The Klein Gordon Equation

Transcription

1 December 30, :35 PM 1. Derivation Let s try to write down the correct relativistic equation of motion for a single particle and then quantize as usual. a. So take (canonical momentum) The Schrödinger equation is now given by and then quantize using b. To get some non-trivial results, introduce an external EM field (we can let e 0 to get a free particle later). The classical equation of motion is which is the Lorentz force (1). To write this equation covariantly, note that (1) Note that the LHS of eqn involves the kinetic momentum ; we reserve for the canonical momentum, which enters QM commutation relations. For a discussion see Goldstein, Classical Mechanics, section 6-4. M. Gericke Phys 7560, Relativistic QM 17

2 January 15, :38 PM and Thus, the spatial component reproduce the equation of motion 1.13, while the time component implies which just says that only electric fields do work (1). So we can combine eqns and 1.15 into which is the covariant equation of motion. (1) One can prove the power law from the eqn. of motion 1.13: Note that here, E (or m ) is the total energy, including the rest-mass m o. M. Gericke Phys 7560, Relativistic QM 18

3 January 15, :52 PM c. Can we derive this result from a relativistic version of Hamilton s equations of motion? Define a Lorentz scalar and write a covariant form of Hamilton s equations as Note that is not the usual hamiltonian H ; we must connect to H later. For now, let s just find an that gives us the equations of motion (1.16). We ll relate eqn to the usual Hamilton s equations later. Claim: We can reproduce eqn by choosing which is the kinetic momentum, in distinction to the canonical momentum p. Which, by equation 1.17, is equal to M. Gericke Phys 7560, Relativistic QM 19

4 January 15, :36 PM But Plug this into eqn. 1.20: So the equations 1.17 and 1.18 reproduce the equation of motion But if a particle obeys the eqns. of motion, then is determined: Note that M. Gericke Phys 7560, Relativistic QM 20

5 January 15, :53 PM Now let and allow the resulting operator to act on the wavefunction, just as we would do non-relativistically. This is the Klein-Gordon equation. Unfortunately, this is 2nd order in the time derivatives. We would like to go back to our old friend, a first-order Schrödinger equation. Connection to the Familiar Hamilton s Equations and definition of H We have again equations 1.17 Let s identify po = H and show that we get the familiar equations of motion. Thus, po = H = H(x,p), while is a constant of the motion. From which it follows that and M. Gericke Phys 7560, Relativistic QM 21

6 12:06 AM But So These are Hamilton s equations. Therefore, from eqn. 1.21, we have Which is the total energy with an electro-static potential. M. Gericke Phys 7560, Relativistic QM 22

7 12:21 AM Discussion i. We could have started from equation 1.25 and worked backwards to prove that we get the covariant equation of motion Since the information content of equation 1.17 is that same as that of eqns. 1.24, our procedure justifies that the Klein-Gordon equation (1.22) is consistent with all the classical dynamics (1). ii. H is hermitian if, but 1) 2) 3) 4) The square root leads to an infinite number of derivatives upon expansion. Which means that we have all powers of acting on. So the theory is non-local. It s very hard to make such theories consistent with causality since, if you want to know the properties of a particle described by eqn at a given point, you have to know it s properties everywhere else. We can t write a simple differential equation for with this H. We could, however, work in momentum space and define things by Fourier transforms. We can t write a first order continuity equation. The theory can be made consistent for free particles (e = 0), since stationary states are eigenfunctions of p, but as soon as you put the particle in a sizable external field A, you re in trouble. iii. Conclusion: Our attempt to extend the Schrödinger equation to the relativistic regime leads immediately to severe problems (2). (1) The relation is simply a way to guarantee that is a Lorentz scalar: The relationship between and p o or ( H ) is the inverse of the relationship between and t. Thus, the eight equations contained in 1.17 embody the dynamics of the seven Hamilton s equations (1.24) (with the definition of p o = H ) plus the input that is a Lorentz scalar. (2) Moreover, there is no justification for just squaring eqn if you don t believe that 1.25 gives a good wave equation. M. Gericke Phys 7560, Relativistic QM 23

8 2:11 PM 2. Probability density and the failure of the Klein-Gordon equation So let s return to the Klein-Gordon Equation and see how far we can get. Rewrite equation 1.22 as Is there a quantity that satisfies the continuity equation, that we can interpret as a probability density? The most general solution to equation 1.26 can always be written as a superposition of plane waves and therefore takes the form Where So using this and taking and So the continuity equation is M. Gericke Phys 7560, Relativistic QM 24

9 10:31 PM The last equality simply uses the K-G equation (1.22). So this means that eqn represents a conserved current, with a probability density given by: However, we can see immediately that this probability density is not positive definite. Since eqn is a second order equation in time, we can (and must) specify and independently. In fact, can be positive at some points and negative at others. This is not a good probability density. We still have problems. The non-relativistic limit What happens in the non-relativistic limit? Shouldn t things reduce to well-known results and still work? Consider a stationary state Then M. Gericke Phys 7560, Relativistic QM 25

10 10:54 PM Thus, as long as the field (potential) A o is small compared to the overall energy (restmass plus kinetic) E, will have a definite sign and it would certainly appear that for free particles (e = 0), we can get a probabilistic interpretation. However, even for free particles the second order equation (1.26) admits solutions with and we are still in trouble with negative energy solutions. The only hope, for now, is to try and interpret the theory by letting j es, where j is a charge and current density (Pauli and Weisskopf 1934). Then, for small fields, we can interpret the signs of as corresponding to particles of opposite charge. However, the appearance of both charges in a supposedly single particle theory is mysterious and inconsistent. So the charge density may be meaningful, but it s still not clear what we are doing! 3. Free-particle solutions and anti-particles The interpretation of j es as a conserved electromagnetic current gives sensible results in the non-relativistic limit, aside from the ever present negative energy terms. Let s take this idea seriously, demanding that and get (1). be continuous and see how far we For free particles eqn reads (1) An excellent discussion of the interpretation of the wave mechanics of the Klein-Gordon equation is contained in the review article by H. Feshbach and F.Villars (Rev. of Mod. Phys. 30 (1958) 24). It s worth reading! (see also Bjorken and Drell, section 9.7) M. Gericke Phys 7560, Relativistic QM 26

11 11:01 PM We look for stationary states producing The solutions to which are plane waves: Which produces the result What are we going to do with the negative energy solutions? We must keep them if we want a complete set of solutions to eqn We want to get an idea about a possible answer to this question from the current. Take E = ± k : M. Gericke Phys 7560, Relativistic QM 27

12 11:07 PM As far as j and are concerned, letting k - k is exactly the same as letting Which means that This property follows from an invariance of the K-G equation that holds even if we put back the E&M fields. Recall(1) Where the are negative energy solutions with E = - k (stationary states). Now take the complex conjugate of both sides and let e -e : (1) Here, we look for stationary states, like scattering states, where k and E = k are fixed, and the solutions have particular boundary conditions. M. Gericke Phys 7560, Relativistic QM 28

13 11:15 PM Thus, the states charge +e. satisfy the same equation as a solution with E = + k and We can call this solution(1) (for stationary state solutions) Note that the charge and current densities are also invariant under this transformation(2). Thus, the K-G equation is invariant under a charge conjugation transformation. (1) Note that although c and satisfy the same differential equation, they are not the same solution. This is because they satisfy different boundary conditions. To see this, set e = 0 (free particles), so that as defined above. Thus, has momentum opposite to that of, as we ve just seen. In the general case (e 0 ) for scattering states, this implies different boundary conditions for the two solutions. The situation for bound bound states is not clear, although it would appear that (2) To lowest order in e: One can verify that the full expressions for and j (to all orders in e) also satisfy these relations. M. Gericke Phys 7560, Relativistic QM 29

14 11:36 PM This means that the form of the K-G equation (1.33) implies that we must simultaneously consider particles that are identical except for their sign in charge. Thus, the combination of relativistic covariance and quantum mechanics leads to the prediction of antiparticles. This also allows us the interpret the complex conjugate of the negative energy solutions for particles with the opposite charge. Since eqn implies Both sides satisfy the same equations and give the same current density j ; I.e. they are the same. But, although this allows us to interpret the negative energy solutions, we still have problems with consistency, since the single particle equation (1.33) apparently describes two species of particles simultaneously! Nevertheless, things should at least make sense for weak fields. So let s push on and see what happens. (Bjorken and Drell section 9.7 makes this discussion precise.) 4. Electromagnetic Gauge Invariance Note that is invariant under the transformation where is a scalar function. This is called a gauge transformation ; Since it doesn t change the E and B fields, it can t affect the physics. However, our quantum wave equation is written in terms of A : M. Gericke Phys 7560, Relativistic QM 30

15 11:43 PM We can make this equation gauge invariant if we demand that for every gauge transformation on A, we change the phase of (x) at every point in space-time: Then This phase transformation of the wave-function remarkably enough, produces no physical effects. To see that, we can check whether or not the transformation leaves the current density (eqn. 1.31) invariant: M. Gericke Phys 7560, Relativistic QM 31

16 11:50 PM Alternatively, we could say that we get the correct wave equation (1.46) by demanding gauge invariance when This necessitates with A transforming as above. M. Gericke Phys 7560, Relativistic QM 32