Bjorken and Drell - Reading notes

1. A physical system is described by a state function \(\Phi\). This is usually written explicitly using coordinates \(\psi(q,s,t)\) which is a complex function of coordinates \(q=(q_{1},\dots,q_{m})\), time \(t\), and any additional degrees of freedom \(s\) (like spin).

   This has no direct physical interpretation. Instead, \(|\psi(q,s,t)|^{2}\geq0\) is interpreted as the probability of the system having values \((q,s)\) at time \(t\). This requires the sum of all \(|\psi(q,s,t)|^{2}\) to be finite.

2. Every physical observable is represented by a linear Hermitian operator. In particular, the canonical momentum \(p_{i}\) the operator in coordinate-space is \(p_{i} \to -\I\hbar\partial/\partial q_{i}\).

3. A physical system is in an eigenstate of the operator \(\Omega\) if

   \begin{equation*}\tag{1.1} \Omega\Phi_{n}=\omega_{n}\Phi_{n} \end{equation*}

   where \(\Phi_{n}\) is the n-th eigenstate corresponding to eigenvalue \(\omega_{n}\). For Hermitian operators, \(\omega_{n}\) is real.

   We can write this in local coordinates as \(\Omega(q,s,t)\psi_{n}(q,s,t)=\omega_{n}\psi_{n}(q,s,t)\).

4. Expansion postulate: an arbitrary wave function for a physical system can be expanded in a complete orthognormal set of eigenfunctions \(\psi_{n}\) of a complete set of commuting operators \(\Omega_{n}\). We write this as \(\psi = \sum_{n}a_{n}\psi_{n}\) where the requirement of orthonormality is \[\sum_{s}\int\psi^{*}_{n}(q_{1},\dots,s,\dots,t)\psi_{m}(q_{1},\dots,s,\dots,t)\,\D q_{1}\dots=\delta_{m,n}\] (this is how they write it, it's a little ambiguous). The coefficients \(a_{n}\), we interpret \(|a_{n}|^{2}\) as the probability of finding the system in this eigenstate.
5. The result of a measurement of a physical observable is any one of its eigenvalues.

6. The development ofa physical system is expressed by the Schrodinger equation

   \begin{equation*}\tag{1.2} \I\hbar\partial_{t}\psi = H\psi \end{equation*}

   where \(\partial_{t} = \partial/\partial t\).

   A closed system has no explicit time dependence, so \(\partial_{t}\psi=0\) for closed systems.

   Corollary (superposition principle): linearity of \(H\) and conservation of probability from the Hermitian property of \(H\):

   \begin{align*}\tag{1.3} &\frac{\D}{\D t}\sum_{s}\int\psi^{*}\psi(\D q_{1}\cdots)\\ &=\frac{\I}{\hbar}\sum_{s}\int[(H\psi)^{\ast}\psi-\psi^{\ast}(H\psi)](\D q_{1}\cdots)\\ &=0 \end{align*}

The free particle has its Hamiltonian be

The quantum counterpart takes

Thus we obtain the nonrelativistic Schrodinger equation

The total energy \(E\) and the momenta \((p_{x},p_{y},p_{z})\) transform as components of a contravariant 4-vector

of invariant length

where \(m\) is the rest mass of the particle, and \(c\) is the speed of light in a vacuum.

The relativistic Hamiltonian for a free particle is

Therefore the relativistic counterpart to the Schrodinger equation would be

where \(\nabla^{2}\) is the Laplacian operator. The right-hand side of Equation (1.9) is non-local: if we take the series expansion of it, we have infinitely many derivatives (which results in non-local behaviour).

The trick is to rewrite Equation (1.9) as

Equivalently, iterate Equation (1.9) using the fact if \([A,B]=0\) then \(A\psi=B\psi\) implies \(A^{2}\psi=B^{2}\psi\), which gives us

We can see this is the classical wave equation

• Note that this is the Klein-Gordon Equation.

Let us study Equations (1.10) and (1.11). Our first task: construct a conserved current. In analogy to the non-relativistic case, multiply \(\psi^{\ast}\) by (1.11), then subtracting \(\psi\) times the complex-conjugate equation, i.e.,

Or equivalently

We would like to interpret

as a probability density. But this is imposible: it is not a positive-definite expression.

We have

where the \(\alpha_{i}\) and \(\beta\) are \(N\times N\) matrices.

• This equation is precisely the Dirac Equation.

1. We should recover the energy-momentum relation \(E^{2}=p^{2}c^{2}+m^{2}c^{4}\)
2. It must allow for a continuity equation and a probability interpretation for the wave function \(\psi\)
3. It must be Lorentz covariant

If we iterate Equation (1.13), we end up with

This requires the four matrices \(\alpha_{i}\), \(\beta\) obey:

• From \(\alpha_{i}^{2}=\beta^{2}=1\) in Eq (1.16), the eigenvalues of \(\alpha_{i}\) and \(\beta\) must be \(\pm1\).
• From the anticommutation properties, it follows that the trace vanishes, i.e., \(\tr(\alpha_{i})=0\) and \(\tr(\beta)=0\).
• Taken together, this means the number of \(-1\) eigenvalues must be equal to the number of \(+1\) eigenvalues, which means we must have \(N\) be an even number.
• When \(N=2\), there are only three anticommuting matrices (the Pauli matrices). We need four anticommuting matrices. So this can't work.

• When \(N=4\), we have the possibility

  \begin{equation*}\tag{1.17} \alpha_{i} = \begin{bmatrix} 0 & \sigma_{i}\\ \sigma_{i} & 0\end{bmatrix},\qquad\beta=\begin{bmatrix}1&0\\ 0&-1\end{bmatrix} \end{equation*}

  where the \(\sigma_{i}\) are the familiar Pauli matrices, and \(\beta\) consists of \(2\times 2\) identity matrices (or its negation).

We introduce the Hermitian conjugate wave functions \(\psi^{\dagger}=(\psi^{\ast}_{1},\dots,\psi^{\ast}_{4})\) and left-multiply Equation (1.13) by \(\psi^{\dagger}\) to obtain

Now form the Hermitian conjugate of Equation (1.13), then multiply on the right by \(\psi\):

where \(\alpha_{i}^{\dagger}=\alpha_{i}\) and \(\beta^{\dagger}=\beta\).

Subtracting (1.19) from (1.18) gives us

or

We identify the probability density

and the probability current with three components:

Integrating Equation (1.20) over all space, and using Green's theorem, we find

which encourages a tentative interpretation of \(\rho\) as a positive-definite probability density.

1. Establish the Lorentz invariance of the Dirac equation, and
2. Check that the equation makes physical sense.

When a free electron is at rest, Equation (1.13) reduces to

since the de Broglie wavelength is infinitely large and the wave function is uniform over all space.

• Let \(O\) and \(O'\) be two observers in different inertial reference frames which describe the same physical event with different spacetime coordinates.

• The rule relating the coordinates \(x^{\mu}\) which observer \(O\) describes the event, to the coordinates \((x^{\mu})'\) which observer \(O'\) describes the same event, is given by the Lorentz transformation between the two sets of coordinates:

  \begin{equation*}\tag{2.1} (x^{\nu})' = \sum^{3}_{\mu=0}{a^{\nu}}_{\mu}x^{\mu}\equiv {a^{\nu}}_{\mu}x^{\mu} \end{equation*}

  This is a linear homogeneous transformation, and the coefficients \({a^{\nu}}_{\mu}\) depend only upon the relative velocities and spatial orientations of the reference frames of \(O\) and \(O'\).

  • Note: Equation (2.1) describes familiar matrix multiplication taught in elementary linear algebra. Here \(x^{\mu}\) describes a column vector, and \({a^{\nu}}_{\mu}\) describes a matrix with \(\mu\) indexing rows, \(\nu\) indexing columns. Einstein summation convention here is just describing familiar matrix multiplication.

• The basic invariant of the Lorentz transformation is the proper time interval

  \begin{equation*}\tag{2.2} \D s^{2}=g_{\mu\nu}\,\D x^{\mu}\D x^{\nu} \end{equation*}

  (Note that \(\D s=c\,\D\tau\) gives us the "infinitesimal" proper time interval \(\D\tau\) explicitly.) This is derived from the physical observation that the velocity of light is constant in a vacuum.

• Now equations (2.1) and (2.2) taken together leads us to the condition on the coefficients that

  \begin{equation*}\tag{2.3} {a_{\mu}}^{\nu}{a^{\mu}}_{\sigma}={\delta^{\nu}}_{\sigma} \end{equation*}

  where \({\delta^{\nu}}_{\sigma}\) is the Kronecker delta (1 when \(\nu=\sigma\), and 0 otherwise).

• There are two classes of Lorentz transformations, since Equations (2.1) and (2.3) implies \(\det({a^{\mu}}_{\nu})=\pm1\). When the determinant is positive, we have Proper Lorentz Transformations; otherwise, the determinant is negative, and we call it an Improper Lorentz Transformation.
• Proper Lorentz transformations.
  • These form a subgroup: if \(A\) is a proper Lorentz transformation, then \(A^{-1}\) is a proper Lorentz transformation; further, if \(A\) and \(B\) are proper Lorentz transformations, then \(AB\) is another proper Lorentz transformation. The identity matrix is a proper Lorentz transformation describing the case when \(O=O'\). (We should check associativity, but we know matrix multiplication is associative.)
• Improper Lorentz transformations are the discrete transformations of space inversion and time inversion (possibly composed with proper Lorentz transformations).
  • Improper Lorentz transformations cannot be built out of infinitesimal Lorentz transformations.

• We have wave functions \(\psi(x)\) used by observer \(O\), and \(\psi'(x')\) used by observer \(O'\).
• We seek a transformation law relating these two wave functions to each other. Specifically, for \(O'\) to compute \(\psi'(x')\) from \(\psi(x)\).
• Lorentz covariance requires this transformation yield wave functions which are solutions of Dirac equations of the same form in the primed as well as unprimed reference frame.

• This invariance expresses the Lorentz invariance of the underlying momentum shell

  \begin{equation*} p^{\mu}p_{\mu}=m^{2}c^{2}. \end{equation*}

• Gamma matrices. Towards this end, we introduce the notation

  \begin{equation*} \gamma^{0}=\beta,\quad\gamma^{i}=\beta\alpha_{i}. \end{equation*}

  Then we have Dirac's equation be

  \begin{equation*}\tag{2.4} \I\hbar\left(\gamma^{0}\partial_{0}+\gamma^{1}\partial_{1}+\dots+\gamma^{3}\partial_{3}\right)\psi-mc\psi=0. \end{equation*}

• We see the defining relations from Equation (1.16) is:

  \begin{equation*}\tag{2.5} \gamma^{\mu}\gamma^{\nu}+\gamma^{\nu}\gamma^{\mu}=2g^{\mu\nu}\mathbf{1} \end{equation*}

  where \(\mathbf{1}\) is the \(4\times 4\) identity matrix

• The \(\gamma^{i}\) are anti-Hermitian and \((\gamma^{i})^{2}=-\mathbf{1}\).
• The \(\gamma^{0}\) is Hermitian.

• The representation from Equation (1.17) gives us

  \begin{equation*}\tag{2.6} \gamma^{i}=\begin{bmatrix}0 & \sigma^{i}\\-\sigma^{i}&0\end{bmatrix},\quad \gamma^{0}=\begin{bmatrix}1 & 0\\ 0 & -1\end{bmatrix}. \end{equation*}

• Feynman slash notation. We have

  \begin{equation*} \cancel{A} = \gamma^{\mu}A_{\mu} = g_{\mu\nu}\gamma^{\mu}A^{\nu} = \gamma^{0}A^{0}-\vec{\gamma}\cdot\vec{A}, \end{equation*}

  so we have in particular

  \begin{equation*} \cancel{\nabla} = \gamma^{\mu}\partial_{\mu} = \frac{\gamma^{0}}{c}\partial_{t}+\vec{\gamma}\cdot\vec{\nabla}. \end{equation*}

• We can then simplify the Dirac Equation (2.4) to:

  \begin{equation*}\tag{2.7} (\I\hbar\cancel{\nabla} - mc)\psi = 0 \end{equation*}

  Or, using \(p^{\mu}=\I\hbar\partial_{\mu}\)

  \begin{equation*}\tag{2.8} (\cancel{p} - mc)\psi = 0 \end{equation*}

• We can add electromagnetic interaction according to the "minimal" substitution from Equation (1.25):

  \begin{equation*} \left(\cancel{p} - \frac{e\cancel{A}}{c} - mc\right)\psi = 0 \end{equation*}

  This does not affect concerns of covariance, since both \(p^{\mu}\) and \(A^{\mu}\) are 4-vectors.

1. We need an explicit way to compute \(\psi'(x')\) for observer \(O'\) given \(\psi(x)\) from observer \(O\), which describes the same physical state.

2. According to the principle of relativity, \(\psi'(x')\) will be a solution of an Equation which takes of the form of (2.7) in the primed system

   \begin{equation*} \left(\I\hbar\widetilde{\gamma}^{\mu}\frac{\partial}{\partial x^{\mu\prime}} - mc\right)\psi'(x') = 0 \end{equation*}

   where \(\widetilde{\gamma}^{\mu}\) satisfy the same anticommutation relations of Equation (2.5), which means \(\widetilde{\gamma}^{0\dagger}=\widetilde{\gamma}^{0}\) and \(\widetilde{\gamma}^{i\dagger}=-\widetilde{\gamma}^{i}\). (This is required for the Hamiltonian operator to be Hermitian.) It turns out these are equivalent up to a unitary transformation \(U\),

   \begin{equation*} \widetilde{\gamma}_{\mu}=U^{\dagger}\gamma_{\mu}U,\quad U^{\dagger}=U^{-1} \end{equation*}

   (it's a straightforward but lengthy proof)

   • These \(U\) matrices have constant components.

   • We can therefore "pull out" the constant \(U\) matrices, and the Dirac equation in \(O'\) frame becomes

     \begin{equation*}\tag{2.9} \left(\I\hbar\gamma^{\mu}\frac{\partial}{\partial x^{\mu\prime}} - mc\right)\psi'(x') = 0 \end{equation*}

     where Bjorken and Drell write the first term as \(\cancel{p}' = \I\hbar\gamma^{\mu}\partial'_{\mu}\)

• To be explicit, we would like

  \begin{equation*}\tag{2.10} \psi'(x')=\psi'(ax)=S(a)\psi(x)=S(a)\psi(a^{-1}x') \end{equation*}

  where \(S(a)\) is a \(4\times 4\) matrix parametrized by \(a\), and \(S(a)\) operates on the four-component column vector \(\psi(x)\). The \(S(a)\) depends only on the relative velocities and spatial orientations of \(O\) and \(O'\).

• Further, \(S\) must have an inverse, so if \(O\) knows \(\psi'(x')\) which \(O'\) uses to describe the observations of a given state, observer \(O\) may construct their own wave function \(\psi(x)\) by

  \begin{equation*}\tag{2.11} \psi(x) = S^{-1}(a)\psi'(x') = S^{-1}(a)\psi'(ax) \end{equation*}

  Then using Equation (2.10), we can rewrite this as

  \begin{equation*} \psi(x) = S(a^{-1})\psi'(ax). \end{equation*}

  Therefore we obtain the identification

  \begin{equation*} S(a^{-1}) = S^{-1}(a). \end{equation*}
• Now…how to find \(S\)?

• Start with the Dirac equation for observer \(O\) (Eq (2.7)) and try to rewrite it in terms of \(\psi'(x')\) using Equation (2.11):

  \begin{equation*}\tag{2.7'} \left(\I\hbar\gamma^{\mu}\frac{\partial}{\partial x^{\mu}}-mc\right)\psi(x)=0 \end{equation*}

  then using Equation (2.11)

  \begin{equation*} \left(\I\hbar\gamma^{\mu}\frac{\partial}{\partial x^{\mu}}-mc\right)S^{-1}(a)\psi'(x')=0. \end{equation*}

  Since \(S^{-1}(a)\) is constant with respect to \(x^{\mu}\), we can pull it inside the parentheses, and multiply on the left by \(S(a)\) to get:

  \begin{equation*} \left(\I\hbar S(a)\gamma^{\mu}S^{-1}(a)\frac{\partial}{\partial x^{\mu}}-mc\right)\psi'(x')=0. \end{equation*}
  • Note that \(S(a)\) and \(S^{-1}(a)\) cannot be pulled past a \(\gamma\) matrix, since they act on gamma matrices.

• We then can use Equation (2.1) to rewrite the partial derivative operator as (using the chain rule)

  \begin{equation*} \frac{\partial}{\partial x^{\mu}} = \frac{\partial x'^{\nu}}{\partial x^{\mu}}\frac{\partial}{\partial x'^{\nu}}={a^{\nu}}_{\mu}\frac{\partial}{\partial x'^{\nu}}. \end{equation*}

• We combine these previous two steps together to obtain the Dirac equation as:

  \begin{equation*} \left(\I\hbar S(a)\gamma^{\mu}S^{-1}(a){a^{\nu}}_{\mu}\frac{\partial}{\partial x'^{\nu}}-mc\right)\psi'(x')=0. \end{equation*}

  This is form-invariant (i.e., identical with Equation (2.9)), provided an \(S\) can be found such that

  \begin{equation*} S(a)\gamma^{\mu}S^{-1}(a){a^{\nu}}_{\mu} = \gamma^{\nu}, \end{equation*}

  or equivalently,

  \begin{equation*}\tag{2.12} S^{-1}(a)\gamma^{\nu}S(a) = {a^{\nu}}_{\mu}\gamma^{\mu}. \end{equation*}
  • Once we find a solution to Equation (2.12), we will have established the covariance of the Dirac equation.
  • Terminology: a wave function transforming according to Equations (2.10) and (2.12) are called a four-component Lorentz spinor.

• We take

  \begin{equation*}\tag{2.13a} {a^{\nu}}_{\mu} = {g^{\nu}}_{\mu} + \Delta{\omega^{\nu}}_{\mu} \end{equation*}

  with

  \begin{equation*}\tag{2.13b} \Delta\omega^{\nu\mu} = -\Delta\omega^{\mu\nu} \end{equation*}

  infinitesimal (so quadratic and higher-order terms of \(\Delta\omega\) are dropped).

• Exercise: prove if \(\mu\), \(\nu=1,\dots,N\) that \(\Delta\omega^{\mu\nu}\) has \(N(N-1)/2\) independent components. (Hint: look at the diagonal, then look at the off-diagonal.)
• Each of the six independent nonvanishing \(\Delta\omega^{\mu\nu}\) generates an infinitesimal Lorentz transformation.
• \(\Delta\omega^{01}=\Delta\beta\) describes a transformation for a coordinate system moving with velocity \(c\,\Delta\beta\) in the \(x\) direction.
• \(\Delta{\omega^{1}}_{2}=\Delta\varphi\) for a rotation through an angle \(\Delta\varphi\) about the $z$-axis, and so on.

• Expanding \(S\) in powers of \(\Delta\omega^{\nu\mu}\), and keeping only the linear terms in the infinitesimal generators, we have

  \begin{equation*}\tag{2.14} S = \mathbf{1} - \frac{\I}{4}\sigma_{\mu\nu}\,\Delta\omega^{\mu\nu},\quad\mbox{and}\quad S^{-1} = \mathbf{1} + \frac{\I}{4}\sigma_{\mu\nu}\,\Delta\omega^{\mu\nu} \end{equation*}

  with \(\sigma_{\mu\nu}=-\sigma_{\nu\mu}\) by (2.13b).

• Each of the six coefficients \(\sigma_{\mu\nu}\) is really a \(4\times 4\) matrix, as are the transformation \(S\) and the unit matrix \(\mathbf{1}\).

• Inserting Equations (2.13) and (2.14) into Equation (2.12) and keeping first-order terms in \(\Delta\omega^{\mu\nu}\), we find

  \begin{equation*} \Delta{\omega^{\nu}}_{\mu}\,\gamma^{\mu} = \frac{-\I}{4}(\Delta\omega)^{\alpha\beta}(\gamma^{\nu}\sigma_{\alpha\beta}-\sigma_{\alpha\beta}\gamma^{\nu}). \end{equation*}

  Careful: the right-hand side does not necessarily vanish, multiplication of matrices is not commutative.

• From the antisymmetry of the generators \(\Delta\omega^{\mu\nu}\) it follows that

  \begin{equation*}\tag{2.15} 2\I({g^{\nu}}_{\alpha}\gamma_{\beta}-{g^{\nu}}_{\beta}\gamma_{\alpha})=[\gamma^{\nu},\sigma_{\alpha\beta}] \end{equation*}

  where the right-hand side is the commutator \([A,B]=AB-BA\).

• The problem of establishing proper Lorentz covariance of the Dirac equation is now reduced to finding six matrices \(\sigma_{\alpha\beta}\) satisfying Equation (2.15).

• Simplest guess:

  \begin{equation*}\tag{2.16} \sigma_{\mu\nu}=\frac{\I}{2}[\gamma_{\mu},\gamma_{\nu}]. \end{equation*}

  • Then using Equation (2.14), \(S\) for an infinitesimal transformation is given by

    \begin{equation*}\tag{2.17} S = \mathbf{1} + \frac{1}{8}[\gamma_{\mu},\gamma_{\nu}]\,\Delta\omega^{\mu\nu} = \mathbf{1} - \frac{\I}{4}\sigma_{\mu\nu}\,\Delta\omega^{\mu\nu} \end{equation*}
  • We can then build our way up to a finite Lorentz transformation by iterating a succession of infinitesimal Lorentz transformations (the trick transforming Lie algebras to Lie groups).

  • I am going to skip the intermediate steps and jump to the conclusion. The trick is that we can rotate the Lorentz boost to be along the $x$-axis, taking

    \begin{equation*}\tag{2.19} {I^{\nu}}_{\mu} = \begin{bmatrix} 0 & -1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*}

    There are 6 different possible \({I^{\nu}}_{\mu}\) matrices (3 for boosts along the \(x\), \(y\), \(z\) axes; and 3 for spatial rotations about each axis). This allows us to write

    \begin{equation*}\tag{2.18} \Delta{\omega^{\nu}}_{\mu} = (\Delta\omega_{(n)}){(I_{(n)})^{\nu}}_{\mu} \end{equation*}

    where \(\Delta\omega_{(n)}\) is the infinitesimal "angle of rotation" parameter, and \(n\) keeps track of which possible \(I\) matrix we're working with. We'll work through one possible case in some detail, the other cases are "obvious".

  • Lorentz transformation for uniform relative $x$-axis motion (using Lie algebra to Lie group relations)

    \begin{align*} x^{\nu'} &= {(\exp(\omega I))^{\nu}}_{\mu}x^{\mu}\\ &= {(\cosh(\omega I) + \sinh(\omega I))^{\nu}}_{\mu}x^{\mu}\\ &= {(\mathbf{1} - I^{2} + I^{2}\cosh(\omega) + I\sinh(\omega))^{\nu}}_{\mu}x^{\mu} \end{align*}

  • The individual components for this:

    \begin{equation*}\tag{2.20} \begin{bmatrix}x^{0\prime}\\x^{1\prime}\\x^{2\prime}\\x^{3\prime}\end{bmatrix} = \begin{bmatrix} \cosh(\omega) & -\sinh(\omega) & 0 & 0\\ -\sinh(\omega) & \cosh(\omega) & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix}x^{0}\\x^{1}\\x^{2}\\x^{3}\end{bmatrix} \end{equation*}

    Or if you prefer:

    \begin{equation*}\tag{2.21} \left.\begin{array}{rcl} x^{0\prime} &=& (\cosh(\omega))(x^{0} - x^{1}\tanh(\omega))\\ x^{1\prime} &=& (\cosh(\omega))(x^{1} - x^{0}\tanh(\omega))\\ x^{2\prime} &=& x^{2}\\ x^{3\prime} &=& x^{3} \end{array}\right\} \end{equation*}

    where \(\tanh(\omega)=\beta\) and \(\cosh(\omega) = 1/\sqrt{1-\beta^{2}}\).

• Finite spinor transformation \(S\) may be determined using Equations (2.14) and (2.18) as

  \begin{align*} \psi'(x') &= S\psi(x)\\ &= \exp\left(\frac{-\I}{4}\omega\sigma_{\mu\nu}I^{\mu\nu}_{(n)}\right)\psi(x)\tag{2.22} \end{align*}