# Rotation of the spin axis

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 14, Exercise 14.3.6.

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Just as orbital angular momentum operator ${\mathbf{L}}$ is the generator of rotations, the spin operator ${\mathbf{S}}$ can also be used as the generator of rotations in spin space by means of the unitary operator

$\displaystyle U\left[R\left(\boldsymbol{\theta}\right)\right]=e^{-i\boldsymbol{\theta}\cdot\mathbf{S}/\hbar}=e^{-i\boldsymbol{\theta}\cdot\boldsymbol{\sigma}/2} \ \ \ \ \ (1)$

where we’ve written the operator in terms of the Pauli matrices ${\boldsymbol{\sigma}}$, the components of which are

$\displaystyle \sigma_{x}=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right];\quad\sigma_{y}=\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right];\quad\sigma_{z}=\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (2)$

For a spin pointing the direction ${\hat{n}}$, where ${\hat{n}}$ is defined in terms of the spherical angles as

$\displaystyle \hat{n}=\sin\theta\cos\phi\hat{\mathbf{x}}+\sin\theta\sin\phi\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (3)$

the corresponding eigenvectors of the operator ${\hat{n}\cdot\mathbf{S}}$ are

 $\displaystyle \left|\hat{n}+\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\frac{\theta}{2}e^{-i\phi/2}\\ \sin\frac{\theta}{2}e^{i\phi/2} \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle \left|\hat{n}-\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} -\sin\frac{\theta}{2}e^{-i\phi/2}\\ \cos\frac{\theta}{2}e^{i\phi/2} \end{array}\right] \ \ \ \ \ (5)$

If we start with spin pointing in the ${+z}$ direction, then it is in the state

$\displaystyle \left|s_{z}=\frac{\hbar}{2}\right\rangle =\frac{\hbar}{2}\left[\begin{array}{c} 1\\ 0 \end{array}\right] \ \ \ \ \ (6)$

then it should be possible to rotate this state into the general state 4 by applying the correct rotation operators in sequence.

Suppose we first rotate the state by an angle ${\theta}$ about the ${y}$ axis. This rotates the axis of spin so that it lies in the ${xz}$ plane in the first quadrant (that is, positive ${x}$ and positive ${z}$), making an angle ${\theta}$ with the ${z}$ axis. We can now rotate again by an angle ${\phi}$ about the (original) ${z}$ axis. The axis of spin now points in the direction given by ${\hat{n}}$ in 3. That is, it should be true that

$\displaystyle \left|\hat{n}+\right\rangle =U\left[R\left(\phi\hat{\mathbf{z}}\right)\right]U\left[R\left(\theta\hat{\mathbf{y}}\right)\right]\left[\begin{array}{c} 1\\ 0 \end{array}\right] \ \ \ \ \ (7)$

In order to verify this by direct calculation, we need an explicit form for ${U}$. This is derived by Shankar in his equation 14.3.44 so we won’t repeat the derivation here. Basically, it uses the fact that ${\left(\hat{n}\cdot\boldsymbol{\sigma}\right)^{2}=I}$ and expands the exponential 1 as a power series, with the result

$\displaystyle U\left[R\left(\boldsymbol{\theta}\right)\right]=\cos\frac{\theta}{2}I-i\sin\frac{\theta}{2}\left(\hat{\theta}\cdot\boldsymbol{\sigma}\right) \ \ \ \ \ (8)$

We can use this formula to do the calculation.

 $\displaystyle U\left[R\left(\theta\hat{\mathbf{y}}\right)\right]\left[\begin{array}{c} 1\\ 0 \end{array}\right]$ $\displaystyle =$ $\displaystyle \left[\cos\frac{\theta}{2}I-i\sin\frac{\theta}{2}\sigma_{y}\right]\left[\begin{array}{c} 1\\ 0 \end{array}\right]\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\frac{\theta}{2}\\ 0 \end{array}\right]-i\sin\frac{\theta}{2}\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\left[\begin{array}{c} 1\\ 0 \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\frac{\theta}{2}\\ \sin\frac{\theta}{2} \end{array}\right] \ \ \ \ \ (11)$

Applying the second rotation we get

 $\displaystyle U\left[R\left(\phi\hat{\mathbf{z}}\right)\right]\left[\begin{array}{c} \cos\frac{\theta}{2}\\ \sin\frac{\theta}{2} \end{array}\right]$ $\displaystyle =$ $\displaystyle \left[\cos\frac{\phi}{2}I-i\sin\frac{\phi}{2}\sigma_{z}\right]\left[\begin{array}{c} \cos\frac{\theta}{2}\\ \sin\frac{\theta}{2} \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\frac{\theta}{2}\cos\frac{\phi}{2}\\ \sin\frac{\theta}{2}\cos\frac{\phi}{2} \end{array}\right]-i\sin\frac{\phi}{2}\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right]\left[\begin{array}{c} \cos\frac{\theta}{2}\\ \sin\frac{\theta}{2} \end{array}\right]\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\frac{\theta}{2}\left(\cos\frac{\phi}{2}-i\sin\frac{\phi}{2}\right)\\ \sin\frac{\theta}{2}\left(\cos\frac{\phi}{2}+i\sin\frac{\phi}{2}\right) \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\frac{\theta}{2}e^{-i\phi/2}\\ \sin\frac{\theta}{2}e^{i\phi/2} \end{array}\right] \ \ \ \ \ (15)$

which agrees with 4.

# Any spinor is an eigenket of the spin operator for some direction of spin

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 14, Exercise 14.3.1.

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The eigenvectors of the spin ${\frac{1}{2}}$ matrix in an arbitrary direction are given by

 $\displaystyle \left|\hat{n}+\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\frac{\theta}{2}e^{-i\phi/2}\\ \sin\frac{\theta}{2}e^{i\phi/2} \end{array}\right]\ \ \ \ \ (1)$ $\displaystyle \left|\hat{n}-\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} -\sin\frac{\theta}{2}e^{-i\phi/2}\\ \cos\frac{\theta}{2}e^{i\phi/2} \end{array}\right] \ \ \ \ \ (2)$

where the direction vector is given by

$\displaystyle \hat{\mathbf{n}}=\sin\theta\cos\phi\hat{\mathbf{x}}+\sin\theta\sin\phi\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (3)$

The corresponding spin operator is given by the matrix

$\displaystyle \hat{\mathbf{n}}\cdot\mathbf{S}=\frac{\hbar}{2}\left[\begin{array}{cc} \cos\theta & \sin\theta e^{-i\phi}\\ \sin\theta e^{i\phi} & -\cos\theta \end{array}\right] \ \ \ \ \ (4)$

Any 2-component normalized spinor is an eigenvector of such a matrix. To see this, suppose we have an arbitrary spinor written as

 $\displaystyle \left|\chi\right\rangle$ $\displaystyle =$ $\displaystyle \rho_{1}e^{i\phi_{1}}\left[\begin{array}{c} 1\\ 0 \end{array}\right]+\rho_{2}e^{i\phi_{2}}\left[\begin{array}{c} 0\\ 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \rho_{1}e^{i\phi_{1}}\\ \rho_{2}e^{i\phi_{2}} \end{array}\right] \ \ \ \ \ (6)$

where ${\rho_{1,2}}$ and ${\phi_{1,2}}$ are arbitrary real numbers (so that the coefficients on the RHS are arbitrary complex numbers). From normalization we have

$\displaystyle \left\langle \chi\left|\chi\right.\right\rangle =1=\left[\begin{array}{cc} \rho_{1}e^{-i\phi_{1}} & \rho_{2}e^{-i\phi_{2}}\end{array}\right]\left[\begin{array}{c} \rho_{1}e^{i\phi_{1}}\\ \rho_{2}e^{i\phi_{2}} \end{array}\right]=\rho_{1}^{2}+\rho_{2}^{2} \ \ \ \ \ (7)$

Thus we can write ${\rho_{1}}$ and ${\rho_{2}}$ as the sine and cosine of some angle, which we’ll call ${\frac{\theta}{2}}$, giving

$\displaystyle \left|\chi\right\rangle =\left[\begin{array}{c} \cos\frac{\theta}{2}e^{i\phi_{1}}\\ \sin\frac{\theta}{2}e^{i\phi_{2}} \end{array}\right] \ \ \ \ \ (8)$

We can put this in the form 1 as follows. Since an overall phase doesn’t affect the physics of the spinor, we can write

$\displaystyle \left|\chi\right\rangle =e^{i\alpha}\left[\begin{array}{c} \cos\frac{\theta}{2}e^{-i\phi/2}\\ \sin\frac{\theta}{2}e^{i\phi/2} \end{array}\right]=\left[\begin{array}{c} \cos\frac{\theta}{2}e^{i\phi_{1}}\\ \sin\frac{\theta}{2}e^{i\phi_{2}} \end{array}\right] \ \ \ \ \ (9)$

We have the conditions

 $\displaystyle \phi_{1}$ $\displaystyle =$ $\displaystyle \alpha-\frac{\phi}{2}\ \ \ \ \ (10)$ $\displaystyle \phi_{2}$ $\displaystyle =$ $\displaystyle \alpha+\frac{\phi}{2} \ \ \ \ \ (11)$

Solving, we get

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \frac{\phi_{1}+\phi_{2}}{2}\ \ \ \ \ (12)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \phi_{2}-\phi_{1} \ \ \ \ \ (13)$

giving

$\displaystyle \left|\chi\right\rangle =e^{i\left(\phi_{1}+\phi_{2}\right)/2}\left[\begin{array}{c} \cos\frac{\theta}{2}e^{-i\left(\phi_{2}-\phi_{1}\right)/2}\\ \sin\frac{\theta}{2}e^{i\left(\phi_{2}-\phi_{1}\right)/2} \end{array}\right] \ \ \ \ \ (14)$

Thus ${\left|\chi\right\rangle }$ as given by 6 is an eigenvector of the operator 4, where

$\displaystyle \hat{\mathbf{n}}=\sin\theta\cos\left(\phi_{2}-\phi_{1}\right)\hat{\mathbf{x}}+\sin\theta\sin\left(\phi_{2}-\phi_{1}\right)\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (15)$

# Dirac spin operator in quantum field theory

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4.

The total spin of a multiparticle Dirac state is a bit trickier to calculate than the total momentum. For the momentum, the result turned out to be

$\displaystyle \mathbf{P}=\sum_{r,\mathbf{p}}\mathbf{p}\left(N_{r}\left(\mathbf{p}\right)+\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (1)$

which is just the sum of the momenta of the particles and antiparticles. This works because a multiparticle state is an eigenstate of all the number operators, with the eigenvalues being just the number of particles in each state with spin ${r}$ and momentum ${\mathbf{p}}$, and ${\mathbf{p}}$ itself is just a 3-vector which multiplies the result. For example, we can operate on a multiparticle state to get the total momentum like this:

$\displaystyle \mathbf{P}\left|\psi_{r_{1}\mathbf{p}_{1}}\psi_{r_{2}\mathbf{p}_{2}}\psi_{r_{1}\mathbf{p}_{3}}\bar{\psi}_{r_{1}\mathbf{p}_{1}}\right\rangle =2\mathbf{p}_{1}+\mathbf{p}_{2}+\mathbf{p}_{3} \ \ \ \ \ (2)$

If we tried something analogous for the spin component ${\Sigma_{j}}$ we would get

$\displaystyle \Sigma_{j,tot}=\sum_{r,\mathbf{p}}\Sigma_{j}\left(N_{r}\left(\mathbf{p}\right)+\bar{N}_{r}\left(\mathbf{p}\right)\right) \ \ \ \ \ (3)$

As with the momentum, a multiparticle state is still an eigenstate of the number operators, but ${\Sigma_{j}}$ is a matrix operator which can operate only on single particle states, that is, states containing a single 4-component spinor. That is, the operation

$\displaystyle \Sigma_{j}\left|\psi_{r_{1}\mathbf{p}_{1}}\psi_{r_{2}\mathbf{p}_{2}}\psi_{r_{1}\mathbf{p}_{3}}\bar{\psi}_{r_{1}\mathbf{p}_{1}}\right\rangle \ \ \ \ \ (4)$

is not defined, so 3 is not a well-defined quantity.

The solution turns out to be defining the total spin operator as

$\displaystyle _{QFT}\Sigma_{j}=\int_{V}\psi^{\dagger}\Sigma_{j}\psi\;d^{3}x \ \ \ \ \ (5)$

where the ${\Sigma_{j}}$ in the integrand is the usual Dirac spin operator, and ${\psi}$ and ${\psi^{\dagger}}$ are the general solutions to the Dirac equation

 $\displaystyle \psi$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[c_{r}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)e^{-ipx}+d_{r}^{\dagger}\left(\mathbf{p}\right)v_{r}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \psi^{+}+\psi^{-}\ \ \ \ \ (7)$ $\displaystyle \psi^{\dagger}$ $\displaystyle =$ $\displaystyle \sum_{r=1}^{2}\sum_{\mathbf{p}}\sqrt{\frac{m}{VE_{\mathbf{p}}}}\left[d_{r}\left(\mathbf{p}\right)v_{r}^{\dagger}\left(\mathbf{p}\right)e^{-ipx}+c_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}^{\dagger}\left(\mathbf{p}\right)e^{ipx}\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \psi^{\dagger+}+\psi^{\dagger-} \ \ \ \ \ (9)$

Klauber evaluates the integral in his section 4.9.1 for the case of ${\psi}$ and ${\psi^{\dagger}}$ containing only ${c}$ and ${c^{\dagger}}$ operators. The integration uses the usual property of such integrals that any term in the integrand containing an exponential ${e^{ipx}}$ goes to zero because of the boundary conditions. We are left with

$\displaystyle _{QFT}\Sigma_{j}=\sum_{r,s,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}u_{r}^{\dagger}\left(\mathbf{p}\right)\Sigma_{j}u_{s}\left(\mathbf{p}\right)c_{r}^{\dagger}\left(\mathbf{p}\right)c_{s}\left(\mathbf{p}\right) \ \ \ \ \ (10)$

If we apply this operator to some state ${\left|\psi_{s^{\prime}\mathbf{p}^{\prime}}\right\rangle }$ then because the ${c_{s}\left(\mathbf{p}\right)}$ operator (an annihiliation operator) is the first one to operate on the state, only operators with ${s=s^{\prime}}$ and ${\mathbf{p}=\mathbf{\mathbf{p}^{\prime}}}$ will produce a non-zero result. In those cases, the ${s^{\prime}\mathbf{\mathbf{p}^{\prime}}}$ particle is annihilated and then replaced with a ${r\mathbf{p}}$ particle because of the creation operator ${c_{r}^{\dagger}\left(\mathbf{p}\right)}$. That is, the sum over ${\mathbf{p}}$ collapses to a single term where ${\mathbf{p}=\mathbf{\mathbf{p}^{\prime}}}$ and the sum over ${s}$ is eliminated:

$\displaystyle _{QFT}\Sigma_{j}\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\right\rangle =\sum_{r}\frac{m}{E_{\mathbf{\mathbf{p}^{\prime}}}}u_{r}^{\dagger}\left(\mathbf{\mathbf{p}^{\prime}}\right)\Sigma_{j}u_{r}\left(\mathbf{p}^{\prime}\right)\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\right\rangle \ \ \ \ \ (11)$

Now suppose we apply this operator to a 2-particle state ${\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\psi_{s^{\prime\prime}\mathbf{p}^{\prime\prime}}\right\rangle }$. We’ll get a term like 11 for each particle, with the result

$\displaystyle _{QFT}\Sigma_{j}\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\psi_{s^{\prime\prime}\mathbf{p}^{\prime\prime}}\right\rangle =\left[\sum_{r}\frac{m}{E_{\mathbf{\mathbf{p}^{\prime}}}}u_{r}^{\dagger}\left(\mathbf{\mathbf{p}^{\prime}}\right)\Sigma_{j}u_{r}\left(\mathbf{p}^{\prime}\right)+\sum_{r}\frac{m}{E_{\mathbf{\mathbf{p}^{\prime\prime}}}}u_{r}^{\dagger}\left(\mathbf{p}^{\prime\prime}\right)\Sigma_{j}u_{r}\left(\mathbf{p}^{\prime\prime}\right)\right]\left|\psi_{s^{\prime}\mathbf{\mathbf{p}^{\prime}}}\psi_{s^{\prime\prime}\mathbf{p}^{\prime\prime}}\right\rangle \ \ \ \ \ (12)$

Thus in general, because only the terms in the sum over ${\mathbf{p}}$ in 10 that correspond to the momenta of the particles in the many particle state will be non-zero when this operator is applied to that many particle state, and the sum over ${s}$ also collapses, we can write 10 as

 $\displaystyle _{QFT}\Sigma_{j}$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}u_{r}^{\dagger}\left(\mathbf{p}\right)\Sigma_{j}u_{r}\left(\mathbf{p}\right)c_{r}^{\dagger}\left(\mathbf{p}\right)c_{r}\left(\mathbf{p}\right)\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{r,\mathbf{p}}\frac{m}{E_{\mathbf{p}}}u_{r}^{\dagger}\left(\mathbf{p}\right)\Sigma_{j}u_{r}\left(\mathbf{p}\right)N_{r}\left(\mathbf{p}\right) \ \ \ \ \ (14)$

[Klauber’s equation 4-113 is a bit sloppy since he applies 10 to a state which he calls ${\left|\psi_{s\mathbf{p}}\right\rangle }$ which contains the two summation variables ${s}$ and ${\mathbf{p}}$, when in fact this state should refer to a specific spin and momentum and not be part of the sum over ${s}$ and ${\mathbf{p}}$. Likewise, he retains the sum over ${\mathbf{p}}$ in 4-114 even though the annihilation operator removes all but one momentum. The final result 4-115 does appear to be correct however.]

The expectation value of the operator ${_{QFT}\Sigma_{j}}$ between two multiparticle states ${\left|\psi_{1}\right\rangle }$ and ${\left|\psi_{2}\right\rangle }$ is therefore

$\displaystyle \left\langle \psi_{1}\left|_{QFT}\Sigma_{j}\right|\psi_{2}\right\rangle \ \ \ \ \ (15)$

Because the term ${u_{r}^{\dagger}\left(\mathbf{p}\right)\Sigma_{j}u_{r}\left(\mathbf{p}\right)}$ is the product of a ${1\times4}$ row vector (${u_{r}^{\dagger}\left(\mathbf{p}\right)}$), a ${4\times4}$ matrix (${\Sigma_{j}}$) and a ${4\times1}$ column vector (${u_{r}\left(\mathbf{p}\right)}$), the result is just a scalar, that is, a number. Also, because the ${u_{r}}$ spinors are eigenspinors of the ${\Sigma_{3}}$ operator, we have ${\Sigma_{3}u_{up}=+\frac{1}{2}u_{up}}$ and ${\Sigma_{3}u_{down}=-\frac{1}{2}u_{down}}$. Finally, the inner product ${u_{r}^{\dagger}\left(\mathbf{p}\right)u_{r}\left(\mathbf{p}\right)=E_{\mathbf{p}}/m}$. So applying 14 to the state ${\left|\psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\right\rangle }$, for example, and calculating the expectation value in that state, we get

 $\displaystyle \left\langle \psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\left|_{QFT}\Sigma_{3}\right|\psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle \psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\left|\left[+\frac{1}{2}+\frac{1}{2}\right]\right|\psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\right\rangle \ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle +1\left|\psi_{up,\mathbf{\mathbf{p}^{\prime}}}\psi_{up,\mathbf{p}^{\prime\prime}}\right\rangle \ \ \ \ \ (17)$

The expectation value of ${_{QFT}\Sigma_{j}}$ between any two different states produces zero because different multiparticle states are orthogonal.

# Helicity operator in the Dirac equation

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.20.

In the Dirac equation, the spin of a particle is described using the spin operator

$\displaystyle \Sigma_{i}=\frac{1}{2}\left[\begin{array}{cc} \sigma_{i} & 0\\ 0 & \sigma_{i} \end{array}\right];\;i=x,y,z \ \ \ \ \ (1)$

(using natural units where ${\hbar=1}$) and where the Pauli matrices are

 $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \sigma_{y}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle \sigma_{z}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (4)$

and the 0 components in 1 are ${2\times2}$ zero matrices. Spin can be oriented in any direction for particles travelling at a velocity ${v<1}$, although at the speed of light (${v=1}$), the spin is always aligned with the velocity due to length contraction effects. The relation between the directions of spin and the particle’s velocity is given by the helicity. If the 3-momentum ${\mathbf{p}}$ and spin both point in the same direction, the helicity has its maximum value (a positive quantity), while if they point in opposite directions, the helicity has its maximum negative value. If ${\mathbf{p}}$ and spin are at right angles, the helicity is zero.

These relations suggest that a helicity operator can be defined in terms of the scalar product of ${\boldsymbol{\Sigma}}$ and ${\mathbf{p}}$. The helicity operator is

 $\displaystyle \Sigma_{\mathbf{p}}$ $\displaystyle \equiv$ $\displaystyle \boldsymbol{\Sigma}\cdot\frac{\mathbf{p}}{\left|\mathbf{p}\right|}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \Sigma_{1}\frac{p^{1}}{\left|\mathbf{p}\right|}+\Sigma_{2}\frac{p^{2}}{\left|\mathbf{p}\right|}+\Sigma_{3}\frac{p^{3}}{\left|\mathbf{p}\right|} \ \ \ \ \ (6)$

Since the ${\Sigma_{i}}$ are each a ${4\times4}$ matrix, the helicity ${\Sigma_{\mathbf{p}}}$ is also a ${4\times4}$ matrix, but with reference to 3-d space, it is a scalar matrix.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (7)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (8)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (9)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (10)$

For a particle moving in the ${z}$ direction, ${p^{3}>0}$, ${p^{1}=p^{2}=0}$ and we’ve seen that the states ${\left|\psi^{\left(1\right)}\right\rangle }$ and ${\left|\psi^{\left(3\right)}\right\rangle }$ are eigenstates of ${\Sigma_{3}}$ with eigenvalue ${\frac{1}{2}}$. In this case, ${p^{3}/\left|\mathbf{p}\right|=1}$ so from 6

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (11)$

Thus for a particle whose spin and velocity both point in the same direction, the maximum helicity value of ${\frac{1}{2}}$ is obtained. Similarly, if ${p^{3}<0}$, ${p^{1}=p^{2}=0}$ and ${p^{3}/\left|\mathbf{p}\right|=-1}$ so

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =-\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (12)$

With velocity and spin pointing in opposite directions, the maximum negative value of ${-\frac{1}{2}}$ is obtained for the helicity.

Now suppose we have a particle in state ${\left|\psi^{\left(2\right)}\right\rangle }$ and that ${p^{1}\ne0}$, ${p^{2}=p^{3}=0}$. This is not a helicity eigenstate, since ${\left|\psi^{\left(2\right)}\right\rangle }$ is an eigenstate of ${\Sigma_{3}}$ with eigenvalue ${-\frac{1}{2}}$, so the spin is in the ${-z}$ direction while the velocity is in the ${x}$ direction, so the spin is not parallel to the velocity.

Mathematically, in order for ${\left|\psi^{\left(2\right)}\right\rangle }$ to be a helicity eigenstate, the spinor in ${\left|\psi^{\left(2\right)}\right\rangle }$ would have to be an eigenstate of ${\Sigma_{1}}$, which is not true, since

 $\displaystyle \Sigma_{1}\left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right]\sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}}{E+m}\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ 0\\ \frac{p^{1}}{E+m} \end{array}\right]e^{-ipx} \ \ \ \ \ (14)$

If ${p^{3}\ne0}$, ${p^{1}=p^{2}=0}$, however, we would expect ${\left|\psi^{\left(2\right)}\right\rangle }$ to be a helicity eigenstate since the spin and velocity are parallel in this case. Since ${\left|\psi^{\left(2\right)}\right\rangle }$ is an eigenstate of ${\Sigma_{3}}$ with eigenvalue ${-\frac{1}{2}}$, we have

$\displaystyle \Sigma_{\mathbf{p}}\left|\psi^{\left(1\right)}\right\rangle =\pm\frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (15)$

with the + corresponding to ${p^{3}>0}$ and the ${-}$ to ${p^{3}<0}$.

# Dirac equation: non-relativistic limit

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.19.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

In the non-relativistic limit, the relative velocity of the particle satisfies ${v\ll1}$, which means that the momentum components all satisfy ${p^{j}\ll E\approx m}$. Thus ${\sqrt{\frac{E+m}{2m}}\approx1}$ and the solutions reduce to

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (5)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (6)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right]e^{ipx}\ \ \ \ \ (7)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right]e^{ipx} \ \ \ \ \ (8)$

Comparing this with the free particle solutions to the non-relativistic Schrödinger equation, we see that the ${e^{\pm ipx}=e^{\pm Et}e^{\mp\mathbf{p}\cdot\mathbf{x}}}$ factor is just what we’d get in that case. The first two components of the spinors in ${\left|\psi^{\left(1\right)}\right\rangle }$ and ${\left|\psi^{\left(2\right)}\right\rangle }$ also correspond to the eigenstates in the non-relativistic spin ${\frac{1}{2}}$ theory.

In the 4-d case, we can operate on these solutions with the 4-d spin operator ${\Sigma_{z}}$ to get

 $\displaystyle \Sigma_{z}\left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (11)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-ipx}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(2\right)}\right\rangle \ \ \ \ \ (14)$

Similarly, we get

 $\displaystyle \Sigma_{z}\left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(3\right)}\right\rangle \ \ \ \ \ (15)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(4\right)}\right\rangle \ \ \ \ \ (16)$

These are the same results that we get by applying the Pauli spin matrices to the 2-d spin space spinors in the non-relativistic theory.

# Dirac equation: spinors near the speed of light

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.18.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

For a particle at rest (${\mathbf{p}=0,\;E=m}$), or for a particle moving in the ${z}$ direction, all four solutions are eigenstates of the spin operator ${\Sigma_{z}}$, with eigenvalues (spins) of ${\pm\frac{1}{2}}$. If the particle is moving in the ${x}$ or ${y}$ direction, the individual spinors above aren’t eigenstates of any of the spin operators. As the speed of the particle approaches ${c}$, however, we can get some eigenstates of ${\Sigma_{x}}$ and ${\Sigma_{y}}$.

First, suppose the particle is moving in the ${x}$ direction at a speed approaching ${c}$, with any motion in the ${y}$ and ${z}$ directions much smaller by comparison. In this case, ${E\rightarrow p^{1}\rightarrow\infty}$ and the spinor components (which we’ll call ${s^{\left(n\right)}}$ for ${n=1,\ldots,4}$) of the solutions above become

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (7)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (8)$

The ${x}$ spin operator is

$\displaystyle \Sigma_{x}=\frac{1}{2}\left[\begin{array}{cccc} 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right] \ \ \ \ \ (9)$

Multiplying ${\Sigma_{x}}$ into ${s^{\left(1\right)}+s^{\left(2\right)}}$, we get

 $\displaystyle \Sigma_{x}\left(s^{\left(1\right)}+s^{\left(2\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(s^{\left(1\right)}+s^{\left(2\right)}\right) \ \ \ \ \ (11)$

Similarly

 $\displaystyle \Sigma_{x}\left(s^{\left(3\right)}+s^{\left(4\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 1\\ 1\\ 1 \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(s^{\left(3\right)}+s^{\left(4\right)}\right) \ \ \ \ \ (13)$

Thus the sums ${u_{1}+u_{2}}$ and ${v_{1}+v_{2}}$ are both eigenstates of ${\Sigma_{x}}$ with eigenvalue ${\frac{1}{2}}$.

Now suppose the particle is moving in the ${y}$ direction with ${v^{y}\rightarrow1}$ so that ${E\rightarrow p^{2}\rightarrow\infty}$. The four spinors now become

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ i \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ -i\\ 0 \end{array}\right]\ \ \ \ \ (15)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ i\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} -i\\ 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (17)$

The ${y}$ spin operator is

$\displaystyle \Sigma_{y}=\frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right] \ \ \ \ \ (18)$

In this case

 $\displaystyle \Sigma_{y}\left(s^{\left(1\right)}+s^{\left(3\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right]\left[\begin{array}{c} 1\\ i\\ 1\\ i \end{array}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ i\\ 1\\ i \end{array}\right]\ \ \ \ \ (20)$ $\displaystyle \Sigma_{y}\left(s^{\left(2\right)}+s^{\left(4\right)}\right)$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{cccc} 0 & -i & 0 & 0\\ i & 0 & 0 & 0\\ 0 & 0 & 0 & -i\\ 0 & 0 & i & 0 \end{array}\right]\left[\begin{array}{c} -i\\ 1\\ -i\\ 1 \end{array}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} -i\\ 1\\ -i\\ 1 \end{array}\right] \ \ \ \ \ (22)$

Thus the sums ${u_{1}+v_{2}}$ and ${v_{1}+u_{2}}$ are both eigenstates of ${\Sigma_{y}}$ with eigenvalue ${\frac{1}{2}}$. [As the ${u_{j}}$ spinors are supposed to represent particles and the ${v_{j}}$ antiparticles, I’m not sure what a mixture of the two is supposed to represent.]

# Dirac equation: spin of a moving particle

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.17.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

For a particle at rest (${\mathbf{p}=0,\;E=m}$), all four solutions are eigenstates of the spin operator ${\Sigma_{z}}$, with eigenvalues (spins) of ${\pm\frac{1}{2}}$. Here, we have a look at what happens if the particle is moving.

First, suppose the particle is moving in the ${x}$ direction, so that ${p^{1}\ne0}$ and ${p^{2}=p^{3}=0}$. In this case, the spinor components (which we’ll call ${s^{\left(n\right)}}$ for ${n=1,\ldots,4}$) of the solutions above become

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ \frac{p^{1}}{E+m} \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}}{E+m}\\ 0 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ \frac{p^{1}}{E+m}\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (7)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \frac{p^{1}}{E+m}\\ 0\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (8)$

The ${z}$ spin operator is

$\displaystyle \Sigma_{z}=\frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (9)$

Multiplying ${\Sigma_{z}}$ into the four ${s^{\left(n\right)}}$ spinors, we get

 $\displaystyle \Sigma_{z}s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 0\\ 0\\ -\frac{p^{1}}{E+m} \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle \Sigma_{z}s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 0\\ -1\\ \frac{p^{1}}{E+m}\\ 0 \end{array}\right]\ \ \ \ \ (11)$ $\displaystyle \Sigma_{z}s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 0\\ -\frac{p^{1}}{E+m}\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle \Sigma_{z}s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} \frac{p^{1}}{E+m}\\ 0\\ 0\\ -1 \end{array}\right] \ \ \ \ \ (13)$

In each case, one of the non-zero components of ${s^{\left(n\right)}}$ has its sign changed, while the other non-zero component remains the same. Thus none of the ${s^{\left(n\right)}}$ spinors is an eigenstate of ${\Sigma_{z}}$.

Now suppose that ${p^{1}=p^{2}=0}$ and ${p^{3}\ne0}$. The spinors are now

 $\displaystyle s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ 0 \end{array}\right]\ \ \ \ \ (14)$ $\displaystyle s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 0\\ -\frac{p^{3}}{E+m} \end{array}\right]\ \ \ \ \ (15)$ $\displaystyle s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \frac{p^{3}}{E+m}\\ 0\\ 1\\ 0 \end{array}\right]\ \ \ \ \ (16)$ $\displaystyle s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right] \ \ \ \ \ (17)$

Multiplying by ${\Sigma_{z}}$ we get

 $\displaystyle \Sigma_{z}s^{\left(1\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ 0 \end{array}\right]=\frac{1}{2}s^{\left(1\right)}\ \ \ \ \ (18)$ $\displaystyle \Sigma_{z}s^{\left(2\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 0\\ -1\\ 0\\ \frac{p^{3}}{E+m} \end{array}\right]=-\frac{1}{2}s^{\left(2\right)}\ \ \ \ \ (19)$ $\displaystyle \Sigma_{z}s^{\left(3\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ 0\\ 1\\ 0 \end{array}\right]=\frac{1}{2}s^{\left(3\right)}\ \ \ \ \ (20)$ $\displaystyle \Sigma_{z}s^{\left(4\right)}$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left[\begin{array}{c} 0\\ \frac{p^{3}}{E+m}\\ 0\\ -1 \end{array}\right]=-\frac{1}{2}s^{\left(4\right)} \ \ \ \ \ (21)$

Thus the ${s^{\left(n\right)}}$ spinors in this case are eigenstates of ${\Sigma_{z}}$.

# Dirac equation: spin of a particle at rest

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.16.

The solutions to the Dirac equation consist of a 4-element column spinor and a spacetime component:

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 1\\ 0\\ \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{1}e^{-ipx}\ \ \ \ \ (1)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} 0\\ 1\\ \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m} \end{array}\right]e^{-ipx}\equiv u_{2}e^{-ipx}\ \ \ \ \ (2)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{3}}{E+m}\\ \frac{p^{1}+ip^{2}}{E+m}\\ 1\\ 0 \end{array}\right]e^{ipx}\equiv v_{2}e^{ipx}\ \ \ \ \ (3)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{E+m}{2m}}\left[\begin{array}{c} \frac{p^{1}-ip^{2}}{E+m}\\ -\frac{p^{3}}{E+m}\\ 0\\ 1 \end{array}\right]e^{ipx}\equiv v_{1}e^{ipx} \ \ \ \ \ (4)$

The last time we found quantum mechanical solutions that contained column vectors, we introduced these solutions to account for spin in non-relativistic quantum mechanics. With the Dirac equation, spin finds its way into the solutions by being a consequence of the nature of the solutions, rather than by being imposed.

An experimental fact about spin is that the intrinsic spin of a particle is not affected by how fast it is moving. That is, an electron always has spin ${\frac{1}{2}}$ whether it is observed at rest or moving close to the speed of light. However, due to length contraction, the direction of an angular momentum vector for a spinning object does vary with velocity relative to the observer. This is explained more fully in Klauber’s Box 4-2, which looks at the effect of length contraction on a classical (non-quantum) rotating disk. In summary, suppose the angular momentum vector ${\mathbf{L}}$ lies in the ${x-z}$ plane at some angle ${\theta}$ to the ${x}$ axis, so that the plane of the disk, being perpendicular to ${\mathbf{L}}$, makes the same angle ${\theta}$ with the ${z}$ axis. Now suppose the disk moves along the ${x}$ axis at some speed ${v}$. As ${v}$ becomes relativistic, the angle ${\theta}$ diminishes, since the size component of the disk in the ${x}$ direction is contracted, while the component in the ${z}$ direction remains unchanged. When ${v\rightarrow c}$, the disk’s ${x}$ component tends to zero, so that ${\mathbf{L}}$ lies along the ${x}$ axis and the disk spins in the ${yz}$ plane.

Because both ${E}$ and ${\mathbf{p}}$ depend ultimately on ${v}$ in the solutions above, the Dirac solutions actually already contain this relativistic effect within the spinor components. To see this, we need the spin operators in the Dirac theory, in analogy to the Pauli matrices for non-relativistic spin ${\frac{1}{2}}$. For now, we will take these operators to be god-given. They are

$\displaystyle \Sigma_{i}=\frac{1}{2}\left[\begin{array}{cc} \sigma_{i} & 0\\ 0 & \sigma_{i} \end{array}\right];\;i=x,y,z \ \ \ \ \ (5)$

(using natural units where ${\hbar=1}$) and where the Pauli matrices are

 $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle \sigma_{y}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (7)$ $\displaystyle \sigma_{z}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (8)$

and the 0 components in 5 are ${2\times2}$ zero matrices.

As a simple example of how the ${\Sigma_{i}}$ operators work, consider the special case of a particle at rest, so that ${\mathbf{p}=0}$ and ${E=m}$. Then the four solutions at the top reduce to

 $\displaystyle \left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 1\\ 0\\ 0\\ 0 \end{array}\right]e^{-imt}\ \ \ \ \ (9)$ $\displaystyle \left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 1\\ 0\\ 0 \end{array}\right]e^{-imt}\ \ \ \ \ (10)$ $\displaystyle \left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 1\\ 0 \end{array}\right]e^{imt}\ \ \ \ \ (11)$ $\displaystyle \left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} 0\\ 0\\ 0\\ 1 \end{array}\right]e^{imt} \ \ \ \ \ (12)$

These four solutions are all eigenstates of ${\Sigma_{z}}$ as we can see by writing out ${\Sigma_{z}}$ explicitly

$\displaystyle \Sigma_{z}=\frac{1}{2}\left[\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right] \ \ \ \ \ (13)$

Then we find

 $\displaystyle \Sigma_{z}\left|\psi^{\left(1\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(1\right)}\right\rangle \ \ \ \ \ (14)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(2\right)}\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(2\right)}\right\rangle \ \ \ \ \ (15)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(3\right)}\right\rangle$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left|\psi^{\left(3\right)}\right\rangle \ \ \ \ \ (16)$ $\displaystyle \Sigma_{z}\left|\psi^{\left(4\right)}\right\rangle$ $\displaystyle =$ $\displaystyle -\frac{1}{2}\left|\psi^{\left(4\right)}\right\rangle \ \ \ \ \ (17)$

Thus ${\left|\psi^{\left(1\right)}\right\rangle }$ and ${\left|\psi^{\left(3\right)}\right\rangle }$ are eigenstates of a particle with ${z}$ spin component ${\frac{1}{2}}$, and ${\left|\psi^{\left(2\right)}\right\rangle }$ and ${\left|\psi^{\left(4\right)}\right\rangle }$ represent a particle with ${z}$ spin component ${-\frac{1}{2}}$. As you might expect, the first two solutions are for particles and the last two are for antiparticles, although we haven’t actually demonstrated this yet.

# Eigenspinors of the Pauli spin matrices

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 4, Problem 4.15.

In non-relativistic quantum mechanics, the spin ${\frac{1}{2}}$ operators are given in terms of the Pauli matrices as

 $\displaystyle S_{i}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\sigma_{i}\ \ \ \ \ (1)$ $\displaystyle \sigma_{x}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (2)$ $\displaystyle \sigma_{y}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle \sigma_{z}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (4)$

We’ve seen that the spinor states ${\left[\begin{array}{c} 1\\ 0 \end{array}\right]}$ and ${\left[\begin{array}{c} 0\\ 1 \end{array}\right]}$ are eigenstates of the ${S_{z}}$ operator, and that these spinors form a basis for the 2-d spinor space. We can find the eigenstates of ${S_{x}}$ and ${S_{y}}$ in the usual way from matrix algebra. For ${\sigma_{x}}$, the eigenvalues are

 $\displaystyle \left|\begin{array}{cc} -\lambda & 1\\ 1 & -\lambda \end{array}\right|$ $\displaystyle =$ $\displaystyle \lambda^{2}-1=0\ \ \ \ \ (5)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \pm1 \ \ \ \ \ (6)$

For ${\lambda=1}$, the eigenvector equation is

$\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\left[\begin{array}{c} a\\ b \end{array}\right]=\left[\begin{array}{c} a\\ b \end{array}\right] \ \ \ \ \ (7)$

which gives

$\displaystyle a=b \ \ \ \ \ (8)$

To normalize the eigenstate, we can choose ${a=b=\frac{1}{\sqrt{2}}}$, so that the state is ${\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ 1 \end{array}\right]}$.

For ${\lambda=-1}$, we get

$\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\left[\begin{array}{c} a\\ b \end{array}\right]=-\left[\begin{array}{c} a\\ b \end{array}\right] \ \ \ \ \ (9)$

so the normalized eigenstate is ${\frac{1}{\sqrt{2}}\left[\begin{array}{c} 1\\ -1 \end{array}\right]}$.

For ${\sigma_{y}}$, we get

 $\displaystyle \left|\begin{array}{cc} -\lambda & -i\\ i & -\lambda \end{array}\right|$ $\displaystyle =$ $\displaystyle \lambda^{2}-1=0\ \ \ \ \ (10)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \pm1 \ \ \ \ \ (11)$

so the eigenvalues are the same as for ${\sigma_{x}}$ and ${\sigma_{z}}$. The eigenvector equations are

$\displaystyle \left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\left[\begin{array}{c} a\\ b \end{array}\right]=\pm\left[\begin{array}{c} a\\ b \end{array}\right] \ \ \ \ \ (12)$

from which we get

$\displaystyle a=\mp ib \ \ \ \ \ (13)$

Thus the two normalized eigenstates are, for ${\lambda=+1,\;-1}$ respectively:

$\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{c} i\\ -1 \end{array}\right],\;\frac{1}{\sqrt{2}}\left[\begin{array}{c} i\\ 1 \end{array}\right] \ \ \ \ \ (14)$

This can be written in terms of the ${\sigma_{z}}$ eigenstates as

$\displaystyle \frac{1}{\sqrt{2}}\left[\begin{array}{c} i\\ -1 \end{array}\right]=\frac{i}{\sqrt{2}}\left[\begin{array}{c} 1\\ 0 \end{array}\right]+\frac{1}{\sqrt{2}}\left[\begin{array}{c} 0\\ -1 \end{array}\right] \ \ \ \ \ (15)$

Since the coefficients of the two ${\sigma_{z}}$ eigenstates are equal in magnitude, this means that if ${\sigma_{z}}$ is measured for a particle in an ${\sigma_{y}}$ eigenstate, it is equally likely to be spin up or spin down. The same applies to a particle in a ${\sigma_{x}}$ eigenstate.

# Electron in a precessing magnetic field

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.2.

For a spin 1/2 particle in a magnetic field ${\mathbf{B}}$, we’ve seen that the hamiltonian is

$\displaystyle \mathsf{H}=-\gamma\mathbf{B}\cdot\mathsf{S} \ \ \ \ \ (1)$

where ${\gamma}$ is the gyromagnetic ratio, which for an electron is ${-e/m}$. Now suppose that the magnetic field’s direction precesses around the ${z}$ axis (sweeps out a cone) with angular speed ${\omega}$, so that ${\mathbf{B}}$ makes an angle ${\alpha}$ with the ${z}$ axis. That is

$\displaystyle \mathbf{B}\left(t\right)=B_{0}\left[\sin\alpha\cos\left(\omega t\right)\hat{\mathbf{x}}+\sin\alpha\sin\left(\omega t\right)\hat{\mathbf{y}}+\cos\alpha\hat{\mathbf{z}}\right] \ \ \ \ \ (2)$

At time ${t}$, the component of ${\mathbf{S}}$ along ${\mathbf{B}}$ is given by

$\displaystyle \textsf{S}_{B}=\frac{\hbar}{2}\left(\begin{array}{cc} \cos\alpha & \sin\alpha e^{-i\omega t}\\ \sin\alpha e^{i\omega t} & -\cos\alpha \end{array}\right) \ \ \ \ \ (3)$

so the hamiltonian is

 $\displaystyle \mathsf{H}$ $\displaystyle =$ $\displaystyle \frac{\hbar\omega_{1}}{2}\left(\begin{array}{cc} \cos\alpha & \sin\alpha e^{-i\omega t}\\ \sin\alpha e^{i\omega t} & -\cos\alpha \end{array}\right)\ \ \ \ \ (4)$ $\displaystyle \omega_{1}$ $\displaystyle \equiv$ $\displaystyle \frac{eB_{0}}{m} \ \ \ \ \ (5)$

If we freeze the system at time ${t}$ and solve the time-independent Schrödinger equation to get the eigenvalues and eigenspinors we get

$\displaystyle \chi_{+}=\left(\begin{array}{c} \cos(\alpha/2)\\ e^{i\omega t}\sin(\alpha/2) \end{array}\right) \ \ \ \ \ (6)$

$\displaystyle \chi_{-}=\left(\begin{array}{c} e^{-i\omega t}\sin(\alpha/2)\\ -\cos(\alpha/2) \end{array}\right) \ \ \ \ \ (7)$

with energies

$\displaystyle E_{\pm}=\pm\frac{\hbar\omega_{1}}{2} \ \ \ \ \ (8)$

Griffiths gives the exact solution to the time-dependent Schrödinger equation for this problem as

$\displaystyle \chi\left(t\right)=\left[\begin{array}{c} \left(\cos\frac{\lambda t}{2}-i\frac{\omega_{1}-\omega}{\lambda}\sin\frac{\lambda t}{2}\right)\cos\frac{\alpha}{2}e^{-i\omega t/2}\\ \left(\cos\frac{\lambda t}{2}-i\frac{\omega_{1}+\omega}{\lambda}\sin\frac{\lambda t}{2}\right)\sin\frac{\alpha}{2}e^{i\omega t/2} \end{array}\right] \ \ \ \ \ (9)$

where

$\displaystyle \lambda\equiv\sqrt{\omega^{2}+\omega_{1}^{2}-2\omega\omega_{1}\cos\alpha} \ \ \ \ \ (10)$

To prove this, we need to show that

$\displaystyle \mathsf{H}\chi=i\hbar\frac{\partial\chi}{\partial t} \ \ \ \ \ (11)$

As usual, I’ll use Maple to help things along, although even Maple requires a bit of help here and there. We’ll start with ${\mathsf{H}\chi}$ which is the matrix product of 4 and 9. After multiplying out the terms and using the trig identities ${\cos\left(a\pm b\right)=\cos a\cos b\mp\sin a\sin b}$ we get

$\displaystyle \mathsf{H}\chi=\frac{\hbar\omega_{1}}{2\lambda}\left[\begin{array}{c} e^{-i\omega t/2}\cos\frac{\alpha}{2}\left[i\left(4\omega\cos^{2}\frac{\alpha}{2}-3\omega-\omega_{1}\right)\sin\frac{\lambda t}{2}+\lambda\cos\frac{\lambda t}{2}\right]\\ e^{i\omega t/2}\sin\frac{\alpha}{2}\left[i\left(4\omega\cos^{2}\frac{\alpha}{2}-\omega-\omega_{1}\right)\sin\frac{\lambda t}{2}+\lambda\cos\frac{\lambda t}{2}\right] \end{array}\right] \ \ \ \ \ (12)$

Now for the RHS. We get after collecting terms

$\displaystyle i\hbar\frac{\partial\chi}{\partial t}=\frac{\hbar}{2\lambda}\left[\begin{array}{c} e^{-i\omega t/2}\cos\frac{\alpha}{2}\left[i\left(\omega^{2}-\omega\omega_{1}-\lambda^{2}\right)\sin\frac{\lambda t}{2}+\lambda\omega_{1}\cos\frac{\lambda t}{2}\right]\\ e^{i\omega t/2}\sin\frac{\alpha}{2}\left[i\left(\omega^{2}+\omega\omega_{1}-\lambda^{2}\right)\sin\frac{\lambda t}{2}+\lambda\omega_{1}\cos\frac{\lambda t}{2}\right] \end{array}\right] \ \ \ \ \ (13)$

The two sides are equal if both the following are true:

 $\displaystyle \left(4\omega\cos^{2}\frac{\alpha}{2}-3\omega-\omega_{1}\right)\omega_{1}$ $\displaystyle =$ $\displaystyle \omega^{2}-\omega\omega_{1}-\lambda^{2}\ \ \ \ \ (14)$ $\displaystyle \left(4\omega\cos^{2}\frac{\alpha}{2}-\omega-\omega_{1}\right)\omega_{1}$ $\displaystyle =$ $\displaystyle \omega^{2}+\omega\omega_{1}-\lambda^{2} \ \ \ \ \ (15)$

Substituting from 10 both equations give the same condition:

 $\displaystyle 4\omega\omega_{1}\cos^{2}\frac{\alpha}{2}$ $\displaystyle =$ $\displaystyle 2\omega\omega_{1}\left(1+\cos\alpha\right)\ \ \ \ \ (16)$ $\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle 2\cos^{2}\frac{\alpha}{2}-1 \ \ \ \ \ (17)$

The last line is a trig identity, so the time-dependent Schrödinger equation is satisfied.

We can also express 9 as a linear combination of 6 and 7. Griffiths gives the answer as

 $\displaystyle \chi\left(t\right)$ $\displaystyle =$ $\displaystyle \left[\cos\frac{\lambda t}{2}-i\frac{\left(\omega_{1}-\omega\cos\alpha\right)}{\lambda}\sin\frac{\lambda t}{2}\right]e^{-i\omega t/2}\chi_{+}\left(t\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle i\left[\frac{\omega}{\lambda}\sin\alpha\sin\frac{\lambda t}{2}\right]e^{\omega t/2}\chi_{-}\left(t\right) \ \ \ \ \ (18)$

This can be verified by direct calculation. Take the top element first and use the cosine of difference of angles formula:

 $\displaystyle \chi_{1}$ $\displaystyle =$ $\displaystyle \left[\cos\frac{\lambda t}{2}-i\frac{\left(\omega_{1}-\omega\cos\alpha\right)}{\lambda}\sin\frac{\lambda t}{2}\right]e^{-i\omega t/2}\cos\frac{\alpha}{2}+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle i\left[\frac{\omega}{\lambda}\sin\alpha\sin\frac{\lambda t}{2}\right]e^{\omega t/2}e^{-i\omega t}\sin\frac{\alpha}{2}\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-i\omega t/2}\left[\cos\frac{\lambda t}{2}\cos\frac{\alpha}{2}+\frac{i}{\lambda}\sin\frac{\lambda t}{2}\left(-\omega_{1}\cos\frac{\alpha}{2}+\omega\cos\frac{\alpha}{2}\right)\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\cos\frac{\lambda t}{2}-i\frac{\omega_{1}-\omega}{\lambda}\sin\frac{\lambda t}{2}\right)\cos\frac{\alpha}{2}e^{-i\omega t/2} \ \ \ \ \ (21)$

The bottom element works much the same way, using the sine of difference of angles formula:

 $\displaystyle \chi_{2}$ $\displaystyle =$ $\displaystyle \left[\cos\frac{\lambda t}{2}-i\frac{\left(\omega_{1}-\omega\cos\alpha\right)}{\lambda}\sin\frac{\lambda t}{2}\right]e^{-i\omega t/2}e^{i\omega t}\sin\frac{\alpha}{2}-\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle i\left[\frac{\omega}{\lambda}\sin\alpha\sin\frac{\lambda t}{2}\right]e^{\omega t/2}\cos\frac{\alpha}{2}\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\cos\frac{\lambda t}{2}-i\frac{\omega_{1}+\omega}{\lambda}\sin\frac{\lambda t}{2}\right)\sin\frac{\alpha}{2}e^{i\omega t/2} \ \ \ \ \ (23)$

Finally, writing ${\chi\left(t\right)=c_{+}\left(t\right)\chi_{+}+c_{-}\left(t\right)\chi_{-}}$ we can check that the coefficients are normalized.

 $\displaystyle \left|c_{+}\right|^{2}+\left|c_{-}\right|^{2}$ $\displaystyle =$ $\displaystyle \cos^{2}\frac{\lambda t}{2}+\frac{\left(\omega_{1}-\omega\cos\alpha\right)^{2}}{\lambda^{2}}\sin^{2}\frac{\lambda t}{2}+\left[\frac{\omega}{\lambda}\sin\alpha\sin\frac{\lambda t}{2}\right]^{2}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\frac{\lambda t}{2}+\sin^{2}\frac{\lambda t}{2}\left[\frac{\left(\omega_{1}-\omega\cos\alpha\right)^{2}}{\lambda^{2}}+\left(\frac{\omega}{\lambda}\sin\alpha\right)^{2}\right]\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\frac{\lambda t}{2}+\sin^{2}\frac{\lambda t}{2}\left[\frac{\omega^{2}+\omega_{1}^{2}-2\omega\omega_{1}\cos\alpha}{\lambda^{2}}\right]\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \cos^{2}\frac{\lambda t}{2}+\sin^{2}\frac{\lambda t}{2}\left[\frac{\omega^{2}+\omega_{1}^{2}-2\omega\omega_{1}\cos\alpha}{\omega^{2}+\omega_{1}^{2}-2\omega\omega_{1}\cos\alpha}\right]\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (28)$