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Pauli Matrices

  • Thread starter blanik
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  • #1
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My other problem is:

Consider now the space of 2x2 complex matrices. Show that the Pauli Matrices
|I>= 1 0
0 1

|sigma x>= 0 1
1 0

|sigma y>= 0 -i
i 0

|sigma z>= 1 0
0 -1

form an orthonormal basis for this space when k=1/2. To spare yourself from having to compute 10 different matrix products, I recommend that you write out what the inner product is for general matrices A and B first.

I'm really lost on this one! Any help getting me started would be greatly appreciated!
 

Answers and Replies

  • #2
AKG
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Do you know what it means for them to be orthonormal? I think it's safe to assume that the underlying field is C (otherwise you're being asked to prove a falsehood), so it suffices to show that these four matrices are orthonormal, and it will follow that they form an orthonormal basis. Do you know how to check if a set is orthonormal? Since I don't know the inner product you're working with, I can't help you, but if you know the inner product and you know the definition of "orthonormal" you should be able to do this.
 
  • #3
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Well, the inner product is defined as |A||B|cos (angle). I don't have numbers for A & B. He says use the inner product for "general" 2x2 matrices A & B.

Also, I must not understand orthonormal very well. If the two vectors are orthonormal, then they are perpendicular. The dot product which I thought was the same as the inner product would be zero.
 
  • #4
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The only inner product on 2x2 matrices that I've ever really used is

[tex] (A, B) \equiv \frac{1}{2} \textrm{Tr}( A B^{\dagger}) [/tex]

Where Tr(A) is the trace of A, and B-dagger is the Hermitian conjugate of B.
 
  • #5
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He did give us that formula, but how do I use that without actual numbers for A & B? Should I make I = A and sigma x = B and then sigma y = C and sigma z = D? Then I have <A|B> = 1/2Trace(AtB) (where t = Hermitian conjugate) and <C|D> = 1/2Trace(CtD)?

So, for the real numbers, i.e. the matrices of I and sigma x & y, the complex conjugate of any real number is that real number. So, the hermitian equivalent for I=A is the same matrix. The trace is 1 and the trace of sigma x = B is zero. So the inner product <A|B> is zero? Which makes sense if they are orthonormal, right?

The Hermitian conjugate of sigma y = C is also the same matrix because you reverse the signs of the (i)'s for the complex conjugate and then you transpose ending up with the same matrix you started with. So, again, the Trace of C = zero and then the inner product <C|D> also is zero.

Am I even remotely on the right track here???
 
  • #6
AKG
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Let A be:

a1 a2
a3 a4

and B be:

b1 b2
b3 b4

Well you should know how to find the hermitian conjugate of B, you should know how to multiply A and B*, and you should know how to find the trace of AB*. So basically you have a formula that's "ready-to-go" and you just need to know the entries of your matrices in order to compute the inner product, you won't have to do the multiplication or look at the diagonal each time.

If you express these matrices with respect to some basis, then you will be expressing it as a 4-tuple. If we have a matrix A, then we can call the corresponding 4-tuple [A]. Then [A]*M[C] defines an inner product so long as M is a 4x4 positive definite matrix. If we take M = 0.5I, then the defined inner product will be like qbert's, but we can make many different choices for M.
 
  • #7
Gokul43201
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blanik said:
He did give us that formula, but how do I use that without actual numbers for A & B?
What do you mean ? You do have actual numbers for all the matrices.
Should I make I = A and sigma x = B and then sigma y = C and sigma z = D? Then I have <A|B> = 1/2Trace(AtB) (where t = Hermitian conjugate) and <C|D> = 1/2Trace(CtD)?
Why are we talking about C, D etc. Yes, you apply the provided definition of the inner product to each pair of matrices and see if they give you zero.

So, for the real numbers, i.e. the matrices of I and sigma x & y, the complex conjugate of any real number is that real number.
[itex]\sigma_y [/itex] is not a real matrix. But the rest is true.

So, the hermitian equivalent for I=A is the same matrix. The trace is 1 and the trace of sigma x = B is zero. So the inner product <A|B> is zero? Which makes sense if they are orthonormal, right?
It should, but that's not the way to find the inner product. You are misapplying the definition. The inner product is (half) the trace of the product, not (half) the product of the traces. First multiply, then find the trace.

The Hermitian conjugate of sigma y = C is also the same matrix because you reverse the signs of the (i)'s for the complex conjugate and then you transpose ending up with the same matrix you started with.
Correct. In fact, all the matrices are Hermitian since in each case [itex]a_{ij} = (a_{ji})^* [/itex]. If this were in a physics course, you will come across more reasons why these particular matrices want to be hermitian (or selfadjoint).

So, again, the Trace of C = zero and then the inner product <C|D> also is zero.
Again, you must reverse the order of those operations.

Am I even remotely on the right track here???
You're getting there.
 
Last edited:
  • #8
CarlB
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This reminds me of one of those facts about the Pauli spin matrices (and more generally, Clifford algebras), that no one ever talks about.

Normally one models the spin of elementary particles with spinors, in this case, those 2x1 vectors, and one treats the matrices as operators. It would be a little cleaner if the matrices were all you needed. It turns out that it is possible to model spin with the matrices alone.

Using the usual basis for the Pauli spin matrices (or any other basis you prefer, or you can refuse to specify a basis and treat it in Clifford algebra form), let [tex]\hat{u} =[/tex] [tex]u_x\hat{x} + u_y \hat{y} + u_z\hat{z}[/tex], where [tex]\hat{u}[/tex] has length one. For a particle with spin in the [tex]+\hat{u}[/tex] direction, associate the particle with a matrix as follows:

[tex]|+1/2_u +1/2 \rangle \equiv (1 + u_x\sigma_x + u_y\sigma_y + u_z\sigma_z)/2.[/tex]

The matrices so defined are "idempotent", which means that when you square them, you get back the same value. To compute the inner product, [tex]\langle A | B \rangle[/tex], compute the usual complex conjugate product and compute the trace. The inner product so computed is not quite the same as the inner product computed from spinors, but in certain ways it is an improvement.

From a physical point of view, what we can measure about spin states are probabilities. For example, the probability that a spin-1/2 particle that is oriented in the +z direction will give +1/2 when measured in some other direction.

When computing the transition probability between two spinors, say A and B, one computes:

[tex]|\langle A | B \rangle |^2[/tex]

That is, one computes the spinor inner product, and then takes the magnitude and squares it.

For the idempotent matrices defined above, one uses the matrix inner product and then takes the magnitude. There is no need to square it. The result is the same, but using the idempotents eliminates the extra squaring operation.

Here's the (somewhat trivial) calculation for the inner product between two matrices, one associated with spin+1/2 in the +z direction, the other with spin+1/2 in the +u direction, with [tex]\hat{u} = \cos(\theta)\hat{z} + \sin(\theta)\hat{x}[/tex]:

[tex]\textrm{tr}(\left(\begin{array}{cc}1 & 0 \\ 0 & 0 \end{array}\right)
\left(\begin{array}{cc}1+\cos(\theta) & \sin(\theta) \\
\sin(\theta) & 1-\cos(\theta) \end{array}\right)/2)[/tex]
[tex]= (1 + \cos(\theta))/2[/tex]

While this is obvious in the above case, it generalizes to any two of these idempotent matrices that represent spin in whatever directions you'd like. (Try it if you don't believe me, or better, use [tex]\sigma_x[/tex] notation and the result will be obvious.)

Perhaps the idempotents are a more basic method of modeling elementary particles than spinors.

Carl
 

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