1. 3 March 2010Modern Physics II Lecture 81 University of San Francisco Modern Physics for Frommies II The Universe of Schrödinger’s Cat Lecture 8

2. 3 March 2010Modern Physics II Lecture 82 Agenda Administrative Matters Applications Quantum Computing and Quantum Cryptography LASERs and Holography Summing Up

3. 3 March 2010Modern Physics II Lecture 83 Administrative Matters Next Physics and Astronomy Colloquium Today, Wednesday, 3 March 2010 at 4 PM Dr. Seth Shostak, SETI Institute Topic: Strategies in the Search for Extraterrestrial Intelligence Refreshments at 3:30 PM Harney Science Center Room 127 The movie on particles and waves shown at the last class is available on line. Waves and Particles, Annenberg, The Mechanical Universe http://www.learner.org/resources/series42.html

4. 3 March 2010Modern Physics II Lecture 84 Quantum Computing Direct use of quantum mechanical phenomena, e.g. superposition and/or entanglement, to perform operations on data. Potential to be exponentially faster than classical computers, i.e computers on your desk today. Do not allow the computation of functions that are not theoretically computable by classical machines, i.e. they do not alter the Church-Turing thesis. The gain is only in efficiency. Experiments have been carried out in which quantum calculations were executed on a very small number of qubits (quantum bits)

5. 3 March 2010Modern Physics II Lecture 85 Q-Computers differ from other computers (transistors, photonics, DNA) in that it uses specifically QM resources like entanglement. It is conjectured that only Q-computers are capable of achieving the exponential advantage mentioned above. What is it: Classical computer: Memory of bits. 0 or 1 Q-computer: 0 or 1 or any superposition of these is a qubit. Two qubits can be in any superposition of four states. In general, n qubits can be in an arbitrary superposition of 2 n states simultaneously. A classical computer can only be in one of these 2 n states at any one time. A Q-computer operates by manipulating qubits through a fixed sequence of quantum logic gates. Sequence = quantum algorithm

6. 3 March 2010Modern Physics II Lecture 86 How does a classical computer add two numbers Two bit registers each having 2 2 = 4 different states BinaryDecimal 000 011 102 113 Employ an arrangement of Boolean logic gates called a full adder.

7. 3 March 2010Modern Physics II Lecture 87 XOR AND Half Adder:

8. 3 March 2010Modern Physics II Lecture 88 ½ adder ½ adder ½ adder  C0C0 C1C1 A0B0A0B0 A1B1A1B1 Full Adder: Add 1 + 1 which in binary is 01 + 01. 01 + 01 =10 which in decimal is 2

9. 3 March 2010Modern Physics II Lecture 89 Bits vs.qubits: Classical computer with a 3-bit state register State of the machine is a probability distribution over 2 3 =8 3-bit strings: 000, 001, 010, 011, 100, 101, 110, 111 Deterministic computer is in exactly one of these states with probability 1 Probabilistic computer can be in any one of several different states. Probabilities given by 8 non-negative numbers a i. Normalization requires Now consider a 3-qubit quantum computer. Similarly described by an 8-D vector (a 1 … a 8 ), sometimes called a ket in Dirac notation.

10. 3 March 2010Modern Physics II Lecture 810 Now the normalization requirement becomes Additionally, the a i are complex numbers, i.e. they have a phase as well as a magnitude. Im Re ii aiai States being added together will undergo interference. This is a key difference between classical and quantum computing.

11. 3 March 2010Modern Physics II Lecture 811 Superposition of 2 states   |a1>|a1> |a2>|a2> Magnitudes and phases determine the superposition Hologram vs. Photograph

12. 3 March 2010Modern Physics II Lecture 812 Back to our 3-qubit string If we measure the 3 qubits the probability of measuring a particular string will equal the sum of the squared magnitudes of that strings coefficients. Our quantum superposition has collapsed to a classical state. Semi-classically, we can think of the system as being one of the 3- bit strings, we just don’t know which one. Quantum mechanically, we need to keep track of all 2 n complex coefficients in order to see how the quantum system evolves in time. This is a staggering task! 300 qubits define a state described by 2 300 complex numbers. 2 300  10 90 > number of atoms in the observable universe.

13. 3 March 2010Modern Physics II Lecture 813 Operating a Quantum vs. a Classical Computer: In both cases the system must be initialized, e.g. into the all zeros string |000> corresponding to the vector (1, 0, 0,0,0, 0, 0, 0). Classical computer logic operations are in general not reversible e.g. consider an AND gate If the output bit is 0, we can’t tell if the input bit pattern was (1,00, (0,1) or (0,0) In quantum computing the manipulations are said to be unitary, i.e. rotations since they preserve that the sum of the squares add up to 1. Since rotations can br undone by counter rotations, quantum computations should be reversible.

14. 3 March 2010Modern Physics II Lecture 814 Finally, when the computational sequence is complete the result must be read out. Classically, we sample the probability distribution on the 3-bit output register to obtain one definite 3-bit string. Quantum mechanically, we measure the 3-qubit state, which collapses the quantum state to a classical distribution followed by sampling from that distribution. This destroys the quantum state. Schrödinger’s cat cannot be restored to its mixed (unopened box) superposition. Many algorithms only give the correct answer with a certain probability. By repeatedly initializing, running and measuring the probability of getting the correct answer can be increased

15. 3 March 2010Modern Physics II Lecture 815 Quantum Decoherence: Avoidance usually requires isolating the system from its environment. Interactions with the environment cause the system to decohere, entropy increases and information is lost. Typical decoherence times, also called dephasing times, range from nanoseconds at room temperature to seconds at L 4 He temperatures. There are a number of proposed schemes to overcome the decoherence problem. Research is ongoing. There are a very large number of quantum computing candidates. This large number is an indication that the topic is still in its infancy. At the same time, the large number is an indication of a vast amount of flexibility.

16. 3 March 2010Modern Physics II Lecture 816 Quantum Cryptography Problem: Find a mechanism for “Alice” to send a secure message to “Bob” while “Eve” might or might not be eavesdropping. The only encoding scheme that has been proven to provide perfect, unbreakable security is the Vernam cipher or one time pad scheme. Alice sends e-mail to Bob Characters are sent in 7-bit clusters, e.g. a = 1100001, 4 = 0110100 etc. Bob’s computer decodes Alice’s bit stream into letters and numbers. Eve’s computer can do the same. Alice produces a random string of bits, called the key, exactly as long as the message.

17. 3 March 2010Modern Physics II Lecture 817 If the nth bit of the key is 0 then she sends the nth bit of her message as is. If the nth bit of the key is 1 then she sends the nth bit of her message reversed (0  1). e.g. character “a” 1100001standard character “a” 0101101key 1001100code for “a” After encoding, Alice sends Bob the message and the key and he decodes the message.. The key must be sent so that Eve cannot intercept it along with the message. A new key must be sent with each message. There are easy ways to break the code if the same key is used even twice. Hence the name “one time pad”. Problem is not the distribution of messages but the distribution of keys.

18. 3 March 2010Modern Physics II Lecture 818 Can quantum mechanics help? Suppose Alice and Bob set up “EPR distant measurements” from the last lecture. + - + - source Alice and Bob record the exits taken as each pair reaches their analyzers. If Alice records ++-+-, then Bob will record exactly the opposite i.e. --+-+ Alice converts + → 1, - → 0 Bob converts + → 0, - → 1 Alice and Bob now have the same key

19. 3 March 2010Modern Physics II Lecture 819 Not so fast! Snoopy Eve can still intercept the key by inserting a vertical analyzer between the source and Bob. Recall the repeated measurement experiments. Eve and Bob both get the key Exp.1: Measurement of m z and m z + - Exp.Exp. - ignore all none m z = + m B at first analysis and at second Instead, Alice and Bob set up the “random distant measurement”

20. 3 March 2010Modern Physics II Lecture 820 source R G A B C Tilting S-G analyzers in 120º orientations A, B and C (+) exit = red light, (-) exit = green light Left detector flashes green  right atom has m z = + m B Right detector flashes red with a probability of ½. See next slide Conclusion: If orientations are the same, flashes are different always. If orientations are ignored, flashes are different with probability ½. This is, in fact, what is observed.

21. 3 March 2010Modern Physics II Lecture 821 When an atom passes through an analyzer, Alice (or Bob) records both the analyzer orientation and the exit taken. Bob and Alice run the experiment for a long time and then send each other the list of their analyzer orientations. (Each list looks something like BBACABBC…). No need to encode. This tells nothing to Eve. Bob and Alice compare lists. Most cases, analyzers were set to different orientations, but in about 1/3 orientations were the same. Discard exit information, ±, for cases with different orientations and use the remaining cases to construct a key as before. Suppose snoopy Eve tries to intercept the key, as he did before, by placing a vertical analyzer between the source and Bob. Now, when atoms arrive at the tilting analyzers (Alice & Bob) they are no longer entangled. m z = +m B for one and –m B for the other.

22. 3 March 2010Modern Physics II Lecture 822 If the analyzers are vertical (A) this makes no difference. The atoms still emerge from opposite exits. If both detectors are in orientation B, there is some probability (3/8) that the atoms will emerge from the same exit. Alice and Bob have a previous arrangement not to use the entire key they generated above. Instead, Bob mails the first half of his key back to Alice. If it matches Alice’s first half then it is safe for Alice to send her message using the second half of the key. If the two first halves don’t match then Alice knows the key is compromised and does not send her message to Bob. Measurement has caused a change in the information.

23. 3 March 2010Modern Physics II Lecture 823 Polarization If we somehow break this symmetry E. M. radiation is linearly polarized E

24. 3 March 2010Modern Physics II Lecture 824 Polaroid: Material with long complex molecules oriented parallel to one another

25. 3 March 2010Modern Physics II Lecture 825 Add another vertical polarizer Nothing changes. This is like the Stern-Gerlach measure m z twice experiment. What happens if we rotate the intermediate polarizer through some angle, say 45°?

26. 3 March 2010Modern Physics II Lecture 826 Horizontal polarization has been “ regenerated” and some transmission occurs. This is like the tilted Stern-Gerlach experiment. Eve’s snooping, in the form of the intermediate polarizer, affects Bob’s reception of Alice’s key.

27. 3 March 2010Modern Physics II Lecture 827 LASERs Light Amplification by Stimulated Emission of Radiation Produces a narrow intense beam of monochromatic coherent light monochromatic  narrow frequency width coherent  phase is constant across beam cross section absorption stimulated emission

28. 3 March 2010Modern Physics II Lecture 828 For lasing a population inversion must occur so that emissiom dominates over absorption. The excited state must be a metastable state (long lifetime) so that stimulated emission dominates over spontaneous,

29. 3 March 2010Modern Physics II Lecture 829

30. 3 March 2010Modern Physics II Lecture 830 Ruby Laser He-Ne Laser

31. 3 March 2010Modern Physics II Lecture 831 Holography Ordinary photography simply records the intensity as a function of position. This yields a 2-D image. A hologram produces 3-D images via interference, without lenses. Illumination of the film with a laser beam produces a 3-D image.

32. 3 March 2010Modern Physics II Lecture 832 Reflected light from every point on the the object reaches every point on the film. Interference between the two beams allows the film to record both the intensity and the relative phase of the light at each point. Recall the discussion of quantum computing where information storage involved both intensity and phase. Note that information is stored in the pattern as a whole. Destruction of part of the hologram does not result in discrete destruction of part of the image but generates blurriness or loss of detail.

33. 3 March 2010Modern Physics II Lecture 833 Summing Up By no means has this been an exhaustive treatment of a very complicated subject. What did we get done? State of classical physics ca. 1900. and clouds on the horizon Blackbody radiation and Planck’s quantum of action. Einstein, the photoelectric effect and the photon Wave particle duality and the double slit Early atomic models The Bohr atom Wave mechanics Probability vs. determinism. Heisenberg’s uncertainty principle Particles and potentials

34. 3 March 2010Modern Physics II Lecture 834 The hydrogen atom in wave mechanics Atoms with multiple electrons Molecules Superposition and entanglement Applications Glaring ommisions Matrix approach Solid state electronics Nuclear physics etc.

35. 3 March 2010Modern Physics II Lecture 835 Some Comments on Quantum Mechanics God does not play dice with the universe. - Albert Einstein Nobody feels really comfortable with it. - Murray Gell-Mann I can safely say that nobody understands quantum mechanics. Any one who claims to understand quantum mechanics is lying. - Richard Feynman If that electron was up your a__, you’d know where it was. - William Woodruff.

36. 3 March 2010Modern Physics II Lecture 836 Here’s to you, Erwin!