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Transcript
Module 36: Uncertainty relation Lecture 36: Uncertainty relation The Heisenberg’s uncertainty talks about the errors involved in the simultaneous measurement of two physical observables in Quantum mechanics. For example it say that the x components of the position and momentum cannot be measured simultaneously with infinite precision or accuracy. If the errors associated are denoted by ∆x and ∆px then ∆x∆px ≥ h̄/2. As shown in the previous chapter it would be true for any two physical variables if the corresponding quantum mechanical operators do not commute. Hence a very accurate measurement of one quauntity will destroy the accuracy in the other. This should not be thought as the incompetance of the observer or the weaknesses of the measuring instruments. One has to beleive that the Nautre is so. This principle can be used for calculating the ground state energies of various quantum mechanical systems. In this chapter we shall study some simple systems and estimate the energies associated with them. 36.1 Particle in a box We think of a box of siz L and a confine a quantum particle of mass m in it and would like to estimate the least enegy of the particle. According to classical mechanics the least energy is zero when the particle is at rest. But that is not possible in quantum mechanics since if the particle were at rest then its momentum would prcisely be zero and since it would be at rest its position could also be measured with infinite accuracy and that violates Heisenberg’s principle. Hence the particle is never at rest inside the box and since it moves it would have certain kinetic energy (in the absence of any potential inside the box). The least enrgy, permitted by the uncertainty, is the ground state energy of the system and is known as the ‘zero point energy’. We shall now try to estimate this energy using the uncertianty, ∆x∆px ∼ h̄. Since the particle is confined inside the box of length L then the maximum possible error in its position measurement cannot exceed the size of the box. Hence we have, ∆x = L. The maximum error in postion would introduce the minimum error in the momemtum measurement, so we have ∆px |min ∼ h̄/L 205 206 CHAPTER 36. UNCERTAINTY RELATION using the uncertainty relation. Now the actual momentum of the particle cannot be less than the error in the momentum, so the least momemtum possible for the particle is px ∼ h̄/L. This least momentum would produce the least energy of the particle, p2x h̄2 E= ∼ . 2m 2mL2 (36.1) When this problem is exactly solved using the Schrödinger equation one obh̄2 π 2 tains the ground state energy as 2mL 2 which is an order of magnitude larger. But the important result is that the simple calculation using uncertainty shows the right dependence of energy on the box size, L, which scales as L−2 . Now we can have some rough quantitative estimates of the energy in couple of situations. i) We can assume an electron trapped in an atom of size L. In this case we use m = 9.1 × 10−31 kg (mass of the electron) and L ∼ 1 A◦ (rough size of an atom). Substituting the value of h̄ = h/2π = 1.05 × 10−34 , we have h̄2 E =∼ ∼ 6.1 × 10−19 J ∼ 4eV. 2 2mL (36.2) We know that the ionisation potential (ground state) of the simplest atom (Hydrogen) is approximately 13.6 eV and it is of the same order of magnitude as that of our rough estimate. ii) The second example we take of a nucleus which has a size of the order of few fermis (1 fm= 10−15 m). We assume an alpha particle trapped in the nucleus and try to estimate its energy using the uncertainty principle. We roughly take the size of the nucleus (box), L = 5 × 10−15 m. The mass of the alpha particle is 2(mn + mp ) ∼ 4 × 1.67 × 10−27 kg. So the rough energy of the alpha inside the nucleus would be h̄2 E =∼ ∼ 0.33 × 10−13 J ∼ 0.2MeV. 2 2mL (36.3) Again we find that the very crude estimates made exploting the uncertainty principle gives us an order of magintude lower than that of typical alpha particle energies which are usually a few MeV. Of course one can check that a slight change in the size of the box would change this energy drasticlly due to the inverse sqare dependence. For example a nucleus of size 3 fm would make the alpha particle energy approximately 1 MeV. 36.2 The Hydrogen atom In the above cases we did not have any potential. We can make our calculations finer by incorporating the potential in the system. In the following we show 207 36.2. THE HYDROGEN ATOM one such example of one dimensional Hydrogen atom. The total energy of the electron in the Hydrogen atom is written as, E = K.E. + P.E. = p2 e2 − . 2m 4πǫ0 r (36.4) Now as argued earlier the maximum possible error in the position r ∼ ∆r and that corresponds to minimum error ∆p ∼ h̄/∆r. The minimum p ∼ ∆p, since it cannot be less than the error associated with it. Substituing these in (36.4) we get, h̄2 e2 E= . (36.5) − 2m(∆r)2 4πǫ0 ∆r Note that we have two competing terms in the energy expression, (36.5). Now for obtaining the least energy we have to optimise the expression (36.5) with ∂E = 0. Minimising the energy we have, respect to the error ∆r, i.e. ∂∆r ∂E h̄2 e2 =− + = 0. ∂∆r m(∆r)3 4πǫ0 (∆r)2 (36.6) The expression (36.6) finds the error ∆r which minimises the energy and which is equal to 4πǫ0 ⇒ ∆r = . (36.7) me2 When we evalute the ∆r by substituing the values of m, e etc. we find that it is equal to 0.528 Å and which is nothing but the Bohr radius defining the size of the Hydrogen atom. Substituting the value of ∆r from (36.7) in the expression of total energy (36.5) we get the ground state energy of the Hydrogen atom as, me4 E=− . (36.8) 2 × (4πǫ0 )2 h̄2 The expression obtained in (36.8) is the exact expression for the ground state energy of the Hydrogen atom. The negative sign shows that the system (electron-proton) is bound and one has to provide at least this much of energy to tear of the system. When evaluated its magnitude is approximately 13.6 eV and which is nothing but the ionisation potenstial of the Hydrogen atom. It is indeed surprising that by the above crude calculation we produce the exact answers to the Hydrogen atom problem which in reality needs a considerable amount of calculations for soving the Schrödinger equation to arrive at correct answers matching with the experimental values. Question: In fact we made two mistakes in the above calculations and by a sheer stroke of luck these two mistakes cancelled each other and produced the right result. The students are encouraged to point out these mistakes. The other uncertainty which has important consequences is the EnergyTime uncertainty, i.e. ∆E∆t ≥ h̄/2. (36.9) 208 CHAPTER 36. UNCERTAINTY RELATION One of its consequences is that the energy conservation could be violated for a very short spell of time. If one has infinite time to measure the energy of a system then there is no error in the energy measurement and the measurements are accurate. But suppose there is very little time available for measuring the energy ( as in the case of atomic transitions where the transitions last for only order of 10−8 secs) then there are large errors associated with these energy measurements as in the case of the atomic levels. So the atomic levels are not blade sharp but they are fuzzy with a mean value of the energy. Problems 1. Estimate the ground state energy of a quantum harmonic oscillator using uncertainty relation for which the total energy is given by, E= p2x 1 + kx2 , 2m 2 (36.10) where m is the mass q of the oscillator and k is the spring constant. [Ans. h̄ω = h̄ k/m. If you use the minimum uncertainty you would produce half the above value and which is again an exact result for the zero point energy of the quantum one dimensional harmonic ocillator.]