Difficult Derivative: Get Input on Taking the Derivative

[topsquark]

What the heck. MathJax is back up and I'm feeling lucky... It's not an easy one. I'm just looking for some input.

How do you take this derivative?
\(\displaystyle \frac{d}{d(\phi ^* \phi )} ( \phi ^* + \phi )\) where * is the complex conjugate and \(\displaystyle \phi \) is complex.

-Dan

 

[Greg]

What about writing stuff in $a+bi$ form? Then you get

$$\frac{d(2a)}{d(a^2+b^2)}$$

(I think) (Thinking)

 

[topsquark]

What about writing stuff in $a+bi$ form? Then you get

$$\frac{d(2a)}{d(a^2+b^2)}$$

(I think) (Thinking)

Yes, I thought of that myself. But do you see the problem? How to deal with the d(a^2 + b^2)? No matter what I do I can't find a way to do the derivative with both a and b. Especially since the b doesn't appear in the 2a and is thus unconstrained. (I think that's what b might be called. I have no idea.)

-Dan

 

[Greg]

May I ask: where did you get this problem?

 

[topsquark]

May I ask: where did you get this problem?

I am trying to minimize the following potential function: \(\displaystyle V = a ( \phi ^* \phi )^2 + b ( \phi ^* \phi ) + c ( \phi ^* + \phi )\). It represents the potential energy function for the Higgs boson under a small fluctuation \(\displaystyle c ( \phi ^* + \phi )\) where c is small. I need the derivative \(\displaystyle \frac{d}{d ( \phi ^* \phi )}\) which is easy enough to do without that last term. I realize it's all nice and finite and the details really don't matter, since c is small and thus the derivative will also be small, but it's nagging me that I can't solve it.

-Dan

 

[I like Serena]

What the heck. MathJax is back up and I'm feeling lucky... It's not an easy one. I'm just looking for some input.

How do you take this derivative?
\(\displaystyle \frac{d}{d(\phi ^* \phi )} ( \phi ^* + \phi )\) where * is the complex conjugate and \(\displaystyle \phi \) is complex.

-Dan

Hey Dan,

How about applying the chain rule together with the inverse function theorem:
$$
\frac{d}{d(\phi ^* \phi )} ( \phi ^* + \phi )
=\frac{\frac{d}{dx} ( \phi ^* + \phi )}{\frac{d}{dx} ( \phi ^*\phi )}
= \frac{\phi' ^* + \phi'}{\phi' ^*\phi + \phi ^*\phi'}
$$

 

[Ackbach]

I am trying to minimize the following potential function: \(\displaystyle V = a ( \phi ^* \phi )^2 + b ( \phi ^* \phi ) + c ( \phi ^* + \phi )\). It represents the potential energy function for the Higgs boson under a small fluctuation \(\displaystyle c ( \phi ^* + \phi )\) where c is small. I need the derivative \(\displaystyle \frac{d}{d ( \phi ^* \phi )}\) which is easy enough to do without that last term. I realize it's all nice and finite and the details really don't matter, since c is small and thus the derivative will also be small, but it's nagging me that I can't solve it.

-Dan

So, you're trying to minimize $V$ with respect to which variable? An independent variable $\phi$ and its conjugate $\phi^*$ are usually considered independent variables one from the other (at least, it's safer to assume that at first). Could you set
\begin{align*}
\pd{V}{\phi}&=0 \\
\pd{V}{\phi^*}&=0?
\end{align*}
That would, I think, be the somewhat more usual procedure, rather than optimizing with respect to the product $\phi^* \phi$.

 

[I like Serena]

So, you're trying to minimize $V$ with respect to which variable? An independent variable $\phi$ and its conjugate $\phi^*$ are usually considered independent variables one from the other (at least, it's safer to assume that at first). Could you set
\begin{align*}
\pd{V}{\phi}&=0 \\
\pd{V}{\phi^*}&=0?
\end{align*}
That would, I think, be the somewhat more usual procedure, rather than optimizing with respect to the product $\phi^* \phi$.

Good point.
How about defining $\phi=r e^{i\theta}$ and setting $\pd V r=\pd V\theta=0$ though?

 

[Ackbach]

Good point.
How about defining $\phi=r e^{i\theta}$ and setting $\pd V r=\pd V\theta=0$ though?

Sure, you could. But the two equations above are quite straight-forward to work with. I get:
\begin{align*}
2a\phi^*\phi+b\phi^*+c&=0 \\
2a\phi^*\phi+b\phi+c&=0 \quad \implies \\
\phi^*&=\phi.
\end{align*}
Here I'm assuming that $b=0$ is not acceptable. Then you can reformulate by minimizing $V=a\phi^4+b\phi^2+2c\phi$ with respect to the real variable $\phi$.

 

[topsquark]

It's been a while since I responded here. I've been trying to use the comments.

Yes, \(\displaystyle \phi\) and \(\displaystyle \phi ^*\) are the usual independent variables. However two points:
1) I have to somehow turn the derivative of \(\displaystyle V(\phi, \phi ^* )\) by the product of the \(\displaystyle \phi\)'s but not independently. There's a reality issue here.

As to I Like Serena's comment we have point 2:

2) The \(\displaystyle \phi\)'s in this case are the independent variables. This is a problem in QFT and the \(\displaystyle \phi\)'s are not dependent on spatial variables. In the full blown QFT problem relating to this (The problem I posted is actually semi-classical) they are the base operators Mathematically representing position operators and \(\displaystyle \phi\) really should be represented as \(\displaystyle < ~ 0 ~ | ~ \phi ~ | ~ 0 ~ >\).

As to the derivative trick, how general is this? The problem I'm doing is simple and essentially one dimensional but if we had more variables... Can we always take \(\displaystyle \frac{df}{dg} = \frac{ \frac{df}{dx} }{ \frac{dg}{dx} }\)? The chain rule says that \(\displaystyle \frac{df}{dg} = \sum_i \frac{df}{dx_i} \frac{dx_i}{dg}\) so we can't always call it \(\displaystyle \frac{1}{\frac{dg}{dx}}\). How would you apply the inverse function rule here?

Thanks all!

-Dan

 

[I like Serena]

As to I Like Serena's comment we have point 2:

2) The \(\displaystyle \phi\)'s in this case are the independent variables. This is a problem in QFT and the \(\displaystyle \phi\)'s are not dependent on spatial variables. In the full blown QFT problem relating to this (The problem I posted is actually semi-classical) they are the base operators Mathematically representing position operators and \(\displaystyle \phi\) really should be represented as \(\displaystyle < ~ 0 ~ | ~ \phi ~ | ~ 0 ~ >\).

I'm a little confused here. To take the derivative of $\phi$, it must be a function of some variable (or more than one) - spatial or not.

As to the derivative trick, how general is this? The problem I'm doing is simple and essentially one dimensional but if we had more variables... Can we always take \(\displaystyle \frac{df}{dg} = \frac{ \frac{df}{dx} }{ \frac{dg}{dx} }\)? The chain rule says that \(\displaystyle \frac{df}{dg} = \sum_i \frac{df}{dx_i} \frac{dx_i}{dg}\) so we can't always call it \(\displaystyle \frac{1}{\frac{dg}{dx}}\). How would you apply the inverse function rule here?

If f and g are scalar functions of the same multiple variables, then the chain rule says:
$$\frac{df}{dg} = \sum_i \pd f {x_i} \frac{dx_i}{dg}$$
Combine with the inverse function theorem to get:
$$\frac{df}{dg} = \sum_i \frac{\pd f {x_i}}{ \frac{dg}{dx_i}}$$

 

[topsquark]

I'm a little confused here. To take the derivative of $\phi$, it must be a function of some variable (or more than one) - spatial or not.

Yes, of course you are right. I've spent a large amount of time with the more advanced stuff that I had temporarily lost sight of the need to use the coordinates to do an actual calculation rather than from a theoretical standpoint. \(\displaystyle \phi = \phi (x^{\mu} )\) where \(\displaystyle x^{\mu}\) is a position 4-vector. For the stuff I've been working with \(\displaystyle \phi\) and it's canonical momentum \(\displaystyle \pi\) can be thought of as basic entities in and of themselves.

-Dan