Orbital Angular Momentum Eigenfunctions

Transcription

1 Obital Angula Moentu Eigenfunctions Michael Fowle 1/11/08 Intoduction In the last lectue we established that the opeatos J Jz have a coon set of eigenkets j J j = j( j+ 1 ) j Jz j = j whee j ae integes o half odd integes and we found the atix eleents of J+ J (and hence those of J x J y ) between these eigenkets. This puely foal stuctue theefoe nails down the allowed values of total angula oentu and of any easued coponent. But thee ae othe things we need to know: fo exaple how is an electon in a paticula angula oentu state in an ato affected by an extenal field? To copute that we need to know the wave function ψ. If a syste has spheical syety such as an electon in the Coulob field of a hydogen nucleus then the Hailtonian H and the opeatos J Jz have a coon set of eigenkets E j. The spheically syetic Hailtonian is unchanged by otation so ust coute with any otation opeato H J = 0 and [ H J z ] = 0. Recall that couting Heitian opeatos can be diagonalized siultaneously and theefoe have a coon set of eigenkets. Fotunately any systes of inteest do have spheical syety at least to a good appoxiation the basic exaple of couse being the hydogen ato so the natual set of basis states is the coon eigenkets of enegy and angula oentu. It tuns out that even when the spheical syety is boken the angula oentu eigenkets ay still be a useful stating point with the syety beaking teated using petubation theoy. Two-Diensional Models As a wa-up execise fo the coplications of the thee-diensional spheically syetic odel it is woth analyzing a two-diensional ciculaly syetic odel that is ψ Hψ ( x y) = + ψ ( x y) + V ( x + y ) ψ x y = E x y. M x y (In this section we ll denote the paticle ass by M to avoid confusion with the angula oentu quantu nube but be waned you ae often going to find used fo both in the sae discussion!) The two-diensional angula oentu opeato is L= p = xpy ypx = i x y. y x It is a staightfowad execise to check that fo the ciculaly-syetic Hailtonian above

2 [ ] H L = 0. Execise: check this. To take advantage of the cicula syety we switch to cicula vaiables ( φ ) 1 = x + y = y x x= y= φ tan / so cos φ sin φ. Tansfoing the Hailtonian and angula oentu into ( φ ) coodinates whee and 1 1 Hψ φ ψ φ V ψ E M φ ( ) = + + ( ) + ( φ) = ψ ( φ ) Execise: check these esults! L= i. φ The angula oentu eigenfunctions ψ φ satisfy Lψ φ i ψ φ ψ φ ( ) = ( ) = ( φ) equivalent to L ψ R ( ) e iφ φ = and fo this to be a single-valued wave function ust be an intege. (This also ensues the heiticity of the opeato the integation-by-pats check has canceling contibutions fo φ = 0 and φ = π. ) =. So i Notice this eans that any function of ultiplied by e φ is an eigenfunction of angula oentu with eigenvalue and in fact any eigenfunction of L with eigenvalue ust be of this fo. So we can facto out the -dependence and wite a coplete set of othonoal eigenfunctions of L noalized by integating aound the cicle: ( φ) Φ = 1 π i e φ φ an intege. It is inteesting to note that this would be a coplete set of wave functions fo a paticle confined to a ing athe like the oiginal Boh obits. In fact nanotech ings in which electons have wave functions like this can now be anufactued. Note also that in such ings one can also have eal wave functions 1/ π sin φ 1/ π cosφ which ae still enegy eigenstates but not

3 3 angula oentu eigenstates since they ae standing waves linea supepositions of waves going aound the ing in opposite diections. The coon eigenstates of the Hailtonian and the angula oentu evidently have the fo E = ψ φ = R Φ φ. E E We should ephasize that although the angula pat of the wave function does not depend on the R does depend on the angula oentu. This adial potential the adial coponent E becoes obvious on putting this ψ ( φ ) into the ( ) E φ vesion of Schödinge s equation R V R ER Φ + Φ = Φ M φ ( φ ) ( φ) ( φ) E E E noting that / φ and canceling out the coon facto Φ ( φ ) to give d 1 d + R E + V RE = ERE M d d E the angula oentu te / M = L /M evidently is equivalent to a epulsive potential. It s called the centifugal baie and is easy to undestand fo classical echanics. To see this conside a classical paticle bound (in two diensions) by an attactive cental foce V(). Split the oentu into a adial coponent p and a coponent in the diection pependicula to the In this one-diensional equation fo the adial wave function R adius enegy p. The angula oentu L= p and is constant (since the foce is cental). The p p p L E = + + V = + +V M M M M substituting p = L/. Since L= the angula pat is exactly equivalent to the above Schödinge equation. But what about the adial pat? Why isn t p just equal to i / and p equal to /? We know the oe coplicated diffeentiation with espect to in the Schödinge equation above ust be coect because it cae fo / x + / y and = x + y 1 φ = tan y/ x..

4 To see why p equal to i / 4 is incoect even though it satisfies [ p ] = i ecall what happens in x-space. We agued thee that p = i / x fo a plane wave because fo the x ipx x / photon analogy acting on the plane wave state Ce this opeato gives the ate of change of phase and theefoe the oentu. But a adial wave is a little diffeent: pictue a photon wave coing out of a single naow slit that is a slit having width fa salle than the photon wavelength. The photon wave will adiate outwads with equal aplitude in all diections (180 ) but the wave aplitude will decease with distance fo the slit to conseve pobability. Fo a long (naow) slit this is essentially a two-diensional poble so the wave function will be ip / ψ Ce /. We know that if we easue the oentu of photons at diffeent distances fo the slit we ll get the sae esult. The wavelength deteines the photon s oentu and it isn t changing. The colo stays the sae. Howeve i / opeating on ψ doesn t just give p : it picks up an exta te fo diffeentiating the so it is obviously not giving us the ight oentu. Fotunately this is easy to fix: we define the opeato pˆ 1 = i + which eliinates the exta te and still satisfies [ p ] = i. Howeve thee is still a sall poble. If we substitute this pˆ in the classical expession fo the enegy following the pocedue we used successfully to find Schödinge s equation in Catesian coodinates we find H p L = + + V M M 1 + L = + + V M M L = + M M + V This is alost but not quite the sae as the equation we found by tansfoing fo Catesian coodinates. The diffeence is the te /8M. So which is ight? Actually ou fist one was ight this second one deived diectly fo the classical Hailtonian does give the sae esult in the classical liit because the diffeence between the vanishes fo 0. We conclude that beginning with the classical Hailtonian and eplacing dynaical vaiables with the appopiate quantu opeatos cannot guaantee that we get the coect quantu Hailtonian: it ight be off by soe te of ode. This would becoe evident in pedicting

5 5 popeties of tuly quantu systes such as atoic enegy levels. Pobles of this kind ae coon in constucting quantu theoies stating fo a classical theoy: essentially in a classical theoy the ode of vaiables in an expession is ielevant but in the quantu theoy thee can only be one coect ode of noncouting vaiables such as / and in any expession. What can we say about the adial wave function RE the oigin ae finite then fo sall E o be discontinuous so R A? If both the enegy and the potential at R A A. Howeve the wave function cannot E. To ake futhe pogess in finding the wave function we need to know the potential. Specific exaples will be analyzed in due couse. It is inteesting to iφ iφ note that the allowed wave functions popotional to e e > 0 ae the coplex functions z * ( z ) if the two-diensional space is apped into the coplex plane. Repesenting any-electon wave functions in the plane in this way was a key to undestanding the quantu Hall effect.