A Hermitian conjugate of the derivative of a wave function

[Gene Naden]

I am continuing to work through Lessons on Particle Physics. The link is
https://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1271v2.pdf
I am on page 22, equation (1.5.58). The authors are deriving the Hermitian conjugate of the Dirac equation (in order to construct the current). I am able to reproduce (1.5.58) except for one difference:
I have $-i\gamma^0 \frac{\partial}{\partial t} \psi^\dagger - i\frac{\partial}{\partial x_k} \psi^\dagger (-\gamma^k)=0$
while the authors have $-i\gamma^0 \frac{\partial}{\partial t} \psi^\dagger - i\frac{\partial}{\partial x^k} \psi^\dagger (-\gamma^k)=0$

I think, but I am not sure, that they are saying $(\frac{\partial \psi}{\partial x_\mu})^\dagger=\frac{\partial \psi^\dagger}{\partial x^\mu}$

I would like clarification on whether this is in fact the root of my error. Thanks.

 

[strangerep]

Am I going blind? I don't see any difference between the 2 equations.

(Btw, I would have thought that the $\gamma^0$ should be on the right of $\psi^\dagger$ in the 1st term.)

 

[stevendaryl]

I am continuing to work through Lessons on Particle Physics. The link is
https://arxiv.org/PS_cache/arxiv/pdf/0906/0906.1271v2.pdf
I am on page 22, equation (1.5.58). The authors are deriving the Hermitian conjugate of the Dirac equation (in order to construct the current). I am able to reproduce (1.5.58) except for one difference:
I have $-i\gamma^0 \frac{\partial}{\partial t} \psi^\dagger - i\frac{\partial}{\partial x_k} \psi^\dagger (-\gamma^k)=0$
while the authors have $-i\gamma^0 \frac{\partial}{\partial t} \psi^\dagger - i\frac{\partial}{\partial x^k} \psi^\dagger (-\gamma^k)=0$

I think, but I am not sure, that they are saying $(\frac{\partial \psi}{\partial x_\mu})^\dagger=\frac{\partial \psi^\dagger}{\partial x^\mu}$

I would like clarification on whether this is in fact the root of my error. Thanks.

Since x^\mu is real,

(\frac{\partial}{\partial x^\mu} \psi)^\dagger = \frac{\partial}{\partial x^\mu} \psi^\dagger

What they're saying is a fact about the gamma matrices (at least in their usual representation):

(\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0

This can be proved by the four facts:

1. (\gamma^0)^2 =1
2. \gamma^0 \gamma^j = - \gamma^j \gamma^0 (with j=1,2,3)
3. (\gamma^0)^\dagger= \gamma^0
4. (\gamma^j)^\dagger = -\gamma^j (with j = 1,2,3)

So start with the Dirac equation:

i \gamma^\mu \frac{\partial}{\partial x^\mu} \psi = m \psi

Take the conjugate:

-i \frac{\partial}{\partial x^\mu} \psi^\dagger (\gamma^\mu)^\dagger = m \psi^\dagger

Rewrite (\gamma^\mu)^\dagger as \gamma^0 \gamma^\mu \gamma^0:

-i \frac{\partial}{\partial x^\mu} \psi^\dagger \gamma^0 \gamma^\mu \gamma^0 = m \psi^\dagger

Multiply on the right by \gamma^0:

-i \frac{\partial}{\partial x^\mu} \psi^\dagger \gamma^0 \gamma^\mu \gamma^0 \gamma^0 = m \psi^\dagger \gamma^0

Now, use that (\gamma^0)^2 = 1

-i \frac{\partial}{\partial x^\mu} \psi^\dagger \gamma^0 \gamma^\mu = m \psi^\dagger \gamma^0

Finally, define \psi^\dagger \gamma^0 \equiv \bar{\psi}

-i \frac{\partial}{\partial x^\mu} \bar{\psi} \gamma^\mu = m \bar{\psi}

(That paper has a mistake in equation 1.5.58. The first term on the left should be -i \partial_t \psi^\dagger \gamma^0, not -i \gamma^0 \partial_t \psi^\dagger.)

 

[Gene Naden]

Hi Stevendaryl,
Thanks for your input.

I am still wondering; what are the relations between $\partial_{x_\mu}$, ($\partial_{x_\mu})^\dagger$ and $\partial^{x_\mu}$?

By multiplying it out, I have satisfied myself that $-i \partial_t \psi^\dagger \gamma^0 = -i \gamma^0 \partial_t \psi^\dagger$. It is because $\gamma^0$ is symmetric and $\partial_t \psi^\dagger$ is a vector rather than a matrix.

 

[Gene Naden]

strangerep,
Thanks for taking an interest in my thread.
There is a difference; I had
$\frac{\partial}{\partial x_k}$
while they had
$\frac{\partial}{\partial x^k}$
I am wondering, however, if this difference matters. But since $x^\mu=g^{\mu\sigma}x_{\sigma}$, it seems to make a difference.

See my earlier post where I argue that $-i \partial_t \psi^\dagger \gamma^0 = -i \gamma^0 \partial_t \psi^\dagger$

 

[stevendaryl]

Hi Stevendaryl,
Thanks for your input.

I am still wondering; what are the relations between $\partial_{x_\mu}$, ($\partial_{x_\mu})^\dagger$ and $\partial^{x_\mu}$?

By multiplying it out, I have satisfied myself that $-i \partial_t \psi^\dagger \gamma^0 = -i \gamma^0 \partial_t \psi^\dagger$. It is because $\gamma^0$ is symmetric and $\partial_t \psi^\dagger$ is a vector rather than a matrix.

In the usual way of doing the Dirac equation, \psi is a column matrix, and \psi^\dagger is a row matrix, and \gamma^0 is a square matrix. So you can't multiply \gamma^0 \psi^\dagger. If A is a matrix with n columns and m rows, and B is a matrix with a columns and b rows, then A B only makes sense if n = b.

 

[Gene Naden]

Yes, Stevendaryl, I see what you mean. Now I need to review my derivation of 1.5.58. I must have gotten confused. Thanks for asserting reality for me!

 

[stevendaryl]

There is something a little confusing about the \dagger operator when applied to the Dirac equation. In the case of the Schrodinger equation, \dagger means Hermitian conjugate, which has the definition:

\langle \psi | A^\dagger \phi\rangle = \langle A \psi|\phi \rangle

So according to this definition, (\partial_x)^\dagger = - \partial_x, because

\langle (-\partial_x \psi) | \phi \rangle= - \int dx \frac{d \psi^*}{dx} \phi = - \int dx [\frac{d}{dx} (\psi^* \phi) - \psi^* \frac{d\phi}{dx}] = + \int dx \psi^* \frac{d\phi}{dx}= \langle \psi| \partial_x \phi\rangle (the term \int dx \frac{d}{dx} (\psi^* \phi) is zero).

But in the case of the Dirac equation, \dagger means the complex conjugate of the transpose of a matrix. Since \partial_x is neither a matrix, nor imaginary, (\partial_x)^\dagger = \partial_x.

Perhaps someone with more mathematical knowledge than me can give more details about the relationship between \dagger for matrices and \dagger for Hilbert space elements?

 

[Gene Naden]

Stevendaryl, I see what you mean that $-\langle \psi | (\partial_x) ^ \dagger \phi \rangle = \langle (-\partial_x \psi) | \phi \rangle= \langle \psi| \partial_x \phi\rangle$based on the definition

⟨ψ|A†ϕ⟩=⟨Aψ|ϕ⟩

Can you please give me some motivation, or a reference, for this definition?

Thanks

 

[stevendaryl]

Stevendaryl, I see what you mean that $-\langle \psi | (\partial_x) ^ \dagger \phi \rangle = \langle (-\partial_x \psi) | \phi \rangle= \langle \psi| \partial_x \phi\rangle$based on the definitionCan you please give me some motivation, or a reference, for this definition?

Thanks

It's described here: https://quantummechanics.ucsd.edu/ph130a/130_notes/node133.html

 

[Gene Naden]

Thank you

 

[strangerep]

There is a difference; I had
$\frac{\partial}{\partial x_k}$
while they had
$\frac{\partial}{\partial x^k}$

Oh crap -- I am going blind.

I am wondering, however, if this difference matters. But since $x^\mu=g^{\mu\sigma}x_{\sigma}$, it seems to make a difference.

Yes, it does matter. The $k$ in $\frac{\partial}{\partial x^k}$ counts as a "downstairs" index. Hence it can be validly contracted with an upstairs index, as in $\gamma^k$.

Otoh, in $\frac{\partial}{\partial x_k}$ the $k$ counts as an upstairs index. You can't validly contract 2 upstairs or 2 downstairs indices.

See my earlier post where I argue that $-i \partial_t \psi^\dagger \gamma^0 = -i \gamma^0 \partial_t \psi^\dagger$

$\gamma^0$ is a 4x4 matrix. $\psi$ is a 4-component column spinor. $\psi^\dagger$ is a 4-component row spinor.

Think about ordinary matrices and vectors. If you take a column vector $v$, what does the transpose $v^T$ look like? Similarly, if you have an ordinary square matrix $M$, what does $(Mv)^T$ look like?

 

[Gene Naden]

Thanks for your response, strangerep.

Yeah, $\gamma^0 \partial_t \psi^\dagger$ is just nonsense, as stevendaryl pointed out. Sorry about that.