# Hermitian conjugation and conserved current in the Dirac equation

1. Nov 24, 2009

### gheremond

Consider the Dirac equation in the ordinary form in terms of $$a$$ and $$\beta$$ matrices
$$i\frac{{\partial \psi }} {{\partial t}} = - i\vec a \cdot \vec \nabla \psi + m\beta \psi$$
The matrices are hermitian, $$\vec a^\dag = \vec a,\beta ^\dag = \beta$$. Daggers denote hermitian conjugation, i.e. complex conjugation followed by transposition for matrix operators. In order to prove the probability current conservation one has to derive from the Dirac equation the conjugate equation according to
$$- i\frac{{\partial \psi ^\dag }} {{\partial t}} = i\vec \nabla \psi ^\dag \cdot \vec a^\dag + m\psi ^\dag \beta ^\dag$$
Using these two equations one can easily show that the probability 4-vector is conserved and $$\rho = \psi ^\dag \psi$$ is the correct definition for the probability density. However, the previous hermitian conjugation operation, as employed, only affects Dirac matrices and the corresponding columns and rows. Notice that
$$\left( { - i\vec a \cdot \vec \nabla \psi } \right)^\dag = i\vec \nabla \psi ^\dag \cdot \vec a^\dag$$
under this operation, so there is no notion of hermiticity of the momentum operator,
$$\vec p^\dag = \left( { - i\vec \nabla } \right)^\dag = i\vec \nabla = - \vec p$$
as one would expect. In the meantime, the Dirac Hamiltonian is (again from the Dirac equation)
$$H_D = - i\vec a \cdot \vec \nabla + m\beta = \vec a \cdot \vec p + m\beta$$
We would expect this to be Hermitian and indeed, using the last expression involving the momentum operator, this seems to be the case, provided though that we now use a hermitian conjugation operation that also affects differential operators, so that we recover
$$\vec p^\dag = \vec p$$
and hermiticity is recovered. Why is this need for two different definitions of the hermitian conjugation in this case? One would expect that the operation affecting both matrix and differential operators should be used throughout, as hermitian conjugation should be defined in terms of the inner product in the full Hilbert space of the theory, involving both summation over spinor components and integration over the position variable. But this would not yield the correct expression for the current conservation. Note that this question applies to the one-particle theory (no QFT arguments) where the momentum is still an operator and a valid observable.

2. Nov 24, 2009

### Avodyne

You are mixing up two different notions of hermitian conjugation. The first acts in the space of components of the Dirac wave function; it takes the column vector $\psi_i(x)$ and turns it into the row vector $\psi^*_i(x)$, and replaces the Dirac matrices with their hermitian conjugates. Differential operators such as $\nabla$ are unaffected by this conjugation. The analog for the nonrelativistic Schrodinger equation is just complex conjugation.

The second sort is hermtian conjugation in the space of square-integrable functions. In this space, $\nabla$ is an antihermitian operator: $\nabla^\dagger=-\nabla$.

It is the first sort that you want to use to derive the continuity equation.

3. Nov 24, 2009

### gheremond

Thanks for the reply! The parallel with complex conjugation in the Schroedinger equation indeed makes some sense. Yet it is not fully formally equivalent. It would be, provided you could define a current and density for the Dirac equation that is valid "component-wise", not as a sum of terms involving all four components. This summation is in fact the same as forming a partial inner product in the Dirac Hilbert space, summing over the spinor components only. This makes the comparison with the Schroedinger case less obvious (since there you did not have to interfere at all with manipulations comparable to taking an inner product).