# UCD PHY 230A - Klein-Gordon Equation (2 pages)

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**View the full content.**## Klein-Gordon Equation

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## Klein-Gordon Equation

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- Pages:
- 2
- School:
- University of California, Davis
- Course:
- Phy 230a - Quantum Theory of Fields

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3 Klein Gordon Equation In non relativistic mechanics the energy for a free particle is E p2 2m 35 To get quantum mechanics we make the following substitutions E i t p i 36 and the Schro dinger equation for a free particle is 2 2 i 2m t 37 In relativistic mechanics the energy of a free particle is p E p 2 c2 m 2 c4 38 Making the same substitution we obtain 2 c2 2 m2 c2 i t 39 It s difficult to interprete the operator on the left hand side so instead we try E 2 p 2 c2 m 2 c4 2 2 c2 2 m2 c4 i t 2 1 m2 c 2 2 or 2 2 2 c t where 1 2 2 c t 2 2 40 41 42 43 Plane wave solutions are readily found by inspection 1 i i exp p x exp Et 44 V p where E 2 p2 c2 m2 c4 and thus E p2 c2 m2 c4 Note that there is a negative energy solution as well as a positive energy solution for each value of p Na vely one should just discard the negative energy solution For a free particle in a positive energy state there is no mechanism for it to make a transition to 6 the negative energy state However if there is some external potential the KleinGordon equation is then altered by the usual replacements e 45 E E e p p A c e i t e 2 c2 i A 2 m2 c4 46 c The solution can always be expressed as a superposition of free particle solutions provided that the latter form a complete set They from a complete set only if the negative energy components are retained so they cannot be simply discarded Recall the probability density and current in Schro dinger equation If we multiply the Schro dinger equation by on the left and multiply the conjugate of the Schro dinger equation by and then take the difference we obtain 2 2 2 i 2m 2 i 2m t Using s js continuity 2mi 47 we then obtain the equation of s js 0 48 t Now we can carry out the same procedure for the free particle Klein Gordon equation m2 c 2 2 m2 c 2 49 2 Taking the difference we obtain 2 2 psi 0 50 This suggests that we can define a probability 4 current j where is a constant 51 and it s conserved j 0 j j 0 j To make j agree with js is chosen to be 2mi So i

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