3.1 The two-state quantum system

对照

3.1.1 The Pauli matrices: observables of the two-state quantum system

For a quantum system with two dimensional state space $\mathcal{H}={\mathbf{C}}^2$ , observables are self-adjoint linear operators on ${\mathbf{C}}^2$ . With respect to a chosen basis of ${\mathbf{C}}^2$ , these are 2 by 2 complex matrices $M$ satisfying the condition $M=M^{†}$ ($M$^† is the conjugate transpose of $M)$ . Any such matrix will be a (real) linear combination of four matrices:

$$M=c_{0}1+c_1{\sigma }_1+c_2{\sigma }_2+c_3{\sigma }_3$$

with $c_j\in \mathbf{R}$ and the standard choice of basis elements given by

$$1=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),{\sigma }_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\sigma }_2=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),{\sigma }_3=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$

where the ${\sigma }_j$ are called the “Pauli matrices”. This choice of basis is a convention, with one aspect of this convention that of taking the basis element in the 3- direction to be diagonal. In common physical situations and conventions, the third direction is the distinguished “up-down” direction in space, so often chosen when a distinguished direction in ${\mathbf{R}}^3$ is needed.

Recall that the basic principle of how measurements are supposed to work in quantum theory says that the only states that have well-defined values for these four observables are the eigenvectors for these matrices, where the value is the eigenvalue, real since the operator is self-adjoint. The first matrix gives a trivial observable (the identity on every state), whereas the last one, ${\sigma }_3$ , has the two eigenvectors

$${\sigma }_3\left(\genfrac{}{}{0ex}{}{1}{0}\right)=\left(\genfrac{}{}{0ex}{}{1}{0}\right)$$

and

$${\sigma }_3\left(\genfrac{}{}{0ex}{}{0}{1}\right)=-\left(\genfrac{}{}{0ex}{}{0}{1}\right)$$

with eigenvalues $+1$ and $-1$ . In quantum information theory, where this is the qubit system, these two eigenstates are labeled $|0⟩$ and $|1⟩$ because of the analogy with a classical bit of information. When we get to the theory of spin in chapter 7, we will see that the observable $\frac{1}{2}{\sigma }_3$ corresponds (in a non-trivial way) to the action of the group $SO(2)=U(1)$ of rotations about the third spatial axis, and the eigenvalues $-\frac{1}{2},+\frac{1}{2}$ of this operator will be used to label the two eigenstates, so

$$|+\frac{1}{2}⟩=\left(\genfrac{}{}{0ex}{}{1}{0}\right)\text{ and }|-\frac{1}{2}⟩=\left(\genfrac{}{}{0ex}{}{0}{1}\right)$$

Such eigenstates $|+\frac{1}{2}⟩$ and $|-\frac{1}{2}⟩$ provide a basis for ${\mathbf{C}}^2$, so an arbitrary vector in $\mathcal{H}$ can be written as

$$|\psi ⟩=\alpha |+\frac{1}{2}⟩+\beta |-\frac{1}{2}⟩$$

for $\alpha ,\beta \in \mathbf{C}$. Only if $\alpha$ or $\beta$ is 0 does the observable ${\sigma }_3$ correspond to a well-defined number that characterizes the state and can be measured. This will be either $\frac{1}{2}$ (if $\beta =0$ so the state is an eigenvector $|+\frac{1}{2}⟩$), or $-\frac{1}{2}$ (if $\alpha =0$ so the state is an eigenvector $|-\frac{1}{2}⟩$).

An easy to check fact is that $|+\frac{1}{2}⟩$ and $|-\frac{1}{2}⟩$ are NOT eigenvectors for the operators ${\sigma }_1$ and ${\sigma }_2$. One can also check that no pair of the three ${\sigma }_j$ commute, which implies that there are no vectors that are simultaneous eigenvectors for more than one ${\sigma }_j$ . This non-commutativity of the operators is responsible for the characteristic paradoxical property of quantum observables: there exist states with a well defined number for the measured value of one observable ${\sigma }_j$ , but such states will not have a well-defined number for the measured value of the other two non-commuting observables.

The physical description of this phenomenon in the realization of this system as a spin $\frac{1}{2}$ particle is that if one prepares states with a well-defined spin component in the $j$-direction, the two other components of the spin can’t be assigned a numerical value in such a state. Any attempt to prepare states that simultaneously have specific chosen numerical values for the 3 observables corresponding to the ${\sigma }_j$ is doomed to failure. So is any attempt to simultaneously measure such values: if one measures the value for a particular observable ${\sigma }_j$ , then going on to measure one of the other two will ensure that the first measurement is no longer valid (repeating it will not necessarily give the same thing). There are many subtleties in the theory of measurement for quantum systems, but this simple two-state example already shows some of the main features of how the behavior of observables is quite different from that of classical physics.

While the basis vectors $\left(\genfrac{}{}{0ex}{}{1}{0}\right)$ and $\left(\genfrac{}{}{0ex}{}{0}{1}\right)$ are eigenvectors of ${\sigma }_3,{\sigma }_1$ and ${\sigma }_2$ take these basis vectors to non-trivial linear combinations of basis vectors. It turns out that there are two specific linear combinations of ${\sigma }_1$ and ${\sigma }_2$ that do something very simple to the basis vectors. Since

$$({\sigma }_1+i{\sigma }_2)=\left(\begin{array}{cc}0& 2\\ 0& 0\end{array}\right)\text{ and }({\sigma }_1-i{\sigma }_2)=\left(\begin{array}{cc}0& 0\\ 2& 0\end{array}\right)$$

we have

$$({\sigma }_1+i{\sigma }_2)\left(\genfrac{}{}{0ex}{}{0}{1}\right)=2\left(\genfrac{}{}{0ex}{}{1}{0}\right)({\sigma }_1+i{\sigma }_2)\left(\genfrac{}{}{0ex}{}{1}{0}\right)=\left(\genfrac{}{}{0ex}{}{0}{0}\right)$$

and

$$({\sigma }_1-i{\sigma }_2)\left(\genfrac{}{}{0ex}{}{1}{0}\right)=2\left(\genfrac{}{}{0ex}{}{0}{1}\right)({\sigma }_1-i{\sigma }_2)\left(\genfrac{}{}{0ex}{}{0}{1}\right)=\left(\genfrac{}{}{0ex}{}{0}{0}\right)$$

$({\sigma }_1+i{\sigma }_2)$ is called a “raising operator”: on eigenvectors of ${\sigma }_3$ it either increases the eigenvalue by 2, or annihilates the vector. $\left({\sigma }_1-i{\sigma }_2\right)$ is called a “lowering operator”: on eigenvectors of ${\sigma }_3$ it either decreases the eigenvalue by 2, or annihilates the vector. Note that these linear combinations are not self-adjoint and are not observables, $({\sigma }_1+i{\sigma }_2)$ is the adjoint of $\left({\sigma }_1-i{\sigma }_2\right)$ and vice-versa.

3.1.2 Exponentials of Pauli matrices: unitary transformations of the two-state system

We saw in chapter 2 that in the $U(1)$ case, knowing the observable operator $Q$ on $\mathcal{H}$ determined the representation of $U(1)$ , with the representation matrices found by exponentiating $i\theta Q$ . Here we will find the representation corresponding to the two-state system observables by exponentiating the observables in a similar way.

Taking the identity matrix first, multiplication by $i\theta$ and exponentiation gives the diagonal unitary matrix

$$e^{i\theta 1}=\left(\begin{array}{cc}{e}^{i\theta }& 0\\ 0& e^{i\theta }\end{array}\right)$$

This is exactly the case studied in chapter 2, for a $U(1)$ group acting on $\mathcal{H}={\mathbf{C}}^2$ with

$$Q=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

This matrix commutes with any other 2 by 2 matrix, so we can treat its action on $\mathcal{H}$ independently of the action of the ${\sigma }_j$

Turning to the other three basis elements of the space of observables, the Pauli matrices, it turns out that since all the ${\sigma }_j$ satisfy ${\sigma }_j^2=1$ , their exponentials also take a simple form.

$$\begin{array}{rl}{e}^{i\theta {\sigma }_j}& =1+i\theta {\sigma }_j+\frac{1}{2}(i\theta {)}^2{\sigma }_j^2+\frac{1}{3!}(i\theta {)}^3{\sigma }_j^3+\dots \\ & =1+i\theta {\sigma }_j-\frac{1}{2}{\theta }^{2}1-i\frac{1}{3!}{\theta }^3{\sigma }_j+\dots \\ & =(1-\frac{1}{2!}{\theta }^2+\dots )1+i(\theta -\frac{1}{3!}{\theta }^3+\dots ){\sigma }_j\\ & =(\cos \theta )1+i{\sigma }_j(\sin \theta )\end{array}\text{\hspace{1em}(3.1)}$$

As $\theta$ goes from $\theta =0$ to $\theta =2\pi$ , this exponential traces out a circle in the space of unitary 2 by 2 matrices, starting and ending at the unit matrix. This circle is a group, isomorphic to $U(1)$ . So, we have found three different $U(1)$

subgroups inside the unitary 2 by 2 matrices, but only one of them (the case $j=3$) will act diagonally on $\mathcal{H}$, with the $U(1)$ representation determined by

$$Q=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$

For the other two cases $j=1$ and $j=2$ , by a change of basis either one could be put in the same diagonal form, but doing this for one value of $j$ makes the other two no longer diagonal. To understand the $SU(2)$ action on $\mathcal{H}.$ , one needs to consider not just the $U(1)$ subgroups, but the full three dimensional $SU(2)$ group one gets by exponentiating general linear combinations of Pauli matrices.

To compute such exponentials, one can check that these matrices satisfy the following relations, useful in general for doing calculations with them instead of multiplying out explicitly the 2 by 2 matrices:

$$[{\sigma }_j,{\sigma }_k{]}_{+}\equiv {\sigma }_j{\sigma }_k+{\sigma }_k{\sigma }_j=2{\delta }_{jk}1\text{\hspace{1em}(3.2)}$$

Here $[\cdot ,\cdot {]}_{+}$ is called the anticommutator. This relation says that all ${\sigma }_j$ satisfy ${\sigma }_j^2=1$ and distinct ${\sigma }_j$ anticommute (e.g., ${\sigma }_j{\sigma }_k=-{\sigma }_k{\sigma }_j$ for $j\ne k$).

Notice that the anticommutation relations imply that, if we take a vector $\mathbf{v}=(v_1,v_2,v_3)\in {\mathbf{R}}^3$ and define a 2 by 2 matrix by

$$\mathbf{v}\cdot \mathbf{\sigma }=v_1{\sigma }_1+v_2{\sigma }_2+v_3{\sigma }_3=\left(\begin{array}{cc}{v}_3& v_1-i{v}_2\\ v_1+i{v}_2& -v_3\end{array}\right)$$

then taking powers of this matrix we find

$$(\mathbf{v}\cdot \sigma {)}^2=(v_1^2+v_2^2+v_3^2)1=|\mathbf{v}{|}^{2}1$$

If $\mathbf{v}$ is a unit vector, we have

$$(\mathbf{v}\cdot \mathbf{\sigma }{)}^n=\{\begin{array}{ll}1& n\text{ even }\\ \mathbf{v}\cdot \mathbf{\sigma }& n\text{ odd }\end{array}$$

Replacing ${\sigma }_j$ by $\mathbf{v}\cdot \mathbf{\sigma }$ , the same calculation as for equation 3.1 gives (for $v$ a unit vector)

$$e^{i\theta \mathbf{v}\cdot \sigma }=(\cos \theta )1+i(\sin \theta )\mathbf{v}\cdot \sigma \text{\hspace{1em}(3.3)}$$

Notice that the inverse of this matrix can easily be computed by taking $\theta$ to $-\theta$

$$(e^{i\theta \mathbf{v}\cdot \sigma }{)}^{-1}=(\cos \theta )1-i(\sin \theta )\mathbf{v}\cdot \sigma$$

We’ll review linear algebra and the notion of a unitary matrix in chapter 4, but one form of the condition for a matrix $M$ to be unitary is

$$M^{†}=M^{-1}$$

so the self-adjointness of the ${\sigma }_j$ implies unitarity of $e^{i\theta \mathbf{v}\cdot \mathbf{\sigma }}$ since

$$\begin{array}{rl}(e^{i\theta \mathbf{v}\cdot \sigma }{)}^{†}& =((\cos \theta )1+i(\sin \theta )\mathbf{v}\cdot \sigma {)}^{†}\\ & =((\cos \theta )1-i(\sin \theta )\mathbf{v}\cdot {\sigma }^{†})\\ & =((\cos \theta )1-i(\sin \theta )\mathbf{v}\cdot \sigma )\\ & =(e^{i\theta \mathbf{v}\cdot \sigma }{)}^{-1}\end{array}$$

The determinant of $e^{i\theta \mathbf{v}\cdot \mathbf{\sigma }}$ can also easily be computed

$$\begin{array}{rl}\det (e^{i\theta \mathbf{v}\cdot \mathbf{\sigma }})& =\det ((\cos \theta )1+i(\sin \theta )\mathbf{v}\cdot \mathbf{\sigma })\\ & =\det \left(\begin{array}{cc}\cos \theta +i(\sin \theta )v_3& i(\sin \theta )(v_1-i{v}_2)\\ i(\sin \theta )(v_1+i{v}_2)& \cos \theta -i(\sin \theta )v_3\end{array}\right)\\ & ={\cos }^2\theta +({\sin }^2\theta )(v_1^2+v_2^2+v_3^2)\\ & =1\end{array}$$

So, we see that by exponentiating $i$ times linear combinations of the self-adjoint Pauli matrices (which all have trace zero), we get unitary matrices of determinant one. These are invertible, and form the group named $SU(2)$ , the group of unitary 2 by 2 matrices of determinant one. If we exponentiated not just $i\theta \mathbf{v}\cdot \sigma .$ , but $i(\varphi 1+\theta \mathbf{v}\cdot \sigma )$ for some real constant $\varphi$ (such matrices will not have trace zero unless $\varphi =0$) , we would get a unitary matrix with determinant $e^{i2\varphi }$ . The group of all unitary 2 by 2 matrices is called $U(2)$ . It contains as subgroups $SU(2)$ as well as the $U(1)$ described at the beginning of this section. $U(2)$ is slightly different from the product of these two subgroups, since the group element

$$\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)$$

is in both subgroups. In chapter 4 we will encounter the generalization to $SU(n)$ and $U(n)$ , groups of unitary $n$ by $n$ complex matrices.

To get some more insight into the structure of the group $SU(2)$ , consider an arbitrary 2 by 2 complex matrix

$$\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)$$

Unitarity implies that the rows are orthonormal. This results from the condition that the matrix times its conjugate-transpose is the identity

$$\left(\begin{array}{cc}\alpha & \beta \\ \gamma & \delta \end{array}\right)\left(\begin{array}{cc}\overline{\alpha }& \overline{\gamma }\\ \overline{\beta }& \overline{\delta }\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

Orthogonality of the two rows gives the relation

$$\gamma \overline{\alpha }+\delta \overline{\beta }=0⟹\delta =-\frac{\gamma \overline{\alpha }}{\overline{\beta }}$$

The condition that the first row has length one gives

$$\alpha \overline{\alpha }+\beta \overline{\beta }=|\alpha {|}^2+|\beta {|}^2=1$$

Using these two relations and computing the determinant (which has to be 1) gives

$$\alpha \delta -\beta \gamma =-\frac{\alpha \overline{\alpha }\gamma }{\overline{\beta }}-\beta \gamma =-\frac{\gamma }{\overline{\beta }}(\alpha \overline{\alpha }+\beta \overline{\beta })=-\frac{\gamma }{\overline{\beta }}=1$$

so one must have

$$\gamma =-\overline{\beta },\delta =\overline{\alpha }$$

and an $SU(2)$ matrix will have the form

$$\left(\begin{array}{cc}\alpha & \beta \\ -\overline{\beta }& \overline{\alpha }\end{array}\right)$$

where $(\alpha ,\beta )\in {\mathbf{C}}^2$ and

$$|\alpha {|}^2+|\beta {|}^2=1$$

The elements of $SU(2)$ are thus parametrized by two complex numbers, with the sum of their length-squareds equal to one. Identifying ${\mathbf{C}}^2={\mathbf{R}}^4$, these are vectors of length one in ${\mathbf{R}}^4$. Just as $U(1)$ could be identified as a space with the unit circle $S^1$ in $\mathbf{C}={\mathbf{R}}^2$, $SU(2)$ can be identified with the unit three-sphere $S^3$ in ${\mathbf{R}}^4$.

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上一节：3 Two-state Systems and SU(2) · 下一节：3.2 Commutation relations for Pauli matrices

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