The Feynman propagator of the scalar field

Transcription

1 The Feynman propagator of the scalar field Let s evaluate the following expression: 0 φ(x 1 )φ(x 2 ) 0 2E p1 d 3 p 2 2E p2 0 a p1 a p 2 0 e i(p 2 x 2 p 1 x 1 ) (1) Using the following relation, we get: 0 a p1 a p [a p1, a p 2 ] 0 (2π) 3 δ 3 ( p 1 p 2 ) (2) 0 φ(x 1 )φ(x 2 ) 0 d 3 p e ip (x 2 x 1 ) As we will see soon, it turns out that it is very useful to consider the time-ordered product T instead of the above one as follows: φ(x 1 )φ(x 2 ) if t 1 > t 2, T {φ(x 1 )φ(x 2 )} : φ(x 2 )φ(x 1 ) if t 1 < t 2. In other words, we arrange the operators in such a manner that the later time always comes to the left and the earlier time always comes to the right. Given this, we get: ( ) 0 T {φ(x 1 )φ(x 2 )} 0 (2π) 3 e ip (x 2 x 1 ) θ(t 1 t 2 ) + e ip (x 1 x 2 ) θ(t 2 t 1 ) ( ) (2π) 3 e i p ( x 1 x 2 ) e ie p(t 1 t 2 ) θ(t 1 t 2 ) + e i p ( x 1 x 2 ) e ie p(t 1 t 2 ) θ(t 2 t 1 ) (5) ( ) (2π) 3 e i p ( x 1 x 2 ) e ie p(t 1 t 2 ) θ(t 1 t 2 ) + e ie p(t 1 t 2 ) θ(t 2 t 1 ) (6) where in the last step, we took p p in the first term of (5), which leaves d 3 p unchanged. Now, we want to express the above quantity in terms of Fourier transformation in a Lorentz invariant way. To this end, consider the following integral, which you proved in our earlier article Application of residue theorem as an exercise: i dp 0 (t 1 t 2 ) 2π (p 0 ) 2 E p 2 + e ie p(t1 t2) θ(t 1 t 2 ) + e ie p(t1 t2) θ(t 2 t 1 ) (7) iɛeip0 (3) (4) 1

2 Figure 1: Time-ordered product Plugging this into (6), we get: 0 T {φ(x 1 )φ(x 2 )} 0 d 4 p i (2π) 4 (p 0 ) 2 p 2 m 2 + iɛ eie p(t 1 t 2 ) i p ( x 1 x 2 ) d 4 p i eip (x 2 x 1 ) (2π) 4 p 2 m 2 + iɛ (8) Depending on textbooks, physicists often use the following notations: D F (x 1, x 2 ) D F (x 1 x 2 ) D(x 1 x 2 ) F (x 1 x 2 ) 0 T {φ(x 1 )φ(x 2 )} 0 This is called Feynman propagator of scalar field and adding the small iɛ in the denominator in the momentum space Feynman propagator is called Feynman prescription. Now, let us explain why physicists usually consider time-ordered product instead of the unordered one. To this end, we will closely follow the quantum field theory textbook by Mandl and Shaw. First, what is the interpretation of 0 φ(x 1 )φ(x 2 ) 0? Recall from our earlier discussions in The scalar field as harmonic oscillators that 0 φ(x 1 ) is something like x 1 and φ(x 2 ) 0 is something like x 2. If you further recall our earlier discussions in Feynman path integral, you will remember x 1 x 2 is the transition amplitude. Therefore, 0 φ(x 1 )φ(x 2 ) 0 has an interpretation of the transition amplitude from x 2 to x 1. Now, let s apply time-ordered product. Then, 0 T {φ(x 1 )φ(x 2 )} 0 denotes the transition amplitude from x 2 to x 1 when t 2 < t 1. See Fig.1.(a). On the other hand, when t 1 > t 2 it denotes the transition amplitude from x 1 to x 2. To see how these transition amplitudes are used in actual physics, we consider nucleon-nucleon scattering. Nucleons experience force by exchanging virtual mesons one another. See Fig.2. (Mesons are scalar particles. We will soon explain what we mean by virtual. ) Before and after the collision, no mesons are present in the nucleons; they are created and annihilated upon the collision. Here, we consider the simplest process, namely, one meson exchange. As you can see in the figure, there are two cases. t 2 < t 1 and t 1 > t 2. 2

3 Figure 2: nucleon-nucleon scattering In the actual calculation, all the values of x 1 and x 2 are integrated over, meaning that the meson exchange can happen on any place and any time. Therefore, our division into two cases is meaningless. Moreover, if the distance between x 1 and x 2 is space-like the division into the two cases is not Lorentz invariant; depending on the frame of reference some will see t 2 < t 1 while others will see t 1 > t 2. These considerations show that we should consider the both cases together, making the use of time-ordered product. At this point, a careful reader may wonder how a meson, a massive particle, can propagate a space-like separation, which would mean that it travels faster than light. It can, because the meson is not real, but virtual. Our earlier assertion that nucleons exchange particles to experience force should not be taken too literally. Instead, everything comes from mathematical calculations, and the diagram like Fig.2. is only a pictorial way of representing these calculations. Actually, this digram is a Feynman diagram, which we introduced in our earlier article! Now, you will surely understand what we mean by our assertion that it is only a pictorial way of representing calculations. Let me elaborate what it means to be virtual for a particle. In Feynman diagram, there are two-types of line: external and internal. The both ends of an internal line are all connected to other lines, while one or both ends of an external line aren t. For example, nucleons in Fig.2. are external particles and they are real, while mesons in Fig.2. are internal particles and virtual, as they are created and destroyed in a short time. Real particle satisfies the mass-shell condition p 2 m 2 where m is the mass of the concerned particle, while the virtual particle doesn t satisfy this condition. Also, we will later see that p in (8) is interpreted as the momentum of the virtual particle. Apparently, it is integrated over all the values (i.e. d 4 p) and it is allowed that the denominator is non-zero. In our case in Fig.2, there is no integration, as the momentum of the virtual meson 3

4 is fixedly given by the momentum difference of the nucleon before and after. As we will see later, this happens as the amplitude corresponding to this particular Feynman diagram is calculated by integrating over all the values of x 1 and x 2, the Dirac-delta functions that enforce the momentum conservation at the points x 1 and x 2 arise. Don t worry if you don t understand. I didn t even explain what the precise amplitude for this Feynman diagram was. We will explain everything in details later. Finally, let s conclude this article with a comment on unequal time commutator of two φs. Let s consider [φ(x 1 ), φ(x 2 )] for (x 1 x 2 ) 2 < 0. In other words, the separation between x 1 ans x 2 is space-like. Now, observe that this commutator doesn t depend on the reference frame since φs are fully Lorentz-invariant quantities. Given this, notice that there exists a frame of reference in which one observes that x 1 and x 2 are at equal time due to their space-like separation. Remember also that equal time commutator of two φs is zero. Therefore, we conclude that the unequal time commutator of two φs with space-like separation is zero. Of course, this argument doesn t work for the ones with time-like separation. This conclusion has some consequences. Recall that, in quantum mechanics, observables are independent from one another if they commute. For example, you can simultaneously measure the y-component of the position of a particle and z-component of the momentum of the same particle, without any limits on their uncertainties, while same cannot be said about the y-component of the position and y-component of the momentum. Similarly, two φs with space-like separation are independent from one another as they commute. This makes sense, since they cannot affect one another; even light signal, the fastest thing, cannot connect them. Problem 1. Show that the virtual meson in Fig.2. doesn t satisfy the mass-shell condition. (Hint: Let the momentum of one of the two nucleons before the collision be p 1 and the one after be p 2. Then p 1 p 2 p is the momentum of the virtual meson. Let the mass of the nucleon be M, and the mass of the meson be m and show that p 2 (p 1 p 2 ) 2 0 which means p 2 m 2. The calculation can be simpler if you use the reference frame that observes p 1 (M, 0, 0, 0).) Problem 2. Let s consider the Feynman propagator in momentum representation as follows: D(p) d 4 xd(x)e ipx (9) Show D(p) i p 2 m 2 + iɛ Problem 3. Prove the following. (Hint: Use (8).) (10) ( x1 + m 2 )D(x 1 x 2 ) iδ 4 (x 1 x 2 ) (11) 4

5 Thus, the Feynman propagator can be interpreted as Green s function. 5