### D.46 De­riva­tion of the WKB ap­prox­i­ma­tion

The pur­pose in this note is to de­rive an ap­prox­i­mate so­lu­tion to the Hamil­ton­ian eigen­value prob­lem

where the clas­si­cal mo­men­tum is a known func­tion for given en­ergy. The ap­prox­i­ma­tion is to be valid when the val­ues of are large. In quan­tum terms, you can think of that as due to an en­ergy that is macro­scop­i­cally large. But to do the math­e­mat­ics, it is eas­ier to take a macro­scopic point of view; in macro­scopic terms, is large be­cause Planck’s con­stant is so small.

Since ei­ther way is a large quan­tity, for the left hand side of the Hamil­ton­ian eigen­value prob­lem above to bal­ance the right hand side, the wave func­tion must vary rapidly with po­si­tion. Some­thing that varies rapidly and non­triv­ially with po­si­tion tends to be hard to an­a­lyze, so it turns out to be a good idea to write the wave func­tion as an ex­po­nen­tial,

and then ap­prox­i­mate the ar­gu­ment of that ex­po­nen­tial.

To do so, first the equa­tion for will be needed. Tak­ing de­riv­a­tives of us­ing the chain rule gives in terms of

Then plug­ging and its sec­ond de­riv­a­tive above into the Hamil­ton­ian eigen­value prob­lem and clean­ing up gives:
 (D.30)

For a given en­ergy, will de­pend on both what is and what is. Now, since is small, math­e­mat­i­cally it sim­pli­fies things if you ex­pand in a power se­ries with re­spect to :

You can think of this as writ­ing as a Tay­lor se­ries in . The co­ef­fi­cients will de­pend on . Since is small, the con­tri­bu­tion of and fur­ther terms to is small and can be ig­nored; only and will need to be fig­ured out.

Plug­ging the power se­ries into the equa­tion for pro­duces

where primes de­note -​de­riv­a­tives and the dots stand for pow­ers of greater than that will not be needed. Now for two power se­ries to be equal, the co­ef­fi­cients of each in­di­vid­ual power must be equal. In par­tic­u­lar, the co­ef­fi­cients of 1 must be equal, , so there are two pos­si­ble so­lu­tions

For the co­ef­fi­cients of 1 to be equal, , or plug­ging in the so­lu­tion for ,

It fol­lows that the -​de­riv­a­tive of is given by

and in­te­grat­ing gives as

where is an in­te­gra­tion con­stant. Fi­nally, now gives the two terms in the WKB so­lu­tion, one for each pos­si­ble sign, with equal to the con­stant or .