I have just begun studying quantum field theory and am following the book by Peskin and Schroeder for that.
So while quantising the Klein Gordon field, we Fourier expand the field and then work only in the momentum space. What is the need for this expansion?
Quantum Field Theory – Why Use Fourier Expansion?
fourier transformklein-gordon-equationquantum-field-theory
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Your notion of "my previous conception of a particle to be something localized in space" is a classical (non- quantum) conception.
So, in Quantum Mechanics, it it a false conception. And it is the same thing in Quantum Field Theory.
In Quantum Field Theory (as in Quantum Mechanics), you are working with operators. The most natural way - because of the relation with the quantum harmonic oscillator - is to use the impulsion-time representation for these operators : $a(\vec k,t) = a(\vec k) \, e^{i k_0t} , \, a^+(\vec k,t) = a^+(\vec k) \, e^{-i k_0t}$, with $k_0^2 = \vec k^2 + m^2$
But you may use the position representation also :
$$A(\vec x, t) = \int d^4k \,\delta(k^2 - m^2) \, (a(\vec k,t) \, e^{i \vec k.\vec x}+ a^+(\vec k,t) \, e^{-i \vec k.\vec x})$$
$$A(\vec x, t) = \int d^4k \,\delta(k^2 - m^2) \, (a(\vec k) \, e^{i k.x}+ a^+( \vec k) \, e^{-i k.x})$$
So you must be very careful when speaking about "particles". In Quantum field theories, we are working with operators whose eigenvalues are fields (as in Quantum mechanics, the eigenvalues of the Position operator are the only possible measurable positions)
Of corse, instead of speaking about operators, you can speak about states, for instance, the state $|\vec k> = a^+(\vec k)|0>$. But it is a 1-particle state, it is not an operator, and it is not a multi-particle state.
You can also build a "wave-packet" state by considering some particular linear combination of 1-particle states, for instance :
$$ |Wave Packet> = \int d^4 k \,\delta(k^2 - m^2) \, f(\vec k) \, |\vec k>$$
And, if you choose correctly the function $f$, this state could be localized in a precise region of space-time.
Rule number 1: we quantize fields after we transition to the canonical picture. So, we start with the Lagrangian for the Klein-Gordon field ($c=1$, $\hbar=1$): $$L = \int \mathrm{d}^3 x \left[\frac{1}{2}\dot\phi^2(\mathbf{x}) - \frac{1}{2}[\nabla\phi(\mathbf{x})]^2 - \frac{m^2}{2}\phi^2(\mathbf{x})\right]$$ We immediately transform into the classical canonical picture, getting our canonically conjugate momentum to $\phi(\mathbf{x})$ from the definition \begin{align} \pi(\mathbf{x}) & \equiv \frac{\delta L}{\delta \dot\phi(\mathbf{x})} \\ & = \dot\phi(\mathbf{x}). \end{align} From there we can transition to our classical Hamiltonian $$ H = \int \mathrm{d}^3 x\left[\frac{1}{2}\pi^2(\mathbf{x}) + \frac{1}{2}[\nabla\phi(\mathbf{x})]^2 + \frac{m^2}{2}\phi^2(\mathbf{x}) \right] $$
Now that we have our non-commuting variables, we can quantize by imposing the equal time commutation relations, \begin{align} [\phi(\mathbf{x}),\phi(\mathbf{y})] & = [\pi(\mathbf{x}),\pi(\mathbf{y})] = 0 \\ [\phi(\mathbf{x}),\pi(\mathbf{y})] & = i \delta(\mathbf{x}-\mathbf{y}). \end{align}
Now, we want to change from real space to mode space. There are multiple ways to do this that have varying degrees of complexity associated with them.
- Fourier transform: pros - momentum eigenfunctions; cons - our operators aren't Hermitian anymore.
- Harteley transform: pros - operators are Hermitian, no constraints on real/imaginary parts; cons - derivatives aren't quite as simple anymore.
- Sine/Cosine transform: pros - operators are Hermitian; cons - you're basically inspecting the guts of the Fourier transform, making a lot of extra work. Sometimes it's needed, though.
Because it's the lingua franca, we'll do the Fourier transform here. Perform the substitution $\phi(\mathbf{x}) = \int \mathrm{d}^3 k \phi(\mathbf{k}) \frac{e^{i\mathbf{k}\cdot\mathbf{x}}}{(2\pi)^{3/2}}$, and similar for $\pi$, you'll find that \begin{align} H & = \int \mathrm{d}^3 k\left[\frac{1}{2}\pi(-\mathbf{k}) \pi(\mathbf{k}) + \frac{\mathbf{k}^2 + m^2}{2}\phi(-\mathbf{k}) \phi(\mathbf{k}) \right], \\ [\phi(\mathbf{k}),\phi(\mathbf{k}')] & = [\pi(\mathbf{k}),\pi(\mathbf{k}')] = 0, \text{ and} \\ [\phi(\mathbf{k}),\pi(\mathbf{k}')] & = i \delta(\mathbf{k}+\mathbf{k}'). \end{align} To get the above into a more familiar form, you'll have to use the fact that $\phi^\dagger(\mathbf{x}) = \phi(\mathbf{x})\Rightarrow \phi^\dagger(\mathbf{k}) = \phi(-\mathbf{k}).$ Leading to \begin{align} H & = \int \mathrm{d}^3 k\left[\frac{1}{2}\pi^\dagger(\mathbf{k}) \pi(\mathbf{k}) + \frac{\mathbf{k}^2 + m^2}{2}\phi^\dagger(\mathbf{k}) \phi(\mathbf{k}) \right], \\ [\phi(\mathbf{k}),\phi(\mathbf{k}')] & = [\pi(\mathbf{k}),\pi(\mathbf{k}')] = 0, \text{ and} \\ [\phi(\mathbf{k}),\pi^\dagger(\mathbf{k}')] & = i \delta(\mathbf{k}-\mathbf{k}'). \end{align}
This implies that $\phi$ is canonically conjugate to $\pi^\dagger$ not $\pi$. This causes a conundrum: how do we build our ladder operators? For the simple harmonic oscillator we had $a = \sqrt{\frac{m\omega}{2}} \left(x + \frac{i}{m\omega} p\right)$. Which do we use \begin{align} a(\mathbf{k}) & = \sqrt{\frac{\omega}{2}} \left(\phi(\mathbf{k}) + \frac{i}{\omega} \pi^\dagger(\mathbf{k})\right) \text{ or} \tag1\\ a(\mathbf{k}) & = \sqrt{\frac{\omega}{2}} \left(\phi(\mathbf{k}) + \frac{i}{\omega} \pi(\mathbf{k})\right)? \tag2 \end{align} The answer is: it depends on what you care about, because those are both valid in one of the mode pictures available to us.
To see why, it's useful to dig in to the guts. The Fourier transform can be written in terms of the cosine and sine transforms to give $$ \phi(\mathbf{k}) = \phi_+(\mathbf{k}) + i \phi_-(\mathbf{k}), $$ Note that the cosine transform produces an even function, and the sine transform an odd one, and that's why I chose the subscript plus and minuses. Interestingly, the constraints on these functions affect their commutation relations. The nontrivial ones are: $$ [\phi_{\pm}(\mathbf{k}),\pi_{\pm}(\mathbf{k}')] = \frac{i}{2}\delta(\mathbf{k}-\mathbf{k}'). $$ The Harteley transform is then $\phi_H(\mathbf{k}) = \phi_+(\mathbf{k}) + \phi_-(\mathbf{k}).$
Now, without a doubt we can define ladder operators for the sine and cosine transforms, since they're Hermitian \begin{align} a_\pm(\mathbf{k}) & = \sqrt{\frac{\omega}{2}} \left(\phi_\pm(\mathbf{k}) + \frac{i}{\omega} \pi_\pm(\mathbf{k})\right) \text{ with} \\ [a_\pm(\mathbf{k}),a_\pm(\mathbf{k}')] & = \frac{1}{2}\delta(\mathbf{k}-\mathbf{k}'). \end{align} Note that each $a_\pm$ is even or odd just like the operators it's built from. Similarly, the ladder operator for the Harteley mode works just fine $a_H(\mathbf{k}) = \sqrt{\frac{\omega}{2}} \left(\phi_H(\mathbf{k}) + \frac{i}{\omega} \pi_H(\mathbf{k})\right) \Rightarrow [a_H(\mathbf{k}),a_H(\mathbf{k}')] =\delta(\mathbf{k}-\mathbf{k}').$
The reason for defining the ladder operators on the standing waves is because it allows us to inspect what (1) and (2) are doing (you can do a change of variables in the integral to show that (1) is what Peskin & Schroder are using). In terms of the standing mode ladder operators $$a_P(\mathbf{k}) = a_+(\mathbf{k}) + i a_-^\dagger(\mathbf{k}).$$ In other words, this operator cannot be rightly called either a creation or annihilation operator - it does both. If you really wanted to annihilate a traveling wave, you'd build your operator as $$a_T(\mathbf{k}) = a_+(\mathbf{k}) \pm i a_-(\mathbf{k}),$$ where, here the $\pm$ here is intended to convey uncertainty over the correct choice (the plus choice is identical to (2), by the way).
Why choose (1), then? I haven't worked it out, myself, but two possibilities occur to me:
- the commutator of $[a_P,a_P^\dagger]$ looks more like the SHO $[a,a^\dagger]=1$, or
- the number density can be more compactly written in terms of $a_P$ than $a_T$.
Best Answer
Free equations are linear, so exponentials are their solutions. Thus we construct a linear superposition of exponentials to embrace a general case.
Interactions are supposed to change amplitudes of particular waves in these superpositions.