# Hermicity of alpha (dirac equation)

• ballzac
In summary, The question is asking to show that \mathbf\alpha\equiv\left[\begin{array}{cc} 0&\mathbf\sigma\\ \mathbf\sigma&0\end{array}\right] is hermitian. The solution involves understanding that the elements of the vector must satisfy the condition a_{ij} = a^*_{ji} to be hermitian. The vector contains three matrices, two of which are real and symmetric, and one of which is equal to its hermitian conjugate but not its complex conjugate. To adhere to the definition of hermitian, \mathbf\alpha must be treated as a 4x4 matrix, with the zeros representing
ballzac

## Homework Statement

Show that
$$\mathbf\alpha\equiv\left[\begin{array}{cc} 0&\mathbf\sigma\\ \mathbf\sigma&0\end{array}\right]$$
is hermitian.

## 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 $$\mathbf\alpha$$?

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

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.

ballzac said:
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.

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

Awesome. Thanks for that.

## 1. What is the meaning of "Hermicity" in the context of the alpha (Dirac) equation?

Hermicity refers to the property of a mathematical operator to be self-adjoint, meaning that the operator and its adjoint (conjugate transpose) are equal. In the context of the alpha (Dirac) equation, Hermicity ensures that the equation is a valid representation of the physical system being described.

## 2. How does the Hermicity of the alpha (Dirac) equation relate to the conservation of probability?

The Hermicity of the alpha (Dirac) equation guarantees that the total probability of a particle being in a specific state remains constant over time. This is because the Hamiltonian operator, which represents the energy of the system, is Hermitian, and the conservation of probability is directly related to the Hermiticity of the Hamiltonian.

## 3. Can the Hermicity of the alpha (Dirac) equation be violated?

No, the Hermicity of the alpha (Dirac) equation cannot be violated without fundamentally changing the nature of the equation. If the Hermicity is violated, the equation would no longer be a valid representation of the physical system and would not accurately describe its behavior.

## 4. How does the Hermicity of the alpha (Dirac) equation impact the solutions to the equation?

The Hermicity of the alpha (Dirac) equation ensures that the solutions to the equation are physically meaningful. This means that the solutions must satisfy certain conditions, such as being square-integrable, in order to accurately represent the behavior of the physical system.

## 5. Is the Hermicity of the alpha (Dirac) equation a necessary condition for its solutions to be physically valid?

Yes, the Hermicity of the alpha (Dirac) equation is a necessary condition for its solutions to be physically valid. Without it, the solutions would not accurately represent the behavior of the physical system and would not be considered valid in the context of quantum mechanics.

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