# A How to derive general solution to the Klein-Gordon equation

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1. Apr 12, 2016

### "Don't panic!"

I understand that the ansatz to $$(\Box +m^{2})\phi(\mathbf{x},t)=0$$ (where $\Box\equiv\partial^{\mu}\partial_{\mu}=\eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$) is of the form $\phi(\mathbf{x},t)=e^{(iE_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}$, where $E_{\mathbf{k}}=\sqrt{\mathbf{k}^{2}+m^{2}}$, but I've seen in several lecture notes that one first represents $\phi(\mathbf{x},t)$ in terms of its spatial Fourier transform $$\phi(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}\tilde{\phi}(\mathbf{k},t)e^{i\mathbf{k}\cdot\mathbf{x}}$$ first and then require that at each instant in time, the Fourier modes $\tilde{\phi}(\mathbf{k},t)$ satisfy the equation $$\left(\partial^{2}_{t}+(\mathbf{k}^{2}+m^{2})\right)\tilde{\phi}(\mathbf{k},t)=0$$ (where $\partial^{2}_{t}\equiv\frac{\partial^{2}}{\partial t^{2}})$.

Is it correct to say that although we represent $\phi(\mathbf{x},t)$ in terms of its spatial Fourier transform at a particular fixed instant in time, $t$, we require that that it should be possible to do this at every instant in time, hence requiring that the Fourier modes, $\tilde{\phi}(\mathbf{k},t)$ are time-dependent and satisfy the equation given above?

If so, then the approach makes sense to me, since the general solution to $\left(\partial^{2}_{t}+(\mathbf{k}^{2}+m^{2})\right)\tilde{\phi}(\mathbf{k},t)=0$ is given by $$\tilde{\phi}(\mathbf{k},t)=N(\mathbf{k})\left(a(\mathbf{k})e^{-iE_{\mathbf{k}}t}+b(\mathbf{k})e^{iE_{\mathbf{k}}t}\right)$$ where $N(\mathbf{k})$ is some normalisation factor to be specified.
The general solution to the Klein-Gordon equation would then be given by $$\phi(\mathbf{x},t)=\int\frac{d^{3}k}{(2\pi)^{3}}N(\mathbf{k})\left(a(\mathbf{k})e^{-iE_{\mathbf{k}}t}+b(\mathbf{k})e^{iE_{\mathbf{k}}t}\right)e^{i\mathbf{k}\cdot\mathbf{x}}=\int\frac{d^{3}k}{(2\pi)^{3}}N(\mathbf{k})\left(a(\mathbf{k})e^{-i(E_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}+b(\mathbf{-k})e^{i(E_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}\right)$$ (assuming that $N(-\mathbf{k})=N(\mathbf{k})$).

Have I understood this correctly or am I missing something?

2. Apr 12, 2016

### Demystifier

Your equations are correct. However, mathematically, $t$ is like any other variable so one can, if one wants, to make a Fourier transform in $t$ as well. In this case one will have $E_{\bf k}\rightarrow k^0$, $d^3k\rightarrow d^4k$, $a({\bf k})\rightarrow a(k)$ etc, where $k^0$ is independent on ${\bf k}$. However, only $k^0= \pm E_{\bf k}$ will contribute to the integral because $N(k)\propto \delta(k^2-m^2)N({\bf k})$.

3. Apr 12, 2016

### "Don't panic!"

That's what I thought would be the procedure, i.e. represent $\phi$ in terms of it's Fourier transformation (Fourier transforming both $\mathbf{x}$ and $t$). However, in several texts they only consider a spatial Fourier transform at a fixed instant in time $t$ and then simply state the solution. I was wondering what the justification is for this? (and also why one introduces the normalisation factor $N(\mathbf{k})$ in the solution?)

4. Apr 12, 2016

### Demystifier

The two approaches are equivalent, i.e. one can be derived from another. The approach you described is simpler because one does not need to integrate over $dk^0 \delta(k^2-m^2)$.

Sometimes it is useful not to specify the factor $N(\mathbf{k})$ explicitly because there are several different useful normalization choices, each with its advantages and disadvantages. But needless to say, they are all equivalent.

5. Apr 12, 2016

### "Don't panic!"

How does one end up with this term in the other approach?

Why is the factor $N(\mathbf{k})$ needed in the first place anyway?

6. Apr 12, 2016

### vanhees71

Now, a lot is very confused. So let's to the calculation. What we need is that the generalized momentum eigenvector, properly normalized to a $\delta$ distribution is given by the plane wave
$$u_p(x)=\langle x|p \rangle=\frac{1}{\sqrt{2 \pi}} \exp(\mathrm{i} px).$$
Here and in the following I set $\hbar=1$ (natural units). Then we have given
$$U(t,x,x')=\langle x|\exp(-\mathrm{i} \hat{p}^2/(2m)) t)|x' \rangle = \sqrt{\frac{m}{2 \pi \mathrm{i} t}} \exp \left (\frac{\mathrm{i} m(x-x')^2}{2t} \right).$$
Now the momentum representation is in fact to calculate in a very trivial way since for the free particle the energy is a function of momentum only and thus we have
$$\tilde{U}(t,p,p')=\langle p|\exp(-\mathrm{i}\hat{p}^2/(2m)) t)|p' \rangle=\exp(-\mathrm{i} p^2/(2m)) \delta(p-p'),$$
but now you want this in the hard way via the position representation. So you introduce a completeness relation in terms of the position basis twice, i.e.,
$$\tilde{U}(t,p,p')=\int_{\mathbb{R}} \mathrm{d} x \int _{\mathbb{R}} \mathrm{d} x' \langle p|x \rangle U(t,x,x') \langle x'|p' \rangle.$$
But that's
$$\tilde{U}(t,p,p')=\int_{\mathbb{R}} \mathrm{d} x \int _{\mathbb{R}} \mathrm{d} x' \frac{1}{2 \pi} \exp[\mathrm{i} (p'x'-px)] \sqrt{\frac{m}{2 \pi \mathrm{i} t}} \exp \left (\frac{\mathrm{i}m(x-x')^2}{2t} \right).$$
Now to regularize the integral you can think of $m$ to have a small positive imaginary part, which at the end of the calculation you let go to $0^+$. Then Substituting $x''=x'-x$ in the integral over $x'$ you get
$$\tilde{U}(t,p,p')=\int_{\mathbb{R}} \mathrm{d} x \exp[\mathrm{i} x (p'-p)] \int _{\mathbb{R}} \mathrm{d} x' \frac{1}{2 \pi} \exp[\mathrm{i} p' x''] \sqrt{\frac{m}{2 \pi \mathrm{i} t}} \exp \left (\frac{\mathrm{i}m(x-x')^2}{2t} \right).$$
Now the integral over $x''$ is just a Gaussian integral, which gives
$$\tilde{U}(t,p,p')=\int_{\mathbb{R}} \mathrm{d} x \exp[\mathrm{i} x (p'-p)] \exp[-\mathrm{i} p^{\prime2} t/(2m)] = \delta(p-p') \exp[-\mathrm{i} p^{\prime2} t/(2m)] = \delta(p-p') \exp[-\mathrm{i} p^{2} t/(2m)],$$
as it must be.

7. Apr 12, 2016

### Demystifier

See e.g. http://www.tep.physik.uni-freiburg.de/lectures/QFT14/qft.pdf
Sec. 3.2.

You can write all the equations without it if you want. Indeed, it is natural to start with a Fourier transform $A(k)$ and $A^*(k)$ without any factor $N(k)$. But then $A(k)$ satisfy some relations which look somewhat ugly. Therefore you can use a trick by writing $A(k)$ as $A(k)\equiv N(k)a(k)$ and choosing $N(k)$ so that the relations for $a(k)$ look simpler than those for $A(k)$, in which case all the "ugliness" is absorbed by $N(k)$.

8. Apr 12, 2016

### "Don't panic!"

Thanks for the link, it looks very useful! Although, I'm unsure as to motivation of the form of $\tilde{\phi}(p)$ in these notes. Is it simply that it is expressed as an arbitrary linear combination of positive and negative energy solutions, $p^{0}=-\sqrt{\mathbf{p}^{2}+m^{2}}<0$ and $p^{0}=\sqrt{\mathbf{p}^{2}+m^{2}}>0$, such that $$\tilde{\phi}(p)=2\pi\delta(p^{2}-m^{2})(\theta(p^{0})a(\mathbf{p})+\theta(-p^{0})b^{\ast}(-\mathbf{p}))$$ with the delta function ensuring that the solutions satisfy $p^{2}=m^{2}$. ( I have to admit, I'm unsure as to why the second arbitrary function is a function of $-\mathbf{p}$ instead of $\mathbf{p}$, and also why they are both not functions of $p^{\mu}$, but only $\mathbf{p}$)?

Ah ok, so the $N(\mathbf{k})$ is simply introduced to "normalise" the Fourier coefficients $A(k)$ and $A^{\ast}(k)$ to remove any "ugly" prefactors then?!

Also, is the reasoning I gave for the spatial decomposition case correct? i.e. that to solve the K-G equation we start by representing $\phi(\mathbf{x},t)$ in terms of its spatial Fourier transform at a particular fixed instant in time, $t$m $$\int\frac{d^{3}k}{(2\pi)^{3}}\tilde{\phi}(\mathbf{k},t)e^i\mathbf{k}\cdot\mathbf{x}\qquad (1)$$. Furthermore, we require that that it should be possible to do this at every instant in time, hence we require that the Fourier modes, $\tilde{\phi}(\mathbf{k},t)$, are time-dependent. Applying the K-G equation to $(1)$ we find that this requires $\tilde{\phi}(\mathbf{k},t)$ to satisfy the equation $$(\partial^{2}_{t}+\mathbf{k}^{2}+m^{2})\tilde{\phi}(\mathbf{k},t)=0$$

Last edited: Apr 12, 2016
9. Apr 13, 2016

### Demystifier

Yes.

$\mathbf{p}$ is a dummy variable, meaning that there is an integral over it so that the full expression does not depend on $\mathbf{p}$. Therefore you can introduce a new integration variable $\mathbf{p}'=-\mathbf{p}$. Since $\mathbf{p}'$ is also a dummy variable, you can rename it as by omitting the prime. By this procedure $\mathbf{p}$ becomes $-\mathbf{p}$. It is a convenient trick to put the whole expression into a more useful form.

Yes.

It is correct. But there are also other ways (chains of arguments) to arrive at the same decomposition.

10. Apr 13, 2016

### "Don't panic!"

I think what worries me slightly is the fact that I've only Fourier decomposed the spatial part and left the time dependence alone. Also, by doing so at a fixed instant in time, how can I then take a time (2nd) derivative of the the mode functions (as I did in my original post)? Is it simply that one is evaluating the time (2nd) derivative at that particular instant, and one wishes to impose that the Fourier mode functions evolve according to this differential equation such that the same decomposition of the field is valid at all times $t$.

Finally, when it comes to the different "pictures" (Schrödinger, Heisenberg and interaction), can one then the use this expression for the field in the Schrödinger picture, but just fix the particular instant in time $t=t_{0}$, such that $$\phi_{S}(\mathbf{x})=\phi(\mathbf{x},t_{0})=\int\frac{d^{3}k}{(2\pi)^{3}}N(\mathbf{k})(a(\mathbf{k})e^{(-iE_{\mathbf{k}}t_{0}+i\mathbf{k}\cdot\mathbf{x})}+ b(\mathbf{k})e^{(iE_{\mathbf{k}}t_{0}-i\mathbf{k}\cdot\mathbf{x})})$$ (where the subscript $S$ denotes the Schrödinger picture)

The reason I ask is that I'm trying to justify how one can then show, when considering interactions, that since in the interaction picture the field operators evolve with respect to the free Hamiltonian, $H_{0}$, $$\phi_{I}(\mathbf{x},t)=e^{iH_{0}(t-t_{0})}\phi_{S}(\mathbf{x},t_{0})e^{-iH_{0}(t-t_{0})}$$ (where the subscripts $S$ and $I$ denote the Schrödinger and interaction pictures respectively) that they can be expressed in terms of their free field expansion, i.e. $\phi_{I}(\mathbf{x},t)=\phi(\mathbf{x},t)$?!

11. Apr 13, 2016

### Demystifier

It doesn't matter if you Fourier transform at fixed time, as long as you expression for the field satisfies the KG equation at all times. And you can be sure that it does, because your $E_{\bf k}$ is not just any function of ${\bf k}$, but precisely $E_{\bf k}=\sqrt{{\bf k}^2+m^2}$.

12. Apr 13, 2016

### "Don't panic!"

Is it valid to use the argument that one Fourier transforms at a particular instant in time, but then also require that this representation off the field should hold for all time and in order for this to be true the Fourier mode functions $\tilde{\phi}(\mathbf{k},t)$ should evolve in time according to the equation $$(\partial^{2}_{t}+\mathbf{k}^{2}+m^{2})\tilde{\phi}(\mathbf{k},t)=0$$

13. Apr 13, 2016

### Demystifier

Yes, that's another way to say the same thing.

14. Apr 13, 2016

### stevendaryl

Staff Emeritus
Isn't that for the Schrodinger equation, rather than Klein-Gordon?

15. Apr 13, 2016

### Demystifier

Technically you are right. But there is a similar Schwinger's proper time method that works for Klein-Gordon. The idea is to introduce a "Hamiltonian"
$$H=p^2+m^2$$
so that KG equation can be written as
$$H\psi(x)=0$$
where $x$ is a spacetime coordinate. Then the propagator can be expressed in an elegant form by studying "Schrodinger" evolution
$$H\psi(x,s)=i\frac{\partial\psi(x,s)}{\partial s}$$
where $s$ is the proper time. In this way you get propagation in unphysical proper time $s$, but appropriate integration over $s$ eliminates the dependence on $s$ and what remains is the desired propagator in physical spacetime. For more details google about Schwinger proper time method.

By the way, Schwinger is full of such strange methods. It has been said that others publish to show others how to do the calculation, while Schwinger publishes to show that only he could do it.

Last edited: Apr 13, 2016
16. Apr 13, 2016

### vanhees71

Yes, sorry. I mixed up two (weird) threads ;-)).

17. Apr 13, 2016

### "Don't panic!"

Ok great.

One last thing. When considering interactions between fields, in the interaction picture one has that $$\phi_{I}(t,\mathbf{x})=e^{iH_{0}(t-t_{0})}\phi_{s}(t_{0},\mathbf{x})e^{-iH_{0}(t-t_{0})}$$ and one finds that $\phi_{I}(t,\mathbf{x})$ evolves with respect to the free-field Hamiltonian, $H_{0}$. Is it correct to justify why we can then use the free-field expansion of the operator in the interaction picture, by noting that $$\left[H_{0},a(\mathbf{k})\right]=-a(\mathbf{k})E_{\mathbf{k}}\, ,\qquad \left[H_{0},a^{\dagger}(\mathbf{k})\right]=a^{\dagger}(\mathbf{k})E_{\mathbf{k}}$$ to deduce $$e^{iH_{0}(t-t_{0})}a(\mathbf{k})e^{-iH_{0}(t-t_{0})}=a(\mathbf{k})e^{-iE_{\mathbf{k}}(t-t_{0})}\qquad e^{iH_{0}(t-t_{0})}a^{\dagger}(\mathbf{k})e^{-iH_{0}(t-t_{0})}=a^{\dagger}(\mathbf{k})e^{iE_{\mathbf{k}}(t-t_{0})}$$ and then use our previously derived expansion for $\phi(t,\mathbf{x})$, but evaluating it at $t=t_{0}$, i.e. $$\phi(t,\mathbf{x})=\int\frac{d^{3}k}{(2\pi)^{3}}\left(a(\mathbf{k})e^{-i(E_{\mathbf{k}}t_{0}-\mathbf{k}\cdot\mathbf{x})}+a^{\dagger}(\mathbf{k})e^{i(E_{\mathbf{k}}t_{0}-\mathbf{k}\cdot\mathbf{x})}\right)$$ Then, putting it altogether we have, $$\phi_{I}(t,\mathbf{x})=e^{iH_{0}(t-t_{0})}\phi_{s}(t_{0},\mathbf{x})e^{-iH_{0}(t-t_{0})}\\ =\int\frac{d^{3}k}{(2\pi)^{3}}\left(e^{iH_{0}(t-t_{0})}a(\mathbf{k})e^{-iH_{0}(t-t_{0})}e^{-i(E_{\mathbf{k}}t_{0}-\mathbf{k}\cdot\mathbf{x})}+e^{iH_{0}(t-t_{0})}a^{\dagger}(\mathbf{k})e^{-iH_{0}(t-t_{0})}e^{i(E_{\mathbf{k}}t_{0}-\mathbf{k}\cdot\mathbf{x})}\right)\\ \qquad =\int\frac{d^{3}k}{(2\pi)^{3}}\left(a(\mathbf{k})e^{-iE_{\mathbf{k}}(t-t_{0})}e^{-i(E_{\mathbf{k}}t_{0}-\mathbf{k}\cdot\mathbf{x})}+a^{\dagger}(\mathbf{k})e^{iE_{\mathbf{k}}(t-t_{0})}e^{i(E_{\mathbf{k}}t_{0}-\mathbf{k}\cdot\mathbf{x})}\right)=\int\frac{d^{3}k}{(2\pi)^{3}}\left(a(\mathbf{k})e^{-i(E_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}+a^{\dagger}(\mathbf{k})e^{i(E_{\mathbf{k}}t-\mathbf{k}\cdot\mathbf{x})}\right)$$ as required.

18. Apr 13, 2016

### Demystifier

But in #15 I found a way to make sense of it.

19. Apr 13, 2016

### Demystifier

I can't see some of your equations, but what I can see looks OK.

20. Apr 13, 2016

### "Don't panic!"

Sorry, have fixed the math processing error now!