled state to produce an exact
replica of a third quantum state at a different
point in space - www.research.ibm.com/quantuminfo/teleportation
- www.quantum.univie.ac.at/research/photonentangle/t
eleport/index.html - Scientific American, April 2000
33Supplement Free particle motion
- It turns out (next slide) that the constructive
interference region for a matter wavepacket moves
at the group velocity v h/lm p/m - So theres a simple correspondence between the
quantum picture and our classical picture of
particles moving around with momentum p mv. - But the quantum packet will spread out in the
long run, since it has a range of p, so the
correspondence is never perfect.
Dx
Position of maximum probability moves and the
width of probability distribution spreads out
34Supplement Group velocity
- Say a wave-packet starts out at x0 at t0.
- meaning each harmonic component has the same
phase there. - After time t
- the harmonic component at w1 will have changed
phase by tw1 - the harmonic component at w2 will have changed
phase by tw2 - The phase difference between these components at
x0 will now be t(w2- w1) To find the point x
where theyre in phase, we need to find where the
phase difference from moving downstream by x
cancels that - t(w2- w1)x(k2-k1) Or for small differences in
w,k tdwxdk - Result
- vgx/tdw/dk
- In this case vgp/m