Dirac spinor

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Solutions to the Dirac equation for free-particles have the form of a plane-wave:

\psi = \omega e^{-i p \cdot x} \,

where ω is a four-component spinor (Dirac spinor) which is not a function of x.

This spinor can be written

\omega = \begin{bmatrix} \phi \\ \frac{\mathbf{\sigma \cdot p}}{E + m} \phi \end{bmatrix} \,
where
\phi \, is a two-spinor,
\sigma \, are the Pauli matrices,
E, m, p are the Energy, mass, and four-momentum of the particle respectively.

Contents

[edit] Derivation from Dirac equation

The Dirac equation has the form

\left(-i \alpha \cdot \nabla + \beta m \right) \psi = i \frac{\partial \psi}{\partial t} \,

In order to derive the form of the four-spinor ω we have to first note the value of the matrices α and β:

\alpha = \begin{bmatrix} \mathbf{0} & \mathbf{\sigma} \\ \mathbf{\sigma} & \mathbf{0} \end{bmatrix} \quad \quad \beta = \begin{bmatrix} \mathbf{I} & \mathbf{0} \\ \mathbf{0} & -\mathbf{I} \end{bmatrix} \,

These two 4x4 matrices are related to the Dirac gamma matrices. Note that 0 and I are 2x2 matrices here.

The next step is to look for solutions of the form,

\psi = \omega e^{-i p \cdot x} ,

While at the same time splitting ω into two two-spinors:

\omega = \begin{bmatrix} \phi \\ \chi \end{bmatrix} \,.

[edit] Results

Using all of the above information to plug into the Dirac equation results in

E \begin{bmatrix} \phi \\ \chi \end{bmatrix} = \begin{bmatrix} m \mathbf{I} & \mathbf{\sigma \cdot p} \\ \mathbf{\sigma \cdot p} & -m \mathbf{I} \end{bmatrix} \begin{bmatrix} \phi \\ \chi \end{bmatrix} \,.

That matrix equation is really two coupled equations:

  • \left(E - m \right) \phi = \left(\mathbf{\sigma \cdot p} \right) \chi \,
  • \left(E + m \right) \chi = \left(\mathbf{\sigma \cdot p} \right) \phi \,

Solve the 2nd equation for \chi \, and then one can then write

\omega = \begin{bmatrix} \phi \\ \chi \end{bmatrix} = \begin{bmatrix} \phi \\ \frac{\mathbf{\sigma \cdot p}}{E + m} \phi \end{bmatrix} \,

Solve the 1st equation for \phi \, one can find

\omega = \begin{bmatrix} \phi \\ \chi \end{bmatrix} = \begin{bmatrix} - \frac{\mathbf{\sigma \cdot p}}{-E + m} \chi \\ \chi \end{bmatrix} \,

This solution is useful for showing the relation between anti-particle and particle.

[edit] Details

[edit] Two-spinors

The most convenient definitions for the two-spinors are:

\phi^1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \quad \quad \phi^2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \,

and

\chi^1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} \quad \quad \chi^2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} \,

[edit] Pauli matrices

The Pauli matrices are

\sigma_1 = \begin{bmatrix} 0&1\\ 1&0 \end{bmatrix} \quad \quad \sigma_2 = \begin{bmatrix} 0&-i\\ i&0 \end{bmatrix} \quad \quad \sigma_3 = \begin{bmatrix} 1&0\\ 0&-1 \end{bmatrix}

Using, these one can calculate:

\mathbf{\sigma \cdot p} = \sigma_1 p_1 + \sigma_2 p_2 + \sigma_3 p_3 = \begin{bmatrix} p_3 & p_1 - i p_2 \\ p_1 + i p_2 & - p_3 \end{bmatrix}

[edit] Four-spinor for particles

Particles are defined as having positive energy. The normalization for the four-spinor ω is chosen so that \omega^\dagger \omega = 2 E \,. These spinors are denoted as u:

u(\mathbf{p}, s) = \sqrt{E+m} \begin{bmatrix} \phi^{(s)}\\ \frac{\mathbf{\sigma} \cdot \mathbf{p} }{E+m} \phi^{(s)} \end{bmatrix} \,
where s = 1 or 2 (spin "up" or "down")

Explicitly,

u(\mathbf{p}, 1) = \sqrt{E+m} \begin{bmatrix} 1\\ 0\\ \frac{p_3}{E+m} \\ \frac{p_1 + i p_2}{E+m} \end{bmatrix} \quad \mathrm{and} \quad u(\mathbf{p}, 2) = \sqrt{E+m} \begin{bmatrix} 0\\ 1\\ \frac{p_1 - i p_2}{E+m} \\ \frac{-p_3}{E+m} \end{bmatrix}

[edit] Four-spinor for anti-particles

Anti-particles having positive energy E are defined as particles having negative energy and propagating backward in time.

Hence changing the sign of E and \mathbf{p} in the four-spinor for particles will give the four-spinor for anti-particles:

v(\mathbf{p},s) = \sqrt{E+m} \begin{bmatrix} \frac{\mathbf{\sigma} \cdot \mathbf{p} }{E+m} \chi^{(s)}\\ \chi^{(s)} \end{bmatrix} \,

Here we choose the χ solutions.

Explicitly,

v(\mathbf{p}, 1) = \sqrt{E+m} \begin{bmatrix} \frac{p_1 - i p_2}{E+m} \\ \frac{-p_3}{E+m} \\ 0\\ 1 \end{bmatrix} \quad \mathrm{and} \quad v(\mathbf{p}, 2) = \sqrt{E+m} \begin{bmatrix} \frac{p_3}{E+m} \\ \frac{p_1 + i p_2}{E+m} \\ 1\\ 0\\ \end{bmatrix}

[edit] Completeness relations

The completeness relations for the four-spinors u and v are

\sum_{s=1,2}{u^{(s)}_p \bar{u}^{(s)}_p} = p\!\!\!/ + m \,
\sum_{s=1,2}{v^{(s)}_p \bar{v}^{(s)}_p} = p\!\!\!/ - m \,
where
p\!\!\!/ = \gamma^\mu p_\mu \,      (see Feynman slash notation)
\bar{u} = u^{\dagger} \gamma^0 \,


[edit] Dirac spinors and the Dirac algebra

The Dirac matrices are a set of four 4x4 matrices that are used as spin and charge operators.

[edit] Conventions

There are several choices of signature and representation that are in common use in the physics literature. The Dirac matrices are typically written as γμ where μ runs from 0 to 3. In this notation, 0 corresponds to time, and 1 through 3 correspond to x, y, and z.

The + - - - signature is sometimes called the west coast metric, while the - + + + is the east coast metric. At this time the + - - - signature is in more common use and our example will use this signature. To switch from one example to the other, multiply all γμ by i.

After choosing the signature, there are many ways of constructing a representation in the 4x4 matrices, and many are in common use. In order to make this example as general as possible we will not specify a representation until the final step. At that time we will substitute in the "chiral" or "Weyl" representation as used in the popular graduate textbook, An Introduction to Quantum Field Theory, by Michael E. Peskin and Daniel V. Schroeder.

[edit] Construction

First we choose a spin direction for our electron or positron. As with the example of the Pauli algebra discussed above, the spin direction is defined by a unit vector in 3 dimensions, (a,b,c). Following the convention of Peskin & Schroeder, the spin operator for spin in the (a,b,c) direction is defined as the dot product of (a,b,c) with the vector (i\gamma^2\gamma^3,\;\;i\gamma^3\gamma^1,\;\;i\gamma^1\gamma^2) =-(\gamma^1,\;\gamma^2,\;\gamma^3)i\gamma^1\gamma^2\gamma^3:

σ(a,b,c) = iaγ2γ3 + ibγ3γ1 + icγ1γ2

Note that the above is a root of unity, that is, it squares to 1. Consequently, we can make a projection operator from it that projects out the subalgebra of the Dirac algebra that has spin oriented in the (a,b,c) direction:

P_{(a,b,c)} = \frac{1 + \sigma_{(a,b,c)}}2

Now we must choose a charge, +1 (positron) or -1 (electron). Following the conventions of Peskin & Schroeder, the operator for charge is Q = − γ0, that is, electron states will take an eigenvalue of -1 with respect to this operator while positron states will take an eigenvalue of +1.

Note that Q is also a square root of unity. Furthermore, Q commutes with σ(a,b,c). They form a complete set of commuting operators for the Dirac algebra. Continuing with our example, we look for a representation of an electron with spin in the (a,b,c) direction. Turning Q into a projection operator for charge = -1, we have

P_{-Q} = \frac{1 - Q}2 = \frac{1 + \gamma^0}2

The projection operator for the spinor we seek is therefore the product of the two projection operators we've found:

P_{(a,b,c)}\;P_{-Q}

The above projection operator, when applied to any spinor, will give that part of the spinor that corresponds to the electron state we seek. Therefore, to write down a 4x1 spinor we take any non zero column of the above matrix. Continuing the example, we put (a,b,c) = (0,0,1) and have

P_{(0,0,1)} = \frac{1+ i\gamma_1\gamma_2}2

and so our desired projection operator is

P = \frac{1+ i\gamma^1\gamma^2}2\cdot\frac{1 + \gamma^0}2 = \frac{1+\gamma^0 +i\gamma^1\gamma^2 + i\gamma^0\gamma^1\gamma^2}4

The 4×4 gamma matrices used in Peskin & Schroeder (Weyl representation) are

\gamma_0 = \begin{bmatrix}0&1\\1&0\end{bmatrix}
\gamma_k = \begin{bmatrix}0&\sigma^k\\ -\sigma^k& 0\end{bmatrix}

for k=1,2,3 and where σi are the usual 2x2 Pauli matrices. Substituting these in for P gives

P = \frac14\begin{bmatrix}1+\sigma^3&1+\sigma^3\\ 1+\sigma^3&1+\sigma^3\end{bmatrix} =\frac12\begin{bmatrix}1&0&1&0\\0&0&0&0\\ 1&0&1&0\\0&0&0&0\end{bmatrix}

Our answer is any non zero column of the above matrix. The division by two is just a normalization. The first and third columns give the same result:

\left|e^-,\, +\frac12\right\rangle = \begin{bmatrix}1\\0\\1\\0\end{bmatrix}

More generally, for electrons and positrons with spin oriented in the (a,b,c) direction, the projection operator is

\frac14\begin{bmatrix} 1+c&a-ib&\pm (1+c)&\pm(a-ib)\\ a+ib&1-c&\pm(a+ib)&\pm (1-c)\\ \pm (1+c)&\pm(a-ib)&1+c&a-ib\\ \pm(a+ib)&\pm (1-c)&a+ib&1-c \end{bmatrix}

where the upper signs are for the electron and the lower signs are for the positron. The corresponding spinor can be taken as any non zero column. Since a2 + b2 + c2 = 1 the different columns are multiples of the same spinor.

[edit] See also

[edit] References

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