Physics 114 Lecture 9

1
But 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
2
Physics Colloquium
Special (Optional) Lecture

• Quantum Information
• One of the most modern applications of QM
• quantum computing
• quantum communication cryptography,
  teleportation
• quantum metrology
• PGK will give a special 214-level lecture on this
  topic
• Sunday, Feb. 17
• 4 pm, 151 Loomis
• Attendance is optional, but encouraged.

• Spin Based Test-beds for Quantum Information
  Processing
• Professor David G. Cory Massachusetts Institute
  of Technology
• Department of Nuclear Science and Engineering
• February 14, 2008400 pm, Room 141 Loomis Lab

3
Superposition Time-Dependent Quantum States
x
4
Overview

• Superposition of states and particle motion
• Packet States in a Box
• Measurement in quantum physics
• Schrödingers Cat
• Time-Energy Uncertainty Principle

5
Time-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)

• eigenstates

Functions that fit (l 2L/n)
Doesnt fit
6
Review 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).
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Lecture 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
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Lecture 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
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Time-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

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
i2 -1

• 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)

10
Motion of a Free Particle

• Example 1 Wavefunction of a free particle.
• A free particle moves without applied forces so
  we set U(x) 0.

Wavefunction of free particle
Traveling wave solution
Prove it. Take the derivatives
From DeBroglie, p h/l hk. Now we
see that E hw hf
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Complex Wavefunctions

• How can imaginary numbers describe a physical
  system?

Wavefunction of a free particle with momentum p
hk and energy E hw

• What we would measure is in the square of
  Y(x,t) namely, the probability distribution.
  Is it real for this wavefunction?

• For a complex wavefunction, Probability equals
  (absolute value)2 ?Y?2 YY , where Y is
  the complex conjugate of Y. (replace i with i)

A real constant.

• We find that an unconfined free particle with
  momentum hk has an equal probability of being
  anywhere on the x-axis. Of course, if we have
  the particle in our macroscopic apparatus of
  dimension L, then the constant A is roughly
  1/L1/2 in order that ? Y Y dx 1.

12
FYI Wavepackets

• The plane-wave wavefunction for a particles is a
  rather extreme view

• It describes a particle with well defined
  momentum, p hk, but completely uncertain
  position.

• 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

13
Time-dependence of Eigenstates

• Example 2 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.

• But, we are mostly interested in what we
  measure, 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

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Time-dependence of Superpositions

• It is possible that a particle can be in a
  superposition of eigenstates with different
  energy.
• Because superpositions are also solutions of the
  time-dependent SEQ!
• What does it mean that a particle is in two
  states. What is its E?

To answer this, lets see how 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)

15
Particle 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.
Qmtime, shtm, coupled pend., eigenstates
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Particle 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
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Particle 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
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Lecture 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.
At some later time t, what is the probability
density at the exact center of the well? (Hint
What is the symmetry of the two states?) a.
0 b. 1 c. It depends on the precise time
t. 2. What is A2? a. 0.5 b. 0.707 c.
0.866
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Lecture 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. At
some later time t, what is the probability
density at the exact center of the well? (Hint
What is the symmetry of the two states?) a.
0 b. 1 c. It depends on the precise time
t. 2. What is A2? a. 0.5 b. 0.707 c.
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.
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Lecture 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.
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. 2. What is A2? a. 0.5 b. 0.707 c.
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.
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
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Measurements

• The important new result concerning
  superpositions of energy eigenstates is that
  these superpositions represent quantum particles
  that are moving. Consider

• What does it mean that the wavefunction involves
  more than one energy? What do we measure if we
  try to measure E?
• We can still only measure one of the allowed
  energies of the system, 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.
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Lecture 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?
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Lecture 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
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Lecture 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)
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Schrödingers 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.

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Schrödingers 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.
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FYI 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!

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DE 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.
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DE 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.
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Supplementary Problem

• A hydrogen atom is about 0.1 nm in diameter.
  Suppose we wanted to measure the position of its
  electron with an accuracy of, say, 0.01 nm by
  scattering a photon off it.How much energy
  would be transferred to the electron if the
  photon lost most of its energy in the scattering?
• The photon would have to have a wavelength of
  about 0.01 nm.
• The energy of the photon is

Since the binding energy of the electron is only
about 10 eV, this particular measurement doesnt
exactly leave the atom unscathed.
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Supplement 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