Pauli Matrix

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hi, anyone can enlighten me why the pauli x matrix in the z basis is given as
0 1
1 0

while it in the x basis, it is given as
1 0
0 -1

is there a formula or something to calculate this? and how does one know that a matrix is in the z basis or the x basis? thanks a lot!
 

Answers and Replies

  • #2
Fredrik
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I would say that the Pauli matrices are specifically the three matrices at the top of this page. You must be talking about the components of the operator Sx in two different bases.

The relationship between linear operators and matrices is described here. You can use that to find the components of Sx in the x and z bases. I recommend that you try it.
 
  • #3
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Observe that with [tex]U=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}[/tex] we have
[tex]U \sigma_3U^*=\sigma_1[/tex]

Therefore U transforms from z basis to x basis. Now calculate [tex]U\sigma_1U^*[/tex].
 
  • #4
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Observe that with [tex]U=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}[/tex] we have
[tex]U \sigma_3U^*=\sigma_1[/tex]

Therefore U transforms from z basis to x basis. Now calculate [tex]U\sigma_1U^*[/tex].
wow by using your formula i can get from 1 basis to another, but what is U and how do i get it? it looks kind of similar to the |+x> normalized eigenvector of the pauli-x matrix. is there some kind of link ?
 
  • #5
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I would say that the Pauli matrices are specifically the three matrices at the top of this page. You must be talking about the components of the operator Sx in two different bases.

The relationship between linear operators and matrices is described here. You can use that to find the components of Sx in the x and z bases. I recommend that you try it.
yes those are the pauli matrix, i don't really know what i am talking about too :X

my lecturer says that the pauli matrix in the Z basis { |+x> , |-x> } is the matrix you stated in your link

but she also says that the pauli matrix in the X basis { |+x> , |-x> } are
1 0
0 -1,

0 i
-i 0

0 1
1 0

which i have no idea what she is talking about. if it is Z basis, shouldn't it be { |+z> , |-z> } instead of { |+x> , |-x> }? in fact, what is { |+x> , |-x> } ? does this mean X basis?

i don't understand your 2nd link too... sry for being too dumb.
 
  • #6
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wow by using your formula i can get from 1 basis to another, but what is U and how do i get it?
You get it by finding the simplest solution of the equation:

[tex]U\sigma_3=\sigma_1U[/tex]

where

[tex]U=\begin{pmatrix}a&b\\c&d\end{pmatrix}[/tex]

and normalizing it requiring [tex]UU^*=I[/tex].
 
  • #7
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You get it by finding the simplest solution of the equation:

[tex]U\sigma_3=\sigma_1U[/tex]

where

[tex]U=\begin{pmatrix}a&b\\c&d\end{pmatrix}[/tex]

and normalizing it requiring [tex]UU^*=I[/tex].
ok, so i work out the abcd and get a lot of simulataneous eqns and finally found
1 1
1 -1

but normalizing it would mean 1/sqrt(1+1+1+1) = 1/2? how come its 1/sqrt 2?

oh, and how do i tell that
0 1
1 0 is in the z basis and thus the other matrix is in the x basis?
 
  • #8
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4
but normalizing it would mean 1/sqrt(1+1+1+1) = 1/2? how come its 1/sqrt 2?
Because you want UU*=I.

oh, and how do i tell that
0 1
1 0 is in the z basis
Because this is one of the Pauli's matrices with [tex]\sigma_z[/tex] being diagonal.

and thus the other matrix is in the x basis?
Because U maps [tex]\sigma_3[/tex] to [tex]\sigma_1[/tex] - it maps eigenvectors of [tex]\sigma_3[/tex] to eigenvectors of [tex]\sigma_1[/tex]. U is a similarity transformation from z basis to x basis.
 
  • #9
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Because you want UU*=I.



Because this is one of the Pauli's matrices with [tex]\sigma_z[/tex] being diagonal.



Because U maps [tex]\sigma_3[/tex] to [tex]\sigma_1[/tex] - it maps eigenvectors of [tex]\sigma_3[/tex] to eigenvectors of [tex]\sigma_1[/tex]. U is a similarity transformation from z basis to x basis.
:X... thanks alot, let me digest them now, i feel so confused.....
 
  • #10
Fredrik
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the pauli matrix in the Z basis
A matrix is just an array of numbers. It doesn't involve a basis.

However, as I explained, every basis (actually every pair of bases) defines a way to associate a matrix with each linear operator. That's why I have to assume that she's talking about the components of a linear operator in different bases, not a about a matrix in different bases. The latter doesn't make sense, at least not without some additional information.

i don't understand your 2nd link too
It's the single most important result from linear algebra, explained in a way that doesn't require you to understand anything more difficult than the terms "vector space", "linear" and "basis". So you should definitely try again.

Edit: You need one more thing: The definition of matrix multiplication. I suspect that's what's missing. If A and B are matrices such that the number of columns of A is the same as the number of rows of B, the product AB is defined by [itex](AB)_{ij}=A_{ik}B_{kj}[/itex]. Note that the right-hand side should be interpreted as [itex]\sum_k A_{ik}B_{kj}[/itex]. (I explained that convention in that other post).
 
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  • #11
Fredrik
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Have you given it another shot yet? If anything there is causing you problems, let me know what, and I'll explain it.
 
  • #12
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You get it by finding the simplest solution of the equation:

[tex]U\sigma_3=\sigma_1U[/tex]

where

[tex]U=\begin{pmatrix}a&b\\c&d\end{pmatrix}[/tex]

and normalizing it requiring [tex]UU^*=I[/tex].

Hi,

I wonder if you can clarify a bit on how you come up with the relation
[tex]U\sigma_3=\sigma_1U[/tex]

I think that there exists a matrix U that maps [tex]\sigma_3[/tex] to [tex]\sigma_1[/tex]

then we can write [tex]U\sigma_1=\sigma_3[/tex]

and hence [tex]U = \sigma_1 \sigma_3^{-1} [/tex]

But I have a feeling that my method is not correct. I prefer your way since it can easily see that [tex]\sigma_i^2=I[/tex] in any basis.

Thanks,
 
  • #13
1,444
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You want to find U such that

[tex]U\sigma^3U^{-1}=\sigma_1[/tex]

Now multiply both sides by U from the right and you end with

[tex]U\sigma_3=\sigma_1 U[/tex]
 
  • #14
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You want to find U such that

[tex]U\sigma^3U^{-1}=\sigma_1[/tex]

Now multiply both sides by U from the right and you end with

[tex]U\sigma_3=\sigma_1 U[/tex]
Oh my question was how can you come up with such transformation; that is finding a U that satisfies [tex]U\sigma^3U^{-1}=\sigma_1[/tex] instead of a U that satisfies [tex]U\sigma^3=\sigma_1[/tex]

Thanks,
 
  • #15
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4
If [itex]\psi[/itex] is an eigenvector of [itex]\sigma_3[/itex], you want [itex]U\psi[/itex] to be an eigenvector of of [itex]\sigma_1[/tex]. Suppose [itex]\sigma_3\psi=\lambda\psi[/itex], then, with my formula, you will get

[tex]\sigma_1U\psi=U\sigma_3U^{-1}U\psi=U\sigma_3\psi=U\lambda\psi=\lambda U\psi[/tex]

That's what you want. U maps eigenvectors of [itex]\sigma_3[/itex] to eigenvectors of [itex]\sigma_1[/itex]. You will not get this result with your formula.
 
  • #16
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Have you given it another shot yet? If anything there is causing you problems, let me know what, and I'll explain it.
.. i think i have to revisit all the lectures videos during the holidays again to understand what QM is about :X thanks for the help though
 
  • #17
Fredrik
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.. i think i have to revisit all the lectures videos during the holidays again to understand what QM is about :X thanks for the help though
We were talking about the relationship between linear operators and matrices. If you find anything that seems difficult in the post I linked to, it has to be because you have expectations that prevent you from seeing that every detail in that post is trivial.

I can relate to that. I once spent an hour not understanding how to prove that an identity element of a Banach algebra must be unique. The proof looks like this: 1'=1'1=1. I can't think of a reason for why I didn't see that immediately other than that I expected the proof to be hard because it was a problem in a difficult book on functional analysis. I'm pretty sure that if you just decide to trust me when I say that there's nothing difficult in that post, you will understand it completely in 10 minutes. If you understand the concepts "matrix multiplication" and "basis of a vector space" well, it might take you less than 2 minutes.
 

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