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So let’s start today with problem set five which is on Pauli matrices and spin 1/2 and
so on and so forth. I mentioned already that the smallest non-zero quantum number for the
angular momentum is a ½. . And spin 1/2 or j = 1/2 implies 2S + 1 = 2.
So everything is in terms of 2 by 2 matrices. and physically, the first example that we
have of 2 dimensional linear vector space is the spin states of the electron which are
described by the 2 ket vectors. We have pointed out the up corresponds to S = ½, Sz = ½;
all in units of h cross
and similarly the down state is |1/2, - ½>. So recall, our notation was j and m where
j is always the quantum number associated with the J squared. This has always got eigenvalue
h cross squared j times j + 1 J squared rather and Jz has eigenvalue h cross m.
So these were 2 mutually commuting observables and my angular momentum states are labeled
by labeling angular momentum quantum number and projection quantum number n. for the spin
1/2 particle like an electron, this number is always a ½. So I don’t really have to
write this we just have to write what Sz is. This corresponds to eigenvalues + 1/2 h cross
- 1/2 h cross and pictorially one would like to say that this corresponds to spin up or
a spin down. so i use this kind of notation for it.
Now we already know that the spin operator of the electron can be written as h cross
over 2 times the 3 Pauli matrices, sigma and these matrices have standard representation.
Sigma 1 = 0 1 1 0, sigma 2 0 - i i 0, sigma 3 is 1 0 0 – 1. That’s a very useful representation
for spin 1/2 because each of these sigma matrices, the square is = the identity matrix. So sigma
i squared = the identity matrix. And there will be interesting commutation relation.
In fact sigma i sigma j = i epsilon ijk sigma k. so sigma 1 sigma 2 is = i times sigma 3
and so on. It’s easy to see that [sigma i, sigma j] = 2 i epsilon ijk sigma k and
the anti-commutator, sigma i sigma j + 0 if i is
not = j and if i = j, then its = 2 I delta ij. What are the eigenvalues of the sigma
S? It’s 1 and – 1. What about sigma 1 and sigma 2? What are their eigenvalues? They
are also 1 and – 1. They all have the same eigenvalues 1 and – 1. The square of each
of them is the identity matrix. They are linearly independent of each other. The 3 sigmas are
linearly independent and no sigma can be written as a linear combination of the other 2. And
you could take any arbitrary matrix A.
Any 2 by 2 matrix A can be uniquely written as a combination of the sigma’s and the
unit matrix. So you can always write this as a 0 I+ a i sigma i summation i = 1 to 3.
This is unique. Any 2 by 2 matrix can be expanded in this form. What’s a 0? What property
of A does a0 reflect? It’s the trace. What about the trace of the sigma matrices? It
is 0. So it’s clear if you take trace on either sides, the trace of A is twice is a0.
So a 0 = 1/2 A. what about ai? If i formally want to invert this and write it, what should
I write?
Suppose you want to find a1, what would you do? I’d multiply both sides by sigma 1 and
then it becomes sigma 1 A on this side = a 0 sigma1 + a1 I+ a 2 times sigma 1 sigma 2 + a 3 times sigma 1 and sigma
3 and then take trace. So it is clear that you end up with a 1 = 1/2 trace A sigma 1
and so on. So formally that’s the inversion. Similarly for a 2, a 3, etc. so its evident
immediately that any 2 by 2 matrix can be uniquely expanded in terms of the sigma matrices.
The great advantage of this expansion is that the unit matrix on the sigma matrices are
all Hermitian matrices whereas the natural basis is not Hermitian.
When i write a normal 2 by 2 matrix the natural basis for a 2 by 2 matrix is given by this.
This basis is not Hermitian. This poses a lot of difficulty. On the other hand, the
sigma matrices are Hermitian. That’s the great advantage. So that’s one of the reasons
for why one uses the sigma matrices. The other thing is that exponentiation becomes very
simple because e to the power i a dot sigma, where a is an ordinary vector; (a1, a2, a3).
this has a very simple expansion in terms of the cosine and sin of the modulus of a.
it’s easy to verify because you expand this and in the first term for example, it is the
identity + i times a1 sigma 1 + a2 sigma 2 + a3 sigma 3 and then the same thing squared.
but the squaring of a dot sigma will result in terms of the form a1 a2 times sigma 1 sigma
2 + sigma 2 sigma 1 and that will vanish because it anti-commutes. So only the terms which
are squares of the sigma matrices would contribute when you exponentiate.
And the result is this 2 is a 2 by 2 matrix. This is also a 2 by 2 matrix. So, it should
also be expandable in terms of the unit matrix and the sigma matrices and the expansion is
cos a times unit matrix + i a dot sigma over a sin a, where a stands for the square root
of a 1 squared + a 2 squared + a 3 squared. It’s a very useful representation. In fact
you can now see what a rotation is going to do. When you rotate the coordinate system,
you could ask what happens to any operator in this space. It will be the representative
of the rotation operator times the operator times the inverse of the rotation operator
and this will be easily calculable using this identity. Now what i have given here in this
problem set is a whole lot of vector identities using sigma matrices. Among them, let me write
down a couple of them because they are useful for what we are going to do later.
We have a dot sigma times b dot sigma = a dot b times the unit matrix + i times a cross
p dot sigma. This is a scalar quantity and that’s a scalar quantity but each of them
is matrix valued. that is the unit matrix + something which depends on the sigma matrices,
the coefficient is a cross b. so the commutator of a dot sigma with b dot sigma is twice i
times a cross b dot c because it would just be the inverted thing and b cross a will be
- a cross b and that cancels the - sign of the commutator. Then the other one that you
need is sigma with a dot sigma = 2 i a cross sigma.
And similarly sigma cross any unit vector sigma dot the same unit vector is - i sigma
- i times sigma dot n n. all these identities are easily proved by using the commutation
relations between the sigma’s. So much for some mathematical aspects of the sigma matrices.
I have already explained where spin of a particle comes from. in fact the particles are really
the wave functions must transform in a given manner under transformations of the inhomogeneous
Lorentz group namely under rotations, velocity transformations, translations of the space
time axes. This implies that all these wave functions are labeled by a certain set of
quantum numbers among which are the rest mass of a particle and the spin of the particle
or the intrinsic angular momentum of the particle. And in that classification, particles like
the electron, the neutron, proton etc have spin quantum number ½. Particles like the
photon have spin quantum number 1.
This is a whole host of particles with other spins. 1/2 integer spin particles are called
fermions because when you put a collection of them together, they obey Fermi Dirac statistics
where as integer spin particles are called bosons because a collection of them obeys
Bose Einstein statistics. And a little later, we will talk about the differences between
the 2 statistics. Right now i would like to point out that the way you prove the spin
of a single particle like an electron is by using a magnetic field and the reason is as
follows. We talked little bit about this earlier. so let me continue on the same lines.
The spin operator of an electron, Se = h cross over 2 sigma. The eigenvalues of any component are guaranteed
to be the eigenvalues of sigma dot n, where n is a unit vector and they are + or - 1 multiplied
by h cross over 2. Now what we do know is that the intrinsic magnetic moment operator
of an electron mu e this is = a factor called the g factor of the electron, multiplied by
the charge of the electron divided by twice the mass of the electron. e over 2 m is a
standard gyromagnetic ratio, if you have a classical orbiting particle of charge e and
mass m. that’s the one proportionality constant between its magnetic moment due to the current
loop that it forms when it revolves in an orbit and the orbital angular momentum.
That relation is generalized to g e over 2 m e times the spin operator of the electron.
So if i put this in, this = g, the charge of the electron twice the mass of the electron
h cross over 2 sigma. Therefore the eigenvalues of any component of the intrinsic magnetic
dipole moment of the electron are given by + or - this quantity because any component
of it has eigenvalues + or – 1. But the g factor of the electron itself turns out
to be 2 from relativistic quantum mechanics. So this is input information. There is no
way of deriving this g factor in classical physics. It’s not a classical concept or
even in non-relativistic quantum mechanics.
It has to come from relativistic quantum mechanics which predicts that g is equal to 2. Actually
the g, if you measure it exactly, is not 2. It has a small correction and this correction
is known to a very large number of places and has been verified. The correction comes
due to what are called radiative corrections due to quantum field theory. We are not concerned
with that right now. We will simply put in the fact that g is = 2 and therefore this
becomes = - modulus of charge of the electron e, h cross over 2 m e times sigma. And this
is = - the Bohr Magneton times sigma. so if you are in Sz = + 1/2 state, which corresponds
to saying sigma 3 = 1, then the magnetic moment in the z direction has values - the Bohr magneton.
Otherwise it has a value + the Bohr magnetic in the other state. And this difference in
the sign between the magnetic moment and the spin comes about obviously because a charge
of the electron is negative. So this is really all one needs to know.
The moment you put this electron in the magnetic field, if it’s a uniform magnetic field,
then it experiences a torque. The dipole moment experiences a torque. And the magnetic part
of the Hamiltonian is = - mu e dot B. that’s the potential energy and if i put this in,
this is = Bohr Magneton and sigma dot B, the - sign goes away and this is your magnetic
Hamiltonian. What are the eigenvalues of this Hamiltonian? Well, sigma dot B is a component
of sigma in some direction multiplied by the magnitude of B. so it’s clear you take this
to be the unit vector and you divide it by its magnitude. You would get the unit vector
and that portion has eigenvalues + or – 1. So it is clear that the eigenvalues are = +
or - B times mu B, where B is the magnitude of the magnetic field. Since we know that
sigma dot any unit vector has eigenvalues + or – 1. It’s + B mu if the spin points
along the direction of B and - if it points in the opposite direction. So at this level
it’s completely trivial. Now what we would like to do is the following.
Here is my Hamiltonian and i start in an arbitrary state of the electron and remember that this
up state here can be represented by 1 0 and the down state is conveniently represented
by 0 1. I choose the z axis as my axis of quantization for the moment. Then the complete
set of spin states of the electron is comprised of just 2 orthonormal states which are the
up state and the down states and they are each normalized.
So now i ask, suppose i start in an arbitrary spin state of the electron and here is my
arbitrary state. Chi = (a b).That’s an arbitrary state of the electron. I would like to start
with a normalized state always. a and b are 2 complex numbers which satisfy mod a squared
+ mod b squared is 1. In this state what is the probability that the Sz component of electron
has eigenvalue + 1/2 h cross?
How would you compute that? Well, it’s clear that the probability we ask for p up; we want
the probability at the spin is up, this is = the modulus squared of this probability
amplitude by definition. This is the probability amplitude that when it’s in the state chi.
it is in state up and the mod squared of it is the probability itself. So what’s this
= mod a squared and similarly p down = b squared. So that’s my initial state. Now I switch
on a magnetic field in some arbitrary direction. What would happen to the state? Would these
probabilities be the same? First let’s ask a very simple question. What if i switch on
field in the z direction itself? Will chi of P change? So in all cases here is the Hamiltonian.
So what would happen to that state? Let’s say I start with 0 state t = 0. What would
happen to this state?
So I put B and this implies the Hamiltonian is B mu B sigma 3. That’s the Hamiltonian.
So I ask chi(t) = e to the - i by h cross Ht chi(0). this is = e to the - i over h cross,
the Hamiltonian is B mu Bohr Magneton, which is modulus e h cross over 2 m e; you just
expand what this mu B is, and then sigma 3, this t as well acting on a b.
so what does this give us? This is an h cross which cancels out and what does this give
us? i need to use that expansion that i wrote down in terms of cosine and sin and so on.
Let’s write that out just for convenience.
e to the i a dot sigma is = cos a times I - i a dot sigma over a sin a. So let’s use
that and what’s a in this problem? a dot sigma, a along a z direction is just 1 component
here. So what does that give you? a in this problem is = modulus e B over 2 m e multiplied
by time. So what would happen here? What’s eB over 2 m? Its 1/2 the cyclotron frequency
because remember there is a difference between a classical orbiting particle and the spinning
electron because this has a g factor = 2. So this constantly there is going to be change
of factor of 2 everywhere. So we have to be careful. this is = cosine omegac t over 2
as a unit operator - i a dot sigma and the a parts simply cancels out and then you get
sigma 3 sine omega c t over 2 acting on a b. now it’s trivial to find out. So what
does this give you? Its e to the - 2 on a and the other 1 is the bottom element because with a - sign so that becomes
a +. Do the probabilities change?
It can’t because your Hamiltonian is also diagonal in sigma 3. So the probabilities
don’t change. There is not flip at all. Mod a squared and mod b squared are exactly
the same at mod a(t) whole squared and mod b (t) whole squared. If i had the field which
had any component other than the 3 component, then of course there would be an off diagonal
element here. For example, there is a portion sigma 1 and a sigma 2, they would have off
diagonal elements here and they would mix up things between these two and then you would
get oscillations. Then the probabilities would indeed change. This means that the spin up
and spin down, the total probability always remains 1 but there are transitions between
the 2 spin states caused by a transverse speed. What we have to ask for here is somewhat more
interesting than and that’s the following.
The idea was to find out how this magnetic dipole behaves in a magnetic field and what
kind of precessional motion it undergoes. Let me point this out carefully. Consider
a classical particle orbiting in a radius r. the charge of this particle is e and the
mass of the particle is m. then what’s the relation between the magnetic moment and the
orbital angular moment in this case? The orbital angular momentum is m r squared times the
angular velocity of this particle which is 2 pi over the time period, T in the direction
of r cross p. So let’s say some unit vector n. the magnetic dipole moment of this particle
by Ampere's law is the area of the current loop multiplied by the current. The area of
the current loop is pi r square and the current is e over T and it’s also in this direction.
So this gives you a relation between L and mu. the gyromagnetic ratio not surprisingly
is e over 2 m. this is a 2 m factor here and there is an e here so e over 2 m is a gyromagnetic
ratio. Mod mu = over 2 m mod L.
What’s dL over dt, the rate of change of angular momentum? It’s the torque. It’s
mu cross B. what’s the Larmor frequency of precision? What we have to do is to convert
this L. So if we call this constant gamma, so we write mu = gamma L in this fashion,
then L is mu by gamma. So this would imply d over dt mu = gamma times mu cross B. This
implies frequency of precession modulus gamma B. that’s trivial to see from this equation.
Once you get an equation like this, you know its precessional motion and the frequency
of motion is just the magnitude of gamma times magnitude of B. so in the classical case this
would become modulus eB over 2 m. what happens if we put a quantum mechanical particle?
Then for the spinning particle, we have to compute what ds over dt. We can’t just write
it down as torque because that’s a classical equation of motion. How do i compute it? This
is the Heisenberg equation of motion. So I have ih cross dS e over dt on the side is
= the commutator of the spin operator with the Hamiltonian. This is = h cross over 2
sigma that’s S e with the Hamiltonian. But for the Hamiltonian we had a very simple expression
in an arbitrary magnetic field. This is a Bohr magneton multiplied by sigma dot B. this
is what we discovered. So this says dS e over dt is = - i, i am going to bring this i to
this side. - i the h cross cancels, the factor 2 remains, muB remains and then the commutator
of sigma with sigma dot B. but we have an convenient expression for it.
Sigma with a dot sigma is 2 i a cross sigma. These are vectors. These stand for all 3 components
together and they don’t commute with each other. so that’s the reason you get non
0 answers here but you could write this as a dot sigma or sigma dot a. that doesn’t
matter because a is an ordinary vector just as the magnetic field B is an ordinary vector.
So you got a 2 i B cross sigma. This is multiplied by 2 i B cross sigma. The 2 is cancelled.
This = + muB B cross sigma. That is = - mu B sigma cross B. so this was my classical equation and now
i would like to write the quantum equation. The quantum equation says dS e over dt is
= - mu B.
So this is - modulus e h cross over 2 m e times sigma cross B which is = e over m e,
the spin operator; S cross B. therefore again you have precessional motion but what’ the
frequency of precession? Its e over m and not e over 2 m and the reason for this difference
was because of the g factor of the electron. So the precessional motion still appears.
It’s exactly the same as before. In fact what was mu e in terms of the spin operator?
It was g e over 2 m 2 m e into the spin. So this was exactly equal other e over m e Se. so
in fact you could write this as mu e cross B. so the point is the following. The classical
equation of motion says rate of change of angular momentum is = the torque. the torque
on a dipole of magnetic moment mu is mu cross B. so it says dl over dt is mu cross B. quantum
mechanically the exact equation of motion for the spin operator which is an angular
momentum operator is rate of change of the angular momentum operator again = mu cross
B. therefore the equation of motion doesn’t change at all. This is another example of
Ehrenfest's theorem because if i took expectation values on both sides, then it says this = that
because those are the operators.
And Ehrenfest's theorem says that quantum mechanical expectation values obey classical
equations of motion for such classes of Hamiltonians and that is exactly what has happened. But
when you want to actually compute what the precession frequency is, you have to convert
this mu into an L or you have to convert this L into a mu because that’s how you get the
precession of mu. So you have to convert this S into a mu and the factor that converts is
in fact the gyromagnetic ratio and that’s e over m here. So the frequency of precession
is = modulus e B over me whereas as the frequency of the precession here = modulus e over twice
me. if you have the way classically you have a magnetic dipole moment is by imagining there
is motion of a current in the form a loop. so i said let’s take the simplest instance
where you have a particle moving in a circular orbit of radius r and so on and found that
in the classical case the gyromagnetic ratio is charged divided by twice the mass.
Here for the internal motion for the spin degree for freedom, the relations between
the magnetic moment and the intrinsic angular momentum has an extra factor g which has to
be 2 for the electron. This implies that event though the expectation value of this angular
momentum undergoes precessional motion exactly as the classical counter part would, the frequency
of precession is eB over m here but its eB over 2 m e here that’s because there was
a extra g factor sitting there. But otherwise the equations of motion are exactly the same
in both cases. Rate of change of angular momentum is magnetic dipole movement cross the magnetic
field. It’s exactly the same equation except in classical mechanics you write that down
from the rules of classical electromagnetism but in quantum mechanics you have to compute
it. given the Hamiltonian you actual you have to calculate this commutator here and you
discover exactly the same equation of motion.
Now what i would like to do is to ask suppose i have a general ket vector of this kind,
how could i interpret this ket vector? We know that if it was 1 0 i would say that corresponds
to an eigenstate of Sz. if it was 0 1, it would be eigenstate of Sz with the opposite
eigenvalue. What happens if it was some a b could i interpreted as the eigenstates corresponding
to + 1/2 h cross of some spin pointing in some arbitrary direction? This is the question
i would like to ask. Suppose i chose my axis of quantization along some arbitrary direction,
then could i interpret this b as the eigenstate corresponding to spin up in that new direction?
So let’s see if that works out. That will be very useful for what we are going to do
next.
So let’s take some arbitrary unit vector n in space with respect to my fixed coordinates
system specified by polar angles theta and phi. What are the components of this n? The
magnitude is 1. So what is it in spherical polar coordinates? Its = sin theta cos phi,
sin theta sin phi cos theta such that mod n squared is n squared is 1. Now we would
like to ask i mean make the question precise. What’s the up state for this? let me denote
this as up state of S dot n or sigma dot n and the opposite arrow wound be the down state
of this. So what should i do? I would like to find that up state.
i expand this up state. Any arbitrary state can be expanded in eigenstates of Sz which
was my original axis of quantization. So let me simply expand.
We can simply write up state here is = some a times 1 0 + b times 0 1. I know that sigma
dot n on this state must be = 1 + 1 times the same state. i require it to be an eigenstate
of + 1/2 h cross. So when i multiply the sigma by h cross over 2, then i am going to get
s dot n and the eigenvalues + 1/2 h cross. So that’s why we put a + 1 here. And we
have to solve this eigenvalue equation. That’s all we have to do. What does this give you?
Sigma dot n is sigma 1 times this. so it’s clear that sigma 1 is off diagonal completely,
sigma 2 is fully off diagonal and sigma 3 has diagonal terms. We want sigma 3 times
n 3 which is cos theta here.
So this will give you cos theta, - cos theta on this side. And then sigma 1 is 0 1 1 0.
So it’s going to give you sin theta, cos phi, and then sin theta, cos phi acting on
a b is = a b. that’s my eigenvalue equation. This operator here is just a b. then i have
sigma 2 which is multiplied by this coefficient but sigma 2 is a – i. so this is = sin theta
cos phi - i sin theta sin phi and then + i sin theta sin phi and then - cos theta. That’s
my eigenvalue equation. My job is to find a and b and normalize it to unity and then
i am guaranteed to get this ket vector a b which is the eigenstate of sigma dot n with
eigenvalues + 1. So this will be give you cos theta - i sin theta. Sin theta comes out
common. So those are the eigenvalue equations. So what does it give you?
a cos theta + b sin theta e to the - i phi = a and similarly a e to the i phi sin theta
- b cos theta = b. either of them would do because this is something way we going to
normalize a and b such that mod a squared + mod b squared is 1.so it says b sin theta
is a into 1 - cos theta and then i normalize it to unity. So this can be written as 2 b
sin theta over 2 cos theta over 2 which is 1 - cos theta which is 2 a sin squared theta
over 2. So the 2 goes away in both sides. a sin theta over 2 goes away and it tells
you b = a and tan theta over 2 e to the i phi. So I have in fact written the answer
down. So let me write the answer down after you normalize it.
This state is = cos theta over 2 and then e to the - i phi sin theta over 2. That’s
the normalized state. Mod a squared mod + mod b square is 1 in this case. It’s worth
remembering this expression because you can then write down things very fast. By the way,
the overall phase factor is irrelevant because you can always multiply this by some e to
i alpha and it will not change any probability. It won’t change any physics at all. So you
could if you make this look a little more symmetric by multiplying by e to the i phi
over 2 in which case you get e to the i phi over 2 cos theta e to the - i phi over 2 sin
theta over 2.
This looks a little more symmetric but this expression is worth remembering because it
keeps appearing over and over again. so you agree that if you give me any arbitrary ket
a b such that mod a squared + mod b squared is 1, i could interpret that ket as the eigenstate
of S dot n with an appropriate n. the polar angles of n are given by the a’s and b’s in this fashion. Their relative
phase difference between a and b is the azimuthal angle. And a is like cos theta over 2, b is
like sin theta over 2. so the ratio of a to b in fact gives you the cot tangent of theta
over 2 of b 2 a mod b over mod a gives you tan theta over 2. So that’s the way to interpret
an arbitrary state a b. now you can quickly check that all the usual results will come
out. For example, if theta 0 that means axis of quantization is z axis itself. i should
get 1 0 and indeed i do. What happens if theta is pi? You get 0 1 apart from a phase factor
which is irrelevant because the phi angle is undefined. So that works. What about this
eigenstate, Sx= +h cross over 2?
What kind of state corresponds to + 1/2 h cross by 2 as the eigenvalue for Sx? All we
have to do is to substitute values of theta and phi into it. What about the x axis? What
does that correspond to? What value of theta does it correspond to? It’s pi over 2. Phi
= 0. so what does this become in the z basis? This is cos phi over 4 which is over root
2. This is gone and that’s also 1 over root 2. So it’s clear this is just 1 over root
2, 1 1 and the same thing is -, what would happen? What would that correspond to interms
of phi? Phi goes from 0 to pi. It’s the the negative x axis. So that becomes + or
– 1. What about Sy = + or - h cross over 2? What would these correspond to? Now you
want to know the coordinates of the positive y axis? Again theta is = pi over 2 but phi
is either pi over 2 or 3 pi over 2 in the other direction. so this is = cos theta over
2 that’s = 1 over root 2 once again comes out always and then this is a 1 and when phi
is = pi over 2, its e to the - i phi over 2 which is - i and e to the - 3 i pi over
2 is e to the + i pi over 2 which is + i. so this is 1, - or + i.
so if we see such states, you incidentally recognize that these corresponds to eigenstates
of the x and y coordinates but a general eigenstates looks like that. I am going to stop here and
then we resume this. to answer this question it says consider an arbitrary initial state chi = a
b. suppose you measure not Sz in which case the probability of + 1/2 h cross would have
a mod a squared and the other 1 is mod b squared but you measure Sx, and the question is what’s
the probability that you obtain values + 1/2 h cross and - 1/2 h cross respectively. What
would you do? you change basis. The obvious thing to do is to change basis. so we start
with a b and write it as a linear combination of these 2 states and then probability they
are also orthonormal. Every one of them has been made orthonormal as a basis and then
identifies the coefficients. So in that case, the answer will turn out to be mod a + b whole
squared over 2 and mod - b squared over 2. Similarly if you measured Sy, the answer would
be a + iB mod squared a - iB mod squared and 1/2 of these quantities give you the probabilities.
So i hope it’s clear what’s happening. you started with an axis of quantization,
by convention z axis and I took an arbitrary ket vector and identified it with the eigenstate
of S dot n, particular n but I use that backwards to write down what the eigenstates are for
any axis of quantization directly. and if you want to know what happens if i take an
arbitrary initial state and measure in any other arbitrary component, what are the probabilities
of getting + or - 1/2 h cross because those are the only 2 answers possible. The answer
is take this ket vector and expand in that basis. So it is just change of basis and take
the mod squared of the coefficients. So let me stop here.