Advanced Quantum Mechanics, Notes based on online course given by Leonard Susskind - Lecture 8

Transcription

1 Advanced Quantum Mechanics, Notes based on online course given by Leonard Susskind - Lecture 8 If neutrinos have different masses how do you mix and conserve energy Mass is energy. The eigenstates of energy are linear superpositions of something else; that something else is called type of neutrino. The eigenstates of energy do not mix with each other, it s just that the eigenstates themselves are mixtures of different states For example, in Quantum Mechanics, consider a symmetrical potential like the above. If the middle hump is large enough then we could have a particle trapped in either the left or right hand well. There is a small possibility of the particle leaking through the barrier (quantum tunnelling) but, ignoring that to begin with, we could consider the situation to be like a particle in just the left well, below (barrier is so large that the particle in the left well is not aware of the right hand well).

2 AQMLecture8.nb Ground state energy in the single well has no node and energy E If the situation is symmetric then we can consider the same thing for the right hand well. So there would be another ground state with apparently the same energy E. But the two lowest states do not have the same energy because these wave functions are not exact eigenstates of the energy. For a symmetric potential like this we know that the wave function must be either symmetric or antisymmetric. But we can see for each wave function on its own reflected in the axis of symmetry C can be neither symmetric or anti-symmetric as there is nothing in the other well. C But we can make a symmetric or anti-symmetric combination.

3 AQMLecture8.nb 3 The Symmetric combination (above) is the true ground state and has an energy slightly lower than for our earlier ground state restricted to one of the wells. Next excited state. Theorem is that the first excited state always has one node (place where wave function zero). The anti-symmetric combination above has an energy slightly higher than for our earlier ground state restricted to one of the wells. Remember the sign of a wave function does not contain any real information. Energy When consider leakage and non-infinite barrier energy levels are split Originally when considered barrier infinitely high and no leakage. First two energy levels appeared to be the same So wave functions are mixtures.

4 4 AQMLecture8.nb Get mixed to: (left + right) and (left - right) ; these are the real energy eigenstates and have slightly different energies. Neither left nor right on their own are true energy eigenstates. Eigenstate one: Ψ L + Ψ R has Enery E - ϵ where E is the energy for the infinite barrier/no leakage approximation Eigenstate two: Ψ L - Ψ R has Enery E + ϵ If you follow the evolution of these states (ie solve the time dependent Schrödinger equation) and name them Ψ + and Ψ - Ψ + = Ψ - = (Ψ L + Ψ R ) e -i(e-ϵ) t (Ψ L - Ψ R ) e -i(e+ϵ) t So at t=0, Ψ L = (Ψ + + Ψ - ) If we put an electron in the left hand side wave and see how it evolves after a time t we get (Ψ + e -i(e-ϵ) t + Ψ - e -i(e+ϵ) t ) = e-iet (Ψ + e +iϵt + Ψ - e -iϵt ) Equation There will be a time t when the phases e +iϵt and e -iϵt have evolved so they are opposite e +iϵt e -iϵt At this time (t) e +iϵt = -e -iϵt e +iϵt = - t = π ϵ ( e +iϵ π ϵ = e +i π = cos π + i sin π= -) At this point Equation becomes

5 AQMLecture8.nb 5 e -iet e -iϵt (Ψ + e +iϵt e +iϵt + Ψ - e -iϵt e +iϵt ) = e-iet e -iϵt (Ψ +.-+ Ψ - ) = -e-iet e -iϵt (Ψ + - Ψ - ) Previously we found that Ψ + - Ψ - related to Ψ R So after time t = π the particle is on the other side of the barrier ie has moved from the left hand ϵ well to the right hand one The phenomenon of mixing goes together with the phenomenon of oscillations (oscillations between the two wells) Neutrinos behave like this: Analog of a left hand wave function is an electron neutrino (from decay involving electron) Analog of a right hand wave function is an mu neutrino (from decay involving muon) Real eigenstates of the energy (mass) of the neutrino are combinations of the electron neutrino and the mu neutrino which are entirely analogous to Ψ + and Ψ -. When a neutrino is made it is made in either a decay involving an electron or a decay involving a muon. Can think of electron neutrino as like putting the particle in the left hand well. If wait a while the neutrino will switch from an electron neutrino to a mu neutrino in the same way that our particle, after a given time, was in the right hand well Keep waiting and it continues to oscillate between being an electron and a mu neutrino. Interesting because and electron neutrino can only undergo processes involving an electron and the mu neutrino processes involving a muon. Can use this to devise an experiment to check that the neutrino switches between being an electron neutrino and a mu neutrino. The neutrino masses are equivalent to the energy levels. There is coupling in the Hamiltonian that takes ν e ν μ Ammonia Example Ammonia NH 3 - tetrahedron shape Nitrogen likes to sit a certain distance from plane of the hydrogens (lowest energy) There is a symmetric position the other side to which the nitrogen can tunnel through the plane of the hydrogens

6 6 AQMLecture8.nb N H H H Spin example Take a spin and consider the up and down states. With no field they have the same energy E Add a weak magnetic field (weak to keep precession slow) in the up/down direction up and down energies now slightly different If put magnetic field along x axes Energy states to consider are left and right, but these can be given as combinations of up and down l> = r> = ( u> + d>) ( u> - d>) If we then start with electron in up state and put it in the field, it precesses perpendicularly to the x axis, after a certain time it becomes a down (ie you definitely get down if you measure it). Magnetic field is a term in the Hamiltonian that mixes up and down. In between the times when the it is definitely up and definitely down the probabilities for up and for down are both non-zero but weighted according to how far precession has gone between the two extremes. Mixing of ν e ν μ Mixing of ν e and ν μ seems to be just a parameter but there is no room for it in the standard model. Is a small violation of the standard model This mixing explains the apparent deficit of solar neutrinos - were looking for a certain type and found fewer than expected as some had converted to the other type by the time they had got here and were measured.

7 AQMLecture8.nb 7 00:4:00 Reference to Scientific American, Is the electron a Sphere? The electric dipole moment of an electron was measured - d + Dipole moment = Charge times distance Represents a sort of off centre charge distribution A charge distribution can have many different multi-pole moments, many different shape parameters What kind of shape corresponds to having a dipole? Imbalance of charge on one side relative to a centre Spheroid and Ellipsoid shapes would not have a dipole moment as they are symmetric about about any plane passing through the centre (if charge is uniformly distrubited). But absence of a dipole moment does not mean it is a sphere. In Quantum Mechanics there is something funny about spherical symmetry. Suppose we have a quantum mechanic dumb-bell eg a molecule with atoms Doesn t look like a sphere! Take the ground state of the molecule ie the angular momentum about the centre (ignoring vibration for now) If the molecule has integer Spin then the ground state will have zero angular momentum Angular momentum is the generator of rotations; angular momentum zero says that the wave function is completely symmetrical with respect to all rotations If QM state is symmetric, could maybe say it s a dumb bell in a superposition of states giving a probability distribution that is completely symmetric

8 8 AQMLecture8.nb Doesn t have a dipole moment Single measurement might give lop-sided charge distribution, but how fast would the camera have to be to detect that the charge is not smeared over a sphere? Eg to measure to angular postion of the dumb bell Measuring the angle of the dumb-bell leaves it in an eigenstate but uncertainity principle says: definite angle uncertain angular momentum What if the energy to the first excited state is much larger than the energy of the measuring photon? Then the measuring apparatus simply cannot resolve the orientation. For big dumb bell energy levels are very close together can measure orientation But for a molecule the energy levels are somewhat spread, for mesons ( quarks) it s an even bigger jump, very much greater than energy of optical photon. So measurements would only give you that on average the ground state wave function is spherical. Only could find out it is not a sphere by probing with something with enough energy to kick it into an excited state Ordinary Bosons in ground state typically have zero angular momentum and look spherical But Fermions have half spin and so cant have zero angular momentum, if not zero then cant be rotationally symmetric 00:54:7 This is nothing to do with charge distribution is just that half spin particles have a spin axis, so the electron is not a sphere

9 AQMLecture8.nb 9 N S Imagine current going round making a little electro-magnet. Once you know the electron has a direction then you can ask relative to that direction is there a charge imbalance eg more charge on the north than the south; a displacement of charge along the N-S direction (black blob above) If you align magnetic field and spin, is there an electric dipole moment along that axis? Talking about correlation between the electric charge displacement and the magnetic direction - that s the thing that would be called the electic dipole moment of the electron. If you orient the magnetic dipole with the spin along some axis then is there an electric dipole moment along that axis? Electric dipole moments are forbidden by certain symmetries eg reflection in a plane Consider mirror image of magnet; since little circle of current goes the same direction then the relected magnet points the same way as the original magnet N S N S

10 0 AQMLecture8.nb But if there were some charge imbalance then reflected imbalance would be the same distance behind the mirror as the original imbalance is in front; so the electron the other side has it s charge imbalance in a different position on the electron ie one has an electric dipole moment in the opposite direction to the other. So there cant be any charge imbalance (ie blob in above diagram not in the centre of the arrow) unless there is more than one type of electron If we have mirror image symmetry(also called parity symmetry or reflection symmetry) ie if the laws of physics dont distinguish between left and right hand then there is no dipole moment. But reflection symmetry is not a good symmetry. There is a distinction between left and right. There is another possible symmetry - time-reversal eg if running movie backwards is also a solution of the theory Let s suppose it is a symmetry - what happens when we reflect in time? Again - see below - if time reversal symmetry were a good symmetry the can not have a dipole moment unless there are two types of electron N S S Time Reversed -> N To find out if electron has an electric dipole moment, put it in a magnetic field and see if the two orientations of the electron have different energy. Do same thing in an electric field and see if the two orientations of the spin of the electron have different energies (due to the field) Time reversal symmetry would be enough to tell you that the electron cant have an electric dipole moment. But is time reversal a real symmetry of nature? NO there are processes in the standard model that violate time reversal symmetry; but the magnitude of the effect is very small can do calculations to work out big the dipole moment would need to be before we would expect to be able to detect it. Is possible that the standard model needs correcting and corrections would make the time reversal discrepancy effect much worse

11 AQMLecture8.nb Have in Physics the simultaneous symmetry CPT ie time reversal plus change sign of charge plus reflect in a mirror..5:45 Second Quantisation Connection with Fourier transformation In ordinary QM have wave functions and the squares are probability densities for finding particles at specific positions. You can Fourier transform the position wave function to get the wave function in the momentum basis ψ(x) ψ * (x)ψ(x) = P(x) ψ (p) ψ * (p)ψ (p) = P(p) (those are ~ over the top) ψ(x) and ψ (p) are fourier conjugates of each other ψ (p) = dx π ψ(x) e -ipx Fourier transform ψ(x) = dp π ψ (p) e -ipx Equation A Second Quantization: we have a field operator Ψ(x) = i a ī ψ i (x) sum over every eigenvector Free particle on an infinite axis Ψ i (x) would be e ipx Ψ(x) = steate) dp π a - (p) e -ipx where a - (p) removes particle of momentum p (if none then annihilates the In equation A, ψ (p) is playing the same role as a - (p) in the above equation Ψ * (x) when it acts in a vacuum creates a particle at position x a - (p) = dx π Ψ(x) e -ipx a + (p) = dx π Ψ (x) e ipx Could have called ψ (p) = a - (p) Particles of given position and given momentum are Fourier conjugates of eachother

12 AQMLecture8.nb Ψ is a field operator for a particular particle. Next [a i +,a j - ] = δ ij [Ψ + (x), Ψ - (y)] = δ(x-y) using definition of Ψ (x) = a ī ψ i (x) all creation operators commute with each other all annihilation operators commute with each other Ψ + + Ψ - and i(ψ + + Ψ - ) are observables ie the real and imaginary parts are observables If two things dont commute cant measure them both at the same time ie measuring one changes the other If commutator weren t zero for two different positions then measuring field at A would change field at B. Since information cant travel faster than speed of light this would violate causality Next Time - Fermions. Where anti-commutators are important