Understanding Pauli X Matrix in Z & X Bases

[quietrain]

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!

 

[Fredrik]

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 S_x in two different bases.

The relationship between linear operators and matrices is described here. You can use that to find the components of S_x in the x and z bases. I recommend that you try it.

 

[arkajad]

Observe that with $$U=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$ we have
$$U \sigma_3U^*=\sigma_1$$

Therefore U transforms from z basis to x basis. Now calculate $$U\sigma_1U^*$$.

 

[quietrain]

Observe that with $$U=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&-1\end{pmatrix}$$ we have
$$U \sigma_3U^*=\sigma_1$$

Therefore U transforms from z basis to x basis. Now calculate $$U\sigma_1U^*$$.

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 ?

 

[quietrain]

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 S_x in two different bases.

The relationship between linear operators and matrices is described here. You can use that to find the components of S_x 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.

 

[arkajad]

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:

$$U\sigma_3=\sigma_1U$$

where

$$U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

and normalizing it requiring $$UU^*=I$$.

 

[quietrain]

You get it by finding the simplest solution of the equation:

$$U\sigma_3=\sigma_1U$$

where

$$U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

and normalizing it requiring $$UU^*=I$$.

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?

 

[arkajad]

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 $$\sigma_z$$ being diagonal.

and thus the other matrix is in the x basis?

Because U maps $$\sigma_3$$ to $$\sigma_1$$ - it maps eigenvectors of $$\sigma_3$$ to eigenvectors of $$\sigma_1$$. U is a similarity transformation from z basis to x basis.

 

[quietrain]

Because you want UU*=I.



Because this is one of the Pauli's matrices with $$\sigma_z$$ being diagonal.



Because U maps $$\sigma_3$$ to $$\sigma_1$$ - it maps eigenvectors of $$\sigma_3$$ to eigenvectors of $$\sigma_1$$. U is a similarity transformation from z basis to x basis.

:X... thanks alot, let me digest them now, i feel so confused...

 

[Fredrik]

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 $(AB)_{ij}=A_{ik}B_{kj}$. Note that the right-hand side should be interpreted as $\sum_k A_{ik}B_{kj}$. (I explained that convention in that other post).

 

[Fredrik]

Have you given it another shot yet? If anything there is causing you problems, let me know what, and I'll explain it.

 

[Shadowz]

You get it by finding the simplest solution of the equation:

$$U\sigma_3=\sigma_1U$$

where

$$U=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$

and normalizing it requiring $$UU^*=I$$.

Hi,

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

I think that there exists a matrix U that maps $$\sigma_3$$ to $$\sigma_1$$

then we can write $$U\sigma_1=\sigma_3$$

and hence $$U = \sigma_1 \sigma_3^{-1}$$

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

Thanks,

 

[arkajad]

You want to find U such that

$$U\sigma^3U^{-1}=\sigma_1$$

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

$$U\sigma_3=\sigma_1 U$$

 

[Shadowz]

You want to find U such that

$$U\sigma^3U^{-1}=\sigma_1$$

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

$$U\sigma_3=\sigma_1 U$$

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

Thanks,

 

[arkajad]

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

$$\sigma_1U\psi=U\sigma_3U^{-1}U\psi=U\sigma_3\psi=U\lambda\psi=\lambda U\psi$$

That's what you want. U maps eigenvectors of $\sigma_3$ to eigenvectors of $\sigma_1$. You will not get this result with your formula.

 

[quietrain]

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

 

[Fredrik]

.. 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.