Lecture 18 Notes: Angular Momentum

5.61 Page 1 Angular Momentum Since L̂2 and L̂z commute, they share common eigenfunctions. These functions are extremely important for the description of angular momentum problems – they determine the allowed values of angular momentum and, for systems like the Rigid Rotor, the energies available to the system. The first things we would like to know are the eigenvalues associated with these eigenfunctions. We will denote the eigenvalues of L̂2 and L̂z by α and β, respectively so that: L̂2Y β (θ , φ ) = α Y β (θ , φ ) Lˆ Y β (θ , φ ) = β Y β (θ , φ ) α α z α α For brevity, in what follows we will omit the dependence of the eigenstates on θ and φ so that the above equations become Lˆ 2Yαβ = α Yαβ Lˆ zYαβ = β Yαβ It is convenient to define the raising and lowering operators (note the similarity to the Harmonic oscillator!): L̂± ≡ L̂ x ± iLˆ y Which satisfy the commutation relations: ⎡ L̂± , L̂2 ⎤ = 0 ⎡ L̂+ , L̂− ⎤ = 2�L̂z ⎡ L̂z , L̂± ⎤ = ± �L̂± ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ These relations are relatively easy to prove using the commutation relations we’ve already derived: ⎡ Lˆ 2 , Lˆ z ⎤ = 0 ⎡ Lˆ x , Lˆ y ⎤ = i�Lˆ z ⎡ Lˆ y , Lˆ z ⎤ = i�Lˆ x ⎡ Lˆ z , Lˆ x ⎤ = i�Lˆ y ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ For example: ⎡ L̂z , L̂± ⎤ = ⎡ L̂z , L̂ x ⎤ ± i ⎡ L̂z , L̂ y ⎤ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ( = i�L y ± i ( −i�L x ) = ± � L x ± iLy ) = ± �L̂± The raising and lowering operators have a peculiar effect on the eigenvalue of L̂z : Lˆ z ( Lˆ ±Yαβ ) = ( ⎡⎣ Lˆ z , Lˆ ± ⎤⎦ + Lˆ ± Lˆ z )Yαβ = ( ± �Lˆ ± + Lˆ ± β )Yαβ = ( β ± � ) ( Lˆ ±Yαβ ) Thus, L̂+ ( L̂− ) raises (lowers) the eigenvalue of L̂z by � , hence the names. Since the raising and lowering operators commute with L̂2 they do not change the value of α and so we can write Lˆ ±Yαβ ∝ Yαβ ±� and so the eigenvalues of L̂z are evenly spaced! What are the limits on this ladder of eigenvalues? Recall that for the harmonic oscillator, we found that there was a minimum eigenvalue and the eigenstates could 5.61 Page 2 be created by successive applications of the raising operator to the lowest state. There is also a minimum eigenvalue in this case. To see this, note that Lˆ2 + Lˆ2 = Lˆ2 + Lˆ2 ≥ 0 x y x y This result simply reflects the fact that if you take any observable operator and square it, you must get back a positive number. To get a negative value for the average value of L̂2x or L̂2y would imply an imaginary eigenvalue of L̂ x or L̂ y , which is impossible since these operators are Hermitian. Besides, what would an imaginary angular momentum mean? We now apply the above equation for the specific wavefunction Yαβ : ( ∫ ) ) ∫ ( = ∫ Yαβ * (α − β 2 ) Yαβ 0 ≤ Yαβ * L̂2x + L̂2y Yαβ = Yαβ * L̂2 − L̂2z Yαβ =α −β2 Hence β 2 ≤ α and therefore − α ≤ β ≤ α . Which means that there are both maximum and minimum values that β can take on for a given α. If we denote these values by βmax and βmin, respectively, then it is clear that Lˆ +Yαβmax = 0 Lˆ −Yαβmin = 0 . We can then use this knowledge and some algebra tricks trick to determine the relationship between α and βmax (or βmin). First note that: Lˆ + Lˆ −Yαβ min = 0 ⇒ Lˆ − Lˆ +Yαβ max = 0 We can expand this explicitly in terms of L̂ x and L̂ x : Lˆ2x + Lˆ2y + i( Lˆ y Lˆ x − Lˆ x Lˆ y ) Yαβ min = 0 ⇒ Lˆ2x + Lˆ2y − i( Lˆ y Lˆ x − Lˆ x Lˆ y ) Yαβ max = 0 ) ( ( ) However, this is not the most convenient form for the operators, because we don’t know what L̂ x or L̂ y gives when acting on Yαβ . However, we can re­write the same expression in terms of L̂2 and L̂z : Lˆ2x + Lˆ2y ± i( Lˆ y Lˆ x − Lˆ x Lˆ y ) ( L̂2 − L̂2z ) −i�Lˆ z So then we have ( ⇒ (α − β ) ⇒ L̂2 − L̂2z − �L̂z Yαβ max = 0 2 max ) − �β max = 0 ( L̂ − L̂ + �L̂ ) Yαβ = 0 ( α − β + �β ) = 0 2 2 z min z 2 min ⇒ α = β max ( β max + �) = β min ( β min − �) ⇒ β max = − β min ≡ �l min 5.61 Page 3 where in the last line we have simply defined a new variable, l, that is dimensionless (notice that � has the units of angular momentum). So combining these minimum and maximum values we have that − �l ≤ β ≤ �l . Further, since we can get from the lowest to the highest eigenvalue in increments of � by successive applications of the raising operator, it is clear that the difference between the highest and lowest values [ �j − ( −�j ) = 2�l ] must be an integer multiple of � . Thus, l itself must either be an integer or a half­integer. Putting all these facts together, we conclude (Define m ≡ β / � ): Lˆ 2Ylm = �2l(l + 1)Ylm l = 0, 12 ,1, 23 ,2... and Lˆ zYlm = m�Ylm m = −l, −l + 1...l − 1, l where we have replaced α with l and β with m so that Yαβ becomes Ylm . Also, in the first equation, we have noted that 0 ≤ L̂2 = � 2l ( l + 1) implies l ≥ 0 . These are the fundamental eigenvalue equations for all forms of angular momentum. Notice that there is a difference here from what we saw for the rigid rotor. There, we had: �2 EJ = J ( J + 1) 2I J = 0, 1, 2,... where, as a reminder, the quantum number J for the rigid rotor is equivalent to the quantum number l define above. Here, the dependence of the energy on J – E ∝ J ( J + 1) ­ is the same as we found in our derivation for L̂2 . The factor of 1 / 2 I simply arises from the fact that the rigid rotor Hamiltonian is L̂2 / 2 I rather than L̂2 . The real difference is that half integer values of J do not appear for the rigid rotor. At first, you might think this means we made a mistake in our derivation above and that l should only be an integer and not a half integer. However, there is no error. The difference arises because our derivation above is valid for any kind of angular momentum. Thus, while certain values of l may not appear for certain types of angular momentum (e.g. they don’t occur for the rigid rotor) we will see later on that they can appear for other types of angular momentum. Most notably, electrons have an intrinsic spin angular momentum with l = 12 . Thus, while individual systems may have additional restrictions on the 5.61 Page allowed values of l, angular momentum states always obey the above eigenvalue relations. 4