Understanding Quantum Mechanics: Equations for Spin Along x and y Axes Explained

[doggydan42]

I am currently reading Leonard Susskind's "Quantum Mechanics the Theoretical Minimum". In chapter 2.3 and 2.4, he defines |r>, |l> and|i>, |o>, for r and l along the x-axis and i and o along the y axis.

The equations are:

$$|r> = \frac{1}{\sqrt{2}}|u>+\frac{1}{\sqrt{2}}|d>$$
Since $<r|l>=0$,
$$|l> = \frac{1}{\sqrt{2}}|u>-\frac{1}{\sqrt{2}}|d>$$

It then follows that,
$$|i> = \frac{1}{\sqrt{2}}|u>+\frac{i}{\sqrt{2}}|d>$$
$$|o> = \frac{1}{\sqrt{2}}|u>-\frac{i}{\sqrt{2}}|d>$$

I understand that if r and l are real, then i and o must have an imaginary component (proof is done in excercise2.3); however, would it be possible to replace r for i and l for o to switch where the imaginary nmber is and maintain the same meaning within the equation.

Also, I understand that for $<r|l>=0$ then l must have $-|d>$ if the d in r is positive. Would it be possible to exchange the negative sign in r and l, or similarly replace l with r and vice versa.

Overall, given the possible ambiguity in choices for the negative sign and the imaginary component, why are these specific equations chosen?

Thank you in advance

 

[Gene Naden]

Would it be possible to exchange the negative sign in r and l, or similarly replace l with r and vice versa.

Does this book talk about the Pauli spin matrices? |r> must be an eigenvector of $\sigma_x$ with eigenvalue 1 and |l> must be an eigenvector of $\sigma_x$ with eigenvalue -1. This is because |r> is directed along the +x axis and |l> is directed along the -x axis. If you exchange the negative sign in r and l then the eigenvalue relations won't hold.

$\sigma_x$ is a two-by-two matrix with zeroes on the diagonal and 1 on the off-diagonal elements.

And |u> is directed along the z axis. It is a column vector of length two with a 1 in the first position and a zero in the second position. |d> is a column vector with a zero in the first position and a 1 in the second position, because it is directed along the -z axis.

 

[doggydan42]

Does this book talk about the Pauli spin matrices? |r> must be an eigenvector of $\sigma_x$ with eigenvalue 1 and |l> must be an eigenvector of $\sigma_x$ with eigenvalue -1. This is because |r> is directed along the +x axis and |l> is directed along the -x axis. If you exchange the negative sign in r and l then the eigenvalue relations won't hold.

$\sigma_x$ is a two-by-two matrix with zeroes on the diagonal and 1 on the off-diagonal elements.

And |u> is directed along the z axis. It is a column vector of length two with a 1 in the first position and a zero in the second position. |d> is a column vector with a zero in the first position and a 1 in the second position, because it is directed along the -z axis.

The book does talk about paulispin matrices. It makes sense now that the relations hold for choice of direction, though why can't we choose $\sigma_x$ to have the imaginary pauli matrix, which would result in $\sigma_y$ becoming the initial pauli matrix for $\sigma_x$?

Thank you

 

[DrClaude]

The book does talk about paulispin matrices. It makes sense now that the relations hold for choice of direction, though why can't we choose $\sigma_x$ to have the imaginary pauli matrix, which would result in $\sigma_y$ becoming the initial pauli matrix for $\sigma_x$?

You could do that, but then you would have a left-handed coordinate system instead of a right-handed one. The latter is chosen by convention.

 

[doggydan42]

You could do that, but then you would have a left-handed coordinate system instead of a right-handed one. The latter is chosen by convention.

If it all comes down to a case of how we set-up the coordinate system, then maybe I am confused about the interpretation of the pauli matrices. I understand that $\sigma_z$ which includes both 1 and -1 where u and d would be in a matrix, but then $\sigma_x$ has ones along the off-diagonals. How does the puali matrix relate the geometry that represent the right-handed coordinate system?

Thank you

 

[Gene Naden]

How does the puali matrix relate the geometry that represent the right-handed coordinate system?

... then you would have a left-handed coordinate system instead of a right-handed one. The latter is chosen by convention.

The Pauli matrices have commutation relations similar to those of angular momentum:
$[\sigma_a,\sigma_b]=\epsilon_{abc}\sigma_c$ For example $[\sigma_x,\sigma_y]=2i\sigma_z$. This may be related to having a right-handed coordinate system because the cross product of a unit vector on the x-axis with a unit vector on the y-axis gives a unit vector on the z axis.

 

[doggydan42]

The Pauli matrices have commutation relations similar to those of angular momentum:
$[\sigma_a,\sigma_b]=\epsilon_{abc}\sigma_c$ For example $[\sigma_x,\sigma_y]=2i\sigma_z$. This may be related to having a right-handed coordinate system because the cross product of a unit vector on the x-axis with a unit vector on the y-axis gives a unit vector on the z axis.

I see the relationship between angular momentum, which is even discussed in the book, but there is a factor of i. Further, the pauli matrices are:

$$\sigma_z = \begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}
$$
$$\sigma_x = \begin{pmatrix}
0 & 1\\
1 & 0
\end{pmatrix}
$$
$$\sigma_y = \begin{pmatrix}
0 & -i\\
i & 0
\end{pmatrix}
$$

Is there any significance to the placement of the 1's and i's within the matrix that can reveal the geometry of the system?

 

[Gene Naden]

The factor of $i$ is part of the regular commutation relations of angular momentum, which are, for example $[L_x,L_y]=iL_z$ where I am using natural units in which hbar is equal to one.

In general, the matrices could be different, I think. The important thing is that they satisfy the commutation relations. These are, as we have said, related to the handedness of the coordinate system.

 

[doggydan42]

The factor of $i$ is part of the regular commutation relations of angular momentum, which are, for example $[L_x,L_y]=iL_z$ where I am using natural units in which hbar is equal to one.

In general, the matrices could be different, I think. The important thing is that they satisfy the commutation relations. These are, as we have said, related to the handedness of the coordinate system.

That makes sense with the relation between the commutator and poisson brackets. Are these unique solutions that satisfy the commutation relations? Is it possible to prove if it is or isn't unique?

Thank you

 

[Gene Naden]

The Pauli matrices are not unique. The form a basis for the generators of rotations. You can have a different basis. But I haven't seen anybody do that. I think you could permute x, y and z cyclically $\sigma_x \rightarrow \sigma_y$, $\sigma_y \rightarrow \sigma_z$, $\sigma_z \rightarrow \sigma_x$
The state vectors would be different if the Pauli matrices were different. For example, if we permute x, y and z cyclically then the column vector with 1 in the first position and 0 in the second position might represent |r>.

 

[doggydan42]

The Pauli matrices are not unique. The form a basis for the generators of rotations. You can have a different basis. But I haven't seen anybody do that. I think you could permute x, y and z cyclically $\sigma_x \rightarrow \sigma_y$, $\sigma_y \rightarrow \sigma_z$, $\sigma_z \rightarrow \sigma_x$
The state vectors would be different if the Pauli matrices were different. For example, if we permute x, y and z cyclically then the column vector with 1 in the first position and 0 in the second position might represent |r>.

I guess my confusion with the pauli matrices lies in i. $\sigma_z$ seems to represent the possible outcomes of measuring the spin along z. However $\sigma_y$ has two imaginary numbers. What is the best way to think about the meaning behind the puali matrices.

Thank you.

 

[stevendaryl]

That makes sense with the relation between the commutator and poisson brackets. Are these unique solutions that satisfy the commutation relations? Is it possible to prove if it is or isn't unique?

Thank you

If you assume that $\sigma_z = \left( \begin{array}\\ 1 & 0 \\ 0 & -1 \end{array} \right)$, then the commutation rules and hermiticity imply that:

$\sigma_x = \left( \begin{array}\\ 0 & e^{i\alpha} \\ e^{-i\alpha} & 0 \end{array} \right)$
$\sigma_y = \left( \begin{array}\\ 0 & -i e^{i\alpha} \\ +i e^{-i\alpha} & 0 \end{array} \right)$

for some phase $\alpha$. It's just a convenience to pick $\alpha = 0$.

 

[Gene Naden]

Well they represent quantum operators, so there can be complex numbers. Remember the operators for angular momentum in the Schrodinger equation:
$L^\vec=\vec r \times \vec p = \vec r \times -i \nabla$
They have i in them.

The possible measurement outcomes are represented by the eigenvalues of the matrices (or operators). These eigenvalues are real. This is because the Pauli matrices are Hermition. See Wikipedia https://en.wikipedia.org/wiki/Hermitian_matrix#Properties