Chapter 4 Free Particle and Dirac Normalization

Chapter 4

Free particle and Dirac normalization

4.1 Fourier modes and the free particle

We have seen that under boundary conditions we obtain discrete values for some physical quantities as energy. Our interest focus now on a particle free to move over the whole space (which we shall consider it one-dimensional). In case in which there are not boundary conditions, the wave function does not depend on discrete values. The entire wave packet is instead described by Fourier integrals as 1 Z ∞ ψ(x) = dκ φ˜(κ)eiκx. (4.1.1) 2π −∞ We can notice that ψ(x) is a superposition of different waves ψ˜(x, k) = φ˜(κ)eiκx for different values of κ or momentum. Consider one of these wave functions. The larger φ˜(κ) is, for a given value of κ, the closer ψ(x) is described by a plane wave eiκx, which has a definite momentum p = ~κ. This should describe a particle with momentum very close to p if measured. For an infinite value of ψ(p), ψ(x) is given exactly by eipx/~.

This means that measuring a momentum p and getting the exact value of p0 (∆p = 0) implies that ψ(x) = eip0x/~, for which ∆x → ∞.

Therefore, we note that Heisenberg uncertainty is inclosed in Fourier decomposition.

Let us do the opposite by localizing a particle in a point x0 in space but by using waves with deﬁnite momentum. For that we need to add other Fourier components close to 2 CHAPTER 4. FREE PARTICLE AND DIRAC NORMALIZATION

momentum p0, such that they add at x = x0 but increase the total wave’s width. At the end, we obtain a wave packet localized in x = x0 but delocalized in momentum. This is ∆x → 0 implies ∆p → ∞. This is the free particle which is a solution of the Schr¨odingerequation

2 − ~ ψ(x)00 = Eψ(x), (4.1.2) 2m

q 2mE i 2 ipx/ which solution is given by ψ(x) = e ~ = e ~.

Now, we have seen that the probability to ﬁnd a particle in position between x and x ± dx is given by |ψ(x)|2dx. How can we compute the corresponding probability to ﬁnd a particle with momentum between p and p ± dp? This is given by the Parseval Theorem, which states that if

1 Z ∞ 1 Z ∞ ψ(x) = √ dκ φ˜(κ)eiκx = √ dp φ(p)eipx/~, (4.1.3) 2π −∞ 2π~ −∞ then Z ∞ Z ∞ dx|ψ(x)|2 = dp|φ(p)|2. (4.1.4) −∞ −∞ From this it follows that the probability to ﬁnd a particle with momentum width dp is given by |φ(p)|2. If ψ(x) is normalized such that the probability to ﬁnd a particle in the whole space equals to one, then φ(p) is automatically normalized as well.

Notice that a plane wave eip0x/~ resembles the motion of a classical

particle with momentum p0 for which we can not determine its

position. By a superposition of diﬀerent waves with a peak on p0, the position will be determined by the corresponding phases. A particle in motion is then just a bunch of waves for which the phases rotate and the position x changes accordingly.

This can be estimated as follows. Consider a bunch of ﬂat waves eipx/~ with diﬀerent values of momentum. The total wave packet is given by 1 Z ψ(x) = √ dpφ˜(p)eipx/~. (4.1.5) 2π~ 4.1. Fourier modes and the free particle 3

˜ For a momentum function φ(p) mainly localized around p0, we can estimate that Z ψ(x) ∼ φ˜(p) eipx/~ dp, (4.1.6) where assuming that the momentum p is constant within that range, i.e., p = c. Then we conclude that Z ψ(x) ∼ A · 2 eicx/~dp ∼ Ae˜ ipx/~, (4.1.7) − which indeed has a momentum uncertainty ∆p = 2 and the wave function ψ(x) is very close to a plane wave. The largest φ(p) is, the closest ψ(x) resembles a plane wave.. In the limit, for ∆p → 0 (by taking → 0), we expect that ∆x → ∞. How can we show that?

What we know by sure, is that Heinsenberg uncertainty is enclosed in Fourier decomposition. Schematically we can think about it: consider a wave packet centered around the position x0. This represents a particle localized in x0. There is an uncertainty in position and momentum. If we want to reduce the uncertainty in position

∆x, it is necessary to add more waves such that the pick around x0 becomes thinner but larger than before. That would imply the addition of waves with more diﬀerent values of momentum. The result is that ∆x is reduced but ∆p is enlarged.

We can also see that a plane wave is related to a free particle by directly solving Schr¨odingerequation. This reads,

2 − ~ ψ(x)00(x) = Eψ(x), (4.1.8) 2m which solution is given by

√ 2mE i x ip0x/ ψ(x) = e ~ = e ~, (4.1.9) for a particle with a momentum p0. The uncertainty in momentum vanishes, while in position is undetermined. In order to reduce ∆x we should increase ∆p by adding more Fourier components,

1 Z ∞ ψ(x) = √ dp φ(p) eipx/~, (4.1.10) 2π~ −∞ 4 CHAPTER 4. FREE PARTICLE AND DIRAC NORMALIZATION which is also a solution of Sch¨odingerequation. Notice that φ(p) is the identity function for a free particle with a ﬁxed-momentum p.

Now, we have learnt how to compute the probability to ﬁnd a particle between x and x + dx. This is given by |ψ(x)|2dx. Given such function, how can we compute the probability to ﬁnd a particle with some momentum range?

The answer comes from Parseval theorem, which states that if 1 Z ∞ ψ(x) = dp eipxφ(p), (4.1.11) 2π~ −∞ then Z ∞ Z ∞ |ψ(x)|2 dx = |φ(p)2 dp, (4.1.12) −∞ −∞ from which it is concluded that R |φ(p)|2dp = 1, and then that |φ(p)|2 dp measures the probability to ﬁnd a particle with momentum between p and p + dp.

4.1.1 Measuring uncertainties

Consider a particle described by the wave packet 4.1.11, such that position and momentum are determined upon some uncertainty ∆x and ∆p. By changing the phases such that at some time t = t0, the wave packet is centered around x0 and at t = t1 it is centered around x = x1. Such wave packet is interpreted a large scales as teh classical particle with momentum p. Notice that the change of phases modiﬁes the position coordinate around which the wave packet has a pick. Momentum continues to have the same uncertainty as far as the wave width in position does not change.

Observation: Measure uncertainty 6= Heisenberg uncertainty. This means that under “exactly” the same conditions, you can measure position and momentum with an experimental error given by ∆xexp, but get different values for both quantities at different times. Heisenberg uncertainty ∆x is the range of possible different values you measure, plus the corresponding experimental uncertainty. 4.2. Wave collapse 5

4.2 Wave collapse

Let us say that ψ(x) is the wave packet associated to some physical system on which we perform a measurement. Let us assume for the moment, a very concise case in which the wave packet is given by

X ψ(x) = Anψn(x), (4.2.1) n where ψn(x) is a wave packet with a pick around xn. assume as well that xi 6= xj for all i, j and that if there exists a position xf such that ψi(xf ) = ψj(xf ), then their value is very small. Therefore, the wavepacket 4.2.1 presents some maxima related to a larger probability to ﬁnd the particle in such position (actually there are n such picks). Now, let us think that an experiment is performed and we get that the particle was localized around (with an experimental error given by ∆xexp) xi. If we perform the experiment one more time, it is impossible that the particle could be localized in some region around another point diﬀerent that xi. We say that the wave function ψ(x) collapses to ψi(x) after measurement.

It should be clear that by softening the above conditions, we can get that the collapse of the wave packet is another packet conformed by a smaller number of components, i.e.,

P ψ(x) −→ j ψj(x), where the index j does not run over all possible values as at the beginning.

Maybe you are now questioning many things. One important question you must have in mind, is the possible implications on Heisenberg uncertainty of the collapse wave by measuring with some experimental uncertainty. In particular, let us think what happens if we measure a position with an experimental uncertainty smaller than the original uncertainty attributed to the particle.

Before the wave collapses, Heisenberg uncertainty is ∆x∆p ≥ ~/2. For the collapsed wave, the uncertainty bound remains, but now with respect to the new function, i.e., 6 CHAPTER 4. FREE PARTICLE AND DIRAC NORMALIZATION

∆˜x∆˜p ≥ ~/2. The uncertainty in the position after measuring, is given by the experimental uncertainty ∆xexp which is much smaller than ∆x, then

∆˜x∆˜p = ∆x ∆˜p ≥ ~, (4.2.2) exp 2 ⇒ ∆˜p ≥ ~ >> ~ = ∆p. (4.2.3) 2∆xexp 2∆x

Therefore, by reducing the uncertainty in position by measuring, we actually increase the uncertainty in momentum with respect to the original value.

4.2.1 Example: Let our brains be twisted!

Consider a wave packet passing through a wall with a single slit. We already saw that the uncertainty in position and momentum are given by ∆x = d and ∆p ≥ ~/2d where d is the slit aperture. Le t us say that, instead of placing a screen (which allow us to measure the diﬀraction pattern) we use a screen, with a same aperture d, at a distance L from the ﬁrst wall. Right after this second wall, we place a screen in order to measure the results.

The wave packet crossing through the second wall collapses, since momentum associated to the wave has “lost” many of its possible values. Actually, by placing the second wall very far away from the ﬁrst wall, i.e., L >> 1, uncertainty in momentum reduces drastically: ∆˜p << ∆p. As you can notice ∆˜p = ∆pexp where the experiment is the fact that we have put the second wall with a single slit. According to the previous argument, ∆˜p∆˜x ≥ ~/2 and

∆˜x ≥ ~ > ~ = ∆x. (4.2.4) 2∆˜p 2∆p

Then, it seems that by placing the second slit, we reduce the uncertainty in momentum and in consequence the uncertainty in x is greater than d. But such uncertainty refers to the uncertainty in position of a particle passing through the second slit. How can be this uncertainty larger than d, if both slits have the same aperture? 4.3. Dirac normalization 7

4.3 Dirac normalization

We already know that a wave function on position ψ(x) is related to its wave function on momentum by a Fourier transform. Using this, we can construct a “function” for the free particle with fixed momentum or fixed position1. Let us take a wave function ψ(x) and substitute the Fourier decomposition of φ(p) in terms of ψ(x), this is, 1 Z ψ(x) = √ dp φ(p)eipx/~ 2π~ 1 Z 0 = dp dx0 e−ip(x−x )/~ψ(x0), (4.3.1) 2π~ from which we conclude that the function ∞ 1 Z 0 δ(x, x0) = dp e−i(x−x )p/~, (4.3.2) 2π~ −∞ called Dirac function, satisfies the property, Z ψ(x) = dx0 δ(x − x0)ψ(x0). (4.3.3)

More properties are given to be proved in the exercise list No. 2. Finally, note that if we associate a Dirac function as the wave function of a particle, that means that it has a deﬁned position x0, and that ∆x = 0. The corresponding momentum wave function must have an undeﬁned uncertainty. Indeed,

1 Z e−ipx0/~ −ipx/~ φ(p) = √ dx δ(x − x0)e = √ , (4.3.4) 2π~ 2π~ which corresponds to a plane wave, for which ∆p → ∞.

1This is an example of the interplay between physics and mathematics. Physicists required the existence of a function for a free particle with ﬁxed momentum or position. However, the required properties it must satisfy were very strange, and certainly they did not correspond to the properties a function could fulﬁll. Mathematicians refused to accept the validity of quantum mechanics due to this reasons: physicists seemed to be faking the construction of new physics, based on unreal objects. However, later on, some mathematicians proved that these “functions” were not functions, but a new type of objects they called distributions. It turned out that everything physicists have done, was correct except the label “function”. Nowadays, Feynman measure on the path integrals, essential tool to compute interaction amplitudes in particle physics, has not proved to be correct from the point of view of mathematics.