# Homework Help: Klien-Gordon equation to solve 4-vector problem? (Particle)

1. Feb 22, 2015

### rwooduk

1. The problem statement, all variables and given/known data

2. Relevant equations
KG Equation

$\left ( \delta ^{2} + \frac{m^{2}c^{2}}{\hbar^{2}}\right )\Psi = 0$

3. The attempt at a solution

To solve this problem I'm not sure whether to use the KG equation OR apply the operator given in the question (I dont know how to do this)

I can use the KG equation on PSI when given say:

$\Psi = R exp(-iwt+ik_{i}x_{i})$

But if I use the KG equation for this problem, so treat Aμ as PSI in the above equation, I will need to write it in non-covarient form. To do this I can simply take the inner product of qv and xv but I dont know what they are equal to, to be able to do this.

I'm a bit lost!

Please could someone suggest a starting point or at least give me an idea of what qv and xv are actually equal to so I can attempt to apply the operator (any hints on how to do this would be appreciated) or the KG equation.

Thanks for any suggestions in advance.

This is a past exam question that I'm attempting to apply my lecture notes to, it's not any assessed piece of work.

2. Feb 22, 2015

### TSny

The components $q_{\nu}$ are just constants. Suppose you apply $\hat{p}_1$ to $A^\mu$. What do you get?

3. Feb 22, 2015

### rwooduk

thats the problem I dont know what these things are, I know

$\delta _{\mu}= (\frac{1}{c}\frac{\delta}{\delta t},\frac{\delta}{\delta x},\frac{\delta}{\delta y},\frac{\delta}{\delta z})$

and

$\rho_{\mu}= (\frac{E}{c},-\rho_{x},-\rho_{y},-\rho_{z})$

and

$A^{\mu}= (\frac{\Phi }{c},A_{x},A_{y},A_{z})$

but what would δk even be?

i'm struggling with what each of the things in the question are actually equal to.

thanks for the help

Last edited: Feb 22, 2015
4. Feb 22, 2015

### TSny

You want to show that $A^{\mu}$ is an eigenfunction of each of the operators $\hat{p}_\kappa$. So, take the case where $\kappa = 1$. Then $\hat{p}_1 =i \hbar \frac{\partial}{\partial x}$. Let this act on $A^{\mu}$. It will help if you explicitly write out $q_{\nu}x^{\nu}$ in the exponential factor of $A^{\mu}$.

Last edited: Feb 23, 2015
5. Feb 23, 2015

### rwooduk

thanks again for the reply. but that's where my problem lies, I don't know what $q_{\nu}x^{\nu}$ explicitly are. what do $q_{\nu}$ and $x^{\nu}$ equal individually? sorry I'm very new to the notation and we have yet to be given any problems in this notation.

6. Feb 23, 2015

### TSny

The notation $q_{\nu}x^{\nu}$ means $\sum_{\nu = 0}^3q_{\nu}x^{\nu} = q_0x^0 +q_1x^1+q_2x^2+q_3x^3$. The summation is implied in the repeated indices of the index $\nu$ in $q_{\nu}x^{\nu}$. This is the "Einstein summation convention".

The various $q$'s are constants whose meaning you will discover in this exercise. $R$ and $\epsilon^{\mu}$ are also constants.

7. Feb 24, 2015

### rwooduk

thanks very much, I will try with this info and see how far I get.

8. Feb 26, 2015

### rwooduk

ok, this is easier than i thought (i think), I'll post where I am incase it helps others.

$\delta ^{\mu} = \frac{\delta }{\delta x^{\mu}}$

and

$\rho ^{v}x_{v} = \rho^{0}x_{0}+\rho^{1}x_{1}+ \rho^{2}x_{2} + ...$

you can see from this that x is linear, therefore for any term of xv the derivative with respect to x will be 1.

therefore

$\delta ^{\mu} \rho ^{v}x_{v} = \frac{\delta }{\delta x^{\mu}} \rho ^{v}x_{v} = \rho^{v} \frac{\delta }{\delta x^{\mu}} x_{v}=\rho^{v}$

edit

hmm so going back to the problem, it would be:

$p_{k}A^{\mu} = i \hbar(R\epsilon ^{\mu})\frac{\delta }{\delta x^{k}}(e^{-iq_{v}x^{v}})$

in this case would

$\frac{\delta }{\delta x^{k}} x^{v} = 1$?

as the v is an indices and doesnt mean x to the power right?

any further advice would be appreciated

Last edited: Feb 26, 2015
9. Feb 26, 2015

### TSny

This is not quite right. Note that the index $\nu$ is a summation index, so the result cannot be $\rho^{v}$. It would be best to raise and lower the $\nu$ index to write it as $\frac{\delta }{\delta x^{\mu}} \rho _{v}x^{v}$. Try to show that $\frac{\delta }{\delta x^{\mu}} x^{v} = \delta_{\mu \nu}$ where $\delta_{\mu \nu}$ is the Kronecker delta. Or, if you have trouble with this, write out the sum over $\nu$ explicitly:

$\frac{\delta }{\delta x^{\mu}} \rho _{v}x^{v} = \frac{\delta }{\delta x^{\mu}} ( \rho _{0}x^{0}+ \rho _{1}x^{1} + \rho _{2}x^{2} + \rho _{3}x^{3})$

It might help to write out the argument of the exponential: $p_{k}A^{\mu} = i \hbar(R\epsilon ^{\mu})\frac{\delta }{\delta x^{k}}(e^{-i( q_0x^0 + q_1x^1 + q_2x^2 + q_3x^3 })$

10. Feb 27, 2015

### rwooduk

damn, though I had it. ok I'll ask a last question and go from there as you have already helped plenty.

$\frac{\delta }{\delta x^{\mu}} x^{v} = \delta_{\mu \nu}$

this would mean that when mu=v the term would be 1 and zero for non equal componants. so

$\frac{\delta }{\delta x^{1}}x^{2} = 0$

im now getting confused with indices and powers, could you confirm that x2 doesnt mean x squared, it's just a label for the 2nd component of xv?

so you could write it $\frac{\delta }{\delta x^{1}}x^{1} = \frac{\delta }{\delta x^{1}}(x^{1})= \frac{\delta }{\delta A}(A)=1$

or likewise $\frac{\delta }{\delta x^{2}}x^{2} = \frac{\delta }{\delta x^{2}}(x^{2})= \frac{\delta }{\delta B}(B)=1$

11. Feb 27, 2015

### TSny

OK. So, for example, $\frac{\partial}{\partial x^k}(q_0x^0+q_1x^1+q_2x^2+q_3x^3) = q_k$ for k = 0, 1, 2, or 3.

12. Feb 28, 2015

### rwooduk

Thanks for all your help. When I finally get the question finished I'll be sure to post here.

13. Mar 1, 2015

### rwooduk

Last edited: Mar 1, 2015
14. Mar 1, 2015

### rwooduk

Ok think I'm there, I'll write it in detail (including the small things I've picked up) edit i've written Au wrong! but I believe the method is correct for the Au i've written, the original question would require a switch of indices

$A^{\mu} = R\epsilon^{\mu} e^{-k^{v}x_{v}}$

$p_{k}=i\hbar\delta _{k}=i\hbar \frac{\delta }{\delta x_{k}}$

note i originally wrote this $p_{k}=i\hbar\delta _{k}=i\hbar \frac{\delta }{\delta x^{k}}$ which would have been $p^{k}$ not $p_{k}$

therefore since the k indices of the differential matches that of the indices of the x componant in the exponential there is no need to switch indices.

$p_{k}A^{\mu} =i\hbar R\epsilon^{\mu}\frac{\delta }{\delta x_{k}} e^{-k^{v}x_{v}}$

$=i\hbar R\epsilon^{\mu} e^{-k^{v}x_{v}}\frac{\delta }{\delta x_{k}} (-k^{v}x_{v})$

$=i\hbar R\epsilon^{\mu} e^{-k^{v}x_{v}} (-ik^{k})\frac{\delta }{\delta x_{k}} (-k^{v}x_{v})$

note that when the differential acts on the term that it changes it's indices to match the indices letter of the differential.

$=i\hbar (-ik^{k}) A^{\mu}=\hbar k^{k} A^{\mu}$

Last edited: Mar 1, 2015
15. Mar 1, 2015

### TSny

Your original way of writing $\hat{p}_k$ was correct. That is $\hat{p}_k = i\hbar\partial_k = i\hbar\frac{\partial}{\partial x^k}$.

Note that the $x^k$ is in the denominator. So, the superscript $k$ on $x^k$ actually acts like a subscript for the overall expression. That is why the $k$ is a subscript on $\hat{p}$ while it is a superscript on $x$.

I don't understand the sudden appearance of $(-ik^{k})$ in the last line.

16. Mar 1, 2015

### rwooduk

hmm I didnt know that, thanks. So what would $\hat{p}^{k}$ be? $\hat{p}^{k} = i\hbar\partial^{k} = i\hbar\frac{\partial}{\partial x_k}$ which would then act on the upper indices?

this is what was given in class:

it doesnt show the steps but you can see that after acting on the exponential the differential brings down a $k^{\rho}$ not a $k^{v}$. So i've applied the same idea to the $k^{v}$ to give $k^{k}$

17. Mar 1, 2015

### TSny

Yes.

$\frac{\partial}{\partial x_k}$ would act on any function of $x_k$ (with lower index on $x$).

That looks good.

You left out the $i$ in the arguments of the exponentials. You are correct that $\frac{\delta }{\delta x_{k}} (-ik^{v}x_{v}) = -ik^k$

In your last line you should not still have the factor $\frac{\delta }{\delta x_{k}} (-k^{v}x_{v})$

18. Mar 2, 2015

### rwooduk

oops, indeed.

I am very happy to leave it there, thanks for teaching me 4-vectors!

There is just one final thing, this statement:

if $\frac{\partial}{\partial x_k}$ acts on the lower indices and $\frac{\partial}{\partial x^k}$ acts on the lower indices, then what's the difference between them? I don't understand that they both act on the lower indices of x.

19. Mar 2, 2015

### TSny

$\frac{\partial}{\partial x_k}$ and $\frac{\partial}{\partial x^k}$ do not act on indices, they act on functions of $x_k$ or $x^k$.

I'm not sure which convention of the metric you are using. But let's suppose you are using the metric where $x_0 = x^0$ and $x_i = - x^i$ for i = 1, 2, or 3. Then, for example, $$\frac{\partial x^1}{\partial x^1} = 1$$ but $$\frac{\partial x^1}{\partial x_1} =\frac{\partial (-x_1)}{\partial x_1} = - \frac{\partial x_1}{\partial x_1} =-1$$

As a little exercise, you can try evaluating $$\frac{\partial }{\partial x^3}(x_{\mu}x^\mu)$$ and $$\frac{\partial }{\partial x_3}(x_{\mu}x^\mu)$$

20. Mar 2, 2015

### rwooduk

Will try that. Thanks once again for all your help!