# Canonical Quantization of KG field

1. Jul 9, 2010

### luksen

Hi,

for my exam i"m re-reading Peskin&Schroeder and stumbled across equations 2.21-2.25 where the canonical quantization of the KG field is done.

P&S start with doing a fourier trf on $$\phi(x,t)=\int\frac{d^3p}{(2\pi)^3}e^{ip\cdot x}\phi(p,t)$$

applying the KG operator in that results in $$(\frac{\partial^2}{\partial t^2}+|p|+m^2)\phi(p,t)=0$$

P&S go on to recollect hte SHO where $$\phi=\frac{1}{\sqrt{2\omega}}(a+a^\dagger)$$

so P&S say that in analogy you arrive at
$$\phi(x,t)=\int\frac{d^3p}{(2\pi)^3}(a_pe^{ip\cdot x}+a_p^\dagger e^{-ip\cdot x})$$

but straight forward subsitution would yield with no negative frequency terms

$$\phi(x,t)=\int\frac{d^3p}{(2\pi)^3}(a_pe^{ip\cdot x}+a_p^\dagger e^{+ip\cdot x})$$

I've seen this expansion derived differently and understand it when i follo w it but using this SHO analogy i can't follow thie last step to 2.25

i'd be grateful for input

2. Jul 9, 2010

### Fredrik

Staff Emeritus
I didn't look at the calculation in the book, or try to understand every detail of what you said, but there's a variable change that's useful in this type of calculation:

You consider the integral over the two terms separately. Define p'=-p and change the integration variable to p' in one of the integrals. Then you just drop the prime because it doesn't matter what you call the integration variable.

3. Jul 19, 2010

### Rick89

Hi,
honestly I think the mode expansion of psi (i.e. writing it in terms of a and a-dagger) is just the definition of a, no hidden physics (the SHO is why this expansion is useful...because you can then write the Hamiltonian so that you see a and a-dagger are ladder operators). Anyway it is a def. of the operator a, and you necessarily need the minus sign in the exp. because psi is an observable, it must be hermitian

4. Jul 22, 2010

### ismaili

I thought the minus sign in the exp. is because that we are considering "real" scalar fields.
The field itself needs not to be an observable.

By the way, I am wondering why we always decompose the expansion of plane waves into positive-frequency and negative-frequency part, the expansion $$\int d^3p \exp(-i\mathbf{p}\cdot\mathbf{x})$$ should contain all the plane waves already, right?
Why do we need additional $$\exp(+i\mathbf{p}\cdot\mathbf{x})$$?
Is it a hindsight?
Thanks!

5. Jul 23, 2010

### Rick89

hi,
well you include the negative frequency terms exactly to make sure the sum is overall hermitian, it is to constrain it to be such. You're right you wouldn't need both terms, you are in practice including each term twice, this allows you to specify that the operator that goes with +p and with -p are related, are not any two, exactly to have it overall hermitian. With the other part, yes and no, it is so because it is a real field but it also has to be an observable (otherwise the connection real-hermitian would not be so obvious) in complex fields you cannot measure the overall field but just real and imaginary part or such pairs, they must be hermitian.

6. Jul 23, 2010

### Rick89

anyway, some physicists are not so sure about the status of fields, I mean if you measure them. It is a part that leaves doubts. In fact for example the Dirac field is such that you can only measure quantities involving even multiples of it, and it isn't clear at all why. (It is required because there are minus signs popping out when you measure spinor fields at spacelike separated events.)

7. Jul 23, 2010

### ismaili

What we measured in experiments are n-point functions which are related to scattering amplitudes.
I still don't think we can measure the "fields" directly.

The Dirac field you mentioned is an example that we cannot observe the fields directly.
If we could measure the Dirac field directly, causality would be even broken.
Because, we have $$\{\psi_A(x) , \bar{\psi}_B(y)\} =0$$ at space-like separation,
thus in order to get the micro-causality condition $$[O(x) , O(y)] = 0$$ (at space-like separation, where $$O$$ is some observable), $$O$$ has to be composed by even number of Dirac fields. It's impossible to observe a single $\psi$ directly.

As for the positive- and negative-frequency expansion, take the complex scalar field as an example,
$$\phi(x) = \int\frac{d^3p}{(2\pi)^3\sqrt{2E_{\mathbf{p}}}} (a_{\mathbf{p}}e^{i\mathbf{p}\cdot\mathbf{x}} + b_{\mathbf{p}}^\dagger e^{-i\mathbf{p}\cdot\mathbf{x}})$$,
we have $$\phi^\dagger(x) \neq \phi(x)$$, so the complex scalar field is NOT hermitian.

So, I think the inclusion of negative-frequency components of the Fourier mode is not due to hermitian property of the fields. (One could only say so in the case of real scalar field.)
It's also not to make real part of $$\phi$$ to be real.
Because this is easily achieved without including of negative-frequency part.

Last edited: Jul 23, 2010
8. Jul 26, 2010

### ismaili

I came up with one of the reason why we need the negative frequency part which seems to be redundant in the Fourier transformation.
The answer is that the causality requires the inclusion of negative frequency part.
The introduction of negative frequency part makes $$[\phi(x) , \phi(y)]$$ vanish in space-like separation of space-time coordinates. Moreover, the operator attached to this negative part plane wave turns out to be the anti-particle.

9. Jul 29, 2010

### matonski

If you look a little later at equation 2.27, you can deduce that the Fourier Transform is:

$$\phi(\mathbf p, t) = \frac{1}{\sqrt{2\omega_{\mathbf p}}} \left(a_{\mathbf p} + a^\dagger_{-\mathbf p}\right)$$

I think the analogy with the harmonic oscillator still works because both $a^\dagger_{\mathbf p}\right)$ and $a^\dagger_{-\mathbf p}\right)$ create particles with the same frequency. The reason this combination is chosen is so that $\phi$ is hermitian.

10. Jul 29, 2010

### ismaili

Yes, Peskin is making analogy to harmonic oscillator.

But the real reason to choose that combination wouldn't be the hermitianity of $\phi$, it's the coincidence that we are considering a real scalar field $\phi$.
If we were considering a complex scalar field, we wouldn't have $$\phi(\mathbf p, t)$$ to be hermitian.

11. Jul 30, 2010

### matonski

But a real classical field corresponds to a hermitian quantum field, and this section is talking about $\phi$ as a quantum field. In other words, we want that combination of $a$'s and $a^\dagger$'s because we want $\phi$ to be a hermitian quantum field. However, we want $\phi$ to be a hermitian quantum field because we are quantizing a real classical field.

12. Jul 30, 2010

### ismaili

That's right. That's what I mean in the case of quantisation of a real scalar field.
In this case, $$\phi$$ is hermitian.

If we ask this question: why do we expand the complex scalar field in the same way? why do we expand the Dirac field in the same way? That is, expand the fields in terms of positive and negative frequency parts. These fields are not hermitian anymore. So, the inclusion of negative frequency modes definitely has nothing to do with hermitianity of fields in general.

13. Jul 30, 2010

### Fredrik

Staff Emeritus
The field breaks up into positive and negative frequency parts when you solve the equation. Look for solutions of the form $\phi(x)=T(t)X(x)Y(y)Z(z)$. (I hope the minor abuse of notation will be forgiven). This turns the equation into

$$\frac{T''(t)}{T(t)}-\frac{X''(x)}{X(x)}-\frac{Y''(y)}{Y(y)}-\frac{Z''(z)}{Z(z)}+m^2=0$$

Every term is explicitly independent of at least three of the four variables, and it's equal to something that's independent of the fourth, so it must be a constant. The equation

$$\frac{X''(x)}{X(x)}=-(p^1)^2$$

has solutions of the form

$$e^{\pm ip^1x}$$

and the Y and Z equations have solutions of the form

$$e^{\pm ip^2y},\ e^{\pm ip^3z}$$

while the T equation has solutions of the form

$$e^{\pm i\sqrt{\vec p\,^2+m^2}t$$

If we define $$E=\sqrt{\vec p\,^2+m^2}$$, the general (separable) solution is

$$\phi(x)=\int d^3p \left(A_+(\vec p\,)e^{iEt}e^{i\vec p\cdot \vec x}+A_-(\vec p\,)e^{-iEt}e^{i\vec p\cdot \vec x}\right)$$

Note that there's no need to include the exp(-ipx) solutions explicitly since the integration is over all $$\vec p\,\in\mathbb R^3$$, but the exp(-iEt) solution is included explicitly because we don't integrate over E. (That's what breaks up the solution into positive and negative frequency parts). E is a function of $$\vec p\,$$ and is defined so that E>0. Now if we write $p^0=E$, we get

(Edit: I changed a sign here after ismaili's comment below.)

$$\phi(x)=\int d^3p\left(A_+(\vec p\,)e^{ipx}+A_-(-\vec p\,)e^{-ipx}\right)$$

$$=\int d^3p\left(a(\vec p\,)e^{ipx}+b(\vec p\,)e^{-ipx}\right)$$

Now you can use $\operatorname{Im}\phi(x)=0$ to see how b is related to a.

Last edited: Jul 30, 2010
14. Jul 30, 2010

### ismaili

Wow, sensible!
This should be the correct answer.
Making use of this chance, allow me to clarify something about separation of variables.
Is the choice of constants in separation of variables here due to physical reason?
I mean, if we choose $$(p^1)^2$$ here instead, we have an exponential decay or increase solution.
How do we discard the choices of exponential decay or increase solutions?

I think the above equation has typo, and is corrected as above.
(If the above equation is going to connect with following equations reasonably.)

I also corrected some typo of the above equation.
(change integration variable $$\mathbf{p}\rightarrow \mathbf{-p}$$ doesn't create a minus sign).
And this solves my question actually, thank you.
The positive frequency and negative frequency part of the fields arise when we solve the field equations. Indeed, we need both positive and negative frequency parts to expand the general solution.

That's right. The real scalar field demands $$b$$ here to be $$a^\dagger$$.

Last edited: Jul 30, 2010
15. Jul 30, 2010

### Fredrik

Staff Emeritus
I don't know the answer to this question.

I was thinking that

$$Et+\vec p\cdot\vec x=p^0t^0+\sum_i p^ix^i=p_0t^0-p_ix^i=px$$

but perhaps I should define $$\vec p\cdot\vec x=p_ix^i$$ instead.

Oops, quite right.

16. Jul 30, 2010

### Demystifier

Such solutions diverge at plus or minus infinity. They seem not to be realized in nature, so we discard them because they do not seem to be physically relevant.

Try to do some calculations with such terms included and find by yourself various sorts of pathologies and infinities.

17. Jul 30, 2010

### matonski

In this section of Peskin and Schroeder, they are working in the Schrodinger picture. The variable $\mathbf p$ refers to just spatial frequency. I was responding to the original question which was regarding the specific argument given by Peskin and Schroeder. As stated, the original poster understands how the minus sign gets there through other arguments. He just doesn't understand this specific argument, which I'll try to summarize more clearly.

Given that the Fourier transform of the Klein Gordon equation is

$$(\frac{\partial^2}{\partial t^2}+|\mathbf p|^2+m^2)\phi(\mathbf p,t)=0$$

which is the same as the equation of motion for the simple harmonic oscillator of frequency

$$\omega_{\mathbf p} ^2= |\mathbf p|^2+m^2$$

The simple harmonic oscillator has the solution

$$\phi=\frac{1}{\sqrt{2\omega}}(a+a^\dagger)$$

Therefore, since the Fourier transform of $\phi(\mathbf x,t)$ is defined to be

$$\phi(\mathbf x,t)=\int\frac{d^3p}{(2\pi)^3}e^{i\mathbf p\cdot \mathbf x}\phi(\mathbf p,t)$$

Why isn't the solution

$$\phi(\mathbf x,t)=\int\frac{d^3p}{(2\pi)^3\sqrt{2\omega_{\mathbf p}}}e^{i\mathbf p\cdot \mathbf x}\left(a_{\mathbf p}+a_{\mathbf p}^\dagger\right)$$

$$\phi(\mathbf x,t)=\int\frac{d^3p}{(2\pi)^3\sqrt{2\omega_{\mathbf p}}}e^{i\mathbf p\cdot \mathbf x}\left(a_{\mathbf p}+a_{-\mathbf p}^\dagger\right)$$

Which is the one given in the book. Both solutions include the negative spatial frequencies.

Last edited: Jul 30, 2010
18. Jul 30, 2010

### matonski

But $\phi$ is an operator and not simply a number. The term real applies to numbers. The corresponding term for operators is hermitian. If $\phi$ were not hermitian (as in the case of charged fields), then $b$ would not have to equal $a^\dagger$.

19. Jul 30, 2010

### Fredrik

Staff Emeritus
By the way, here's another correct answer to the same question. I don't know why the books always start with a Fourier expansion when a straightforward solution by separation of variables is conceptually so much simpler, but since they do, I want to answer the question using an approach similar to theirs.

If we insert

$$\phi(x)=\int d^4p f(p)e^{ipx}$$

into

$$\left(\frac{\partial^2}{\partial t^2}-\nabla^2+m^2\right)\phi(x)=0$$

and define E=p0, we get

$$\int d^4p f(p)\left(-E^2+\vec p\,^2+m^2\right)e^{ipx}=0$$

This can only be true if f(p)=0 when the quantity in parentheses is ≠0, so f(p) can be written as

$$f(p)=g(\vec p\,)\left(\delta(E-\sqrt{\vec p\,^2+m^2})+\delta(E+\sqrt{\vec p\,^2+m^2})\right)$$

and this clearly breaks up the solution into two parts. I prefer the other approach because it doesn't involve any delta "functions" and doesn't even require the reader to understand Fourier transforms.

20. Jul 30, 2010

### ismaili

Thank you for the discussion!
And this clarified many details.

Yes, this is another correct viewpoint for the positive- and negative-frequency expansion of fields.
Now I think both methods are sensible because their origin is mass-shell condition.

From separation of variables, the choice of temporal constant is constrained, this means that in the expansion of a general solution into plane wave basis, we have to only sum over spatial momentum, and hence results in two possible choice of signs of exponents in temporal plane wave.

From the Fourier transform viewpoint, the EoM encodes mass-shell condition naturally, and this quadratic mass-shell condition has positive- and negative- energy solutions; where, if we insisted the energy to be positive, we say we get positive- and negative- frequency parts.

However, according to Peskin's logic, he worked in the Schodinger picture in the beginning, so the Fourier transform involves only spatial integral of momentum. So, the second viewpoint is somehow not obvious...