# Hermicity of alpha (dirac equation)

1. Nov 29, 2009

### ballzac

1. The problem statement, all variables and given/known data
Show that
$$\mathbf\alpha\equiv\left[\begin{array}{cc} 0&\mathbf\sigma\\ \mathbf\sigma&0\end{array}\right]$$
is hermitian.

3. The attempt at a solution
My first instinct was to say that $$\mathbf\sigma$$ must be equal to its complex conjugate (as it would if it was a scalar and not a matrix). This is not the case, but I noticed that $$\sigma_y$$ is hermitian. The only way I can see it working (using the traditional definition of hermitian conjugate) is to treat $$\mathbf\alpha$$ as a set of three 4x4 matrices, two of which are real (and hence hermitian), and the other also being hermitian. Is this a valid interpretation of $$\mathbf\alpha$$?

2. Dec 1, 2009

### clamtrox

Yup, you are asked if the elements of the vector are hermitian. Also remember what being hermitian means; it means that the elements satisfy $$a_{ij} = a^*_{ji}$$.

3. Dec 1, 2009

### ballzac

Thank you. I think I need to clarify something though. There are three matrices. We can ignore $$\left[\begin{array}{cc}0&\sigma_x\\\sigma_x&0\end{array}\right]$$ and $$\left[\begin{array}{cc}0&\sigma_z\\\sigma_z&0\end{array}\right]$$ because they are real and symmetric and hence hermitian, and focus on $$\left[\begin{array}{cc}0&\sigma_y\\\sigma_y&0\end{array}\right]$$
If I treat it as a 2x2 matrix, and use the definition of hermicity (that you stated), then $$\sigma_y$$ must be equal to its complex conjugate. This is not the case. However, it IS equal to its hermitian conjugate, but that's not what the definition of hermicity states!

Thus, to agree with the definition, $$\mathbf\alpha$$ must be a 4x4 matrix. This would mean that the zeros in $$\mathbf\alpha$$ are really 2x2 null matrices.

The only other way I can see it working is if the definition of hermicity is only valid when the elements of the matrix are all scalars. In this case, the definition would need to be modified to say that the hermitian conjugate is the transposed matrix of hermitian conjugates, rather than complex conjugates. This would obviously require the condition that the hermitian conjugate of a scalar is the same as its complex conjugate, to avoid the argument being circular.

To sum up, the main point of my question is whether $$\mathbf\alpha$$ as defined in my first post is actually shorthand for a 4x4 matrix.

4. Dec 2, 2009

### clamtrox

It is. The notation would not make any sense otherwise.

5. Dec 2, 2009

### ballzac

Awesome. Thanks for that.