10.2: Angular Momentum in Hydrogen Atom
( \newcommand{\kernel}{\mathrm{null}\,}\)
In a hydrogen atom, the wavefunction of an electron in a simultaneous eigenstate of L2 and Lz has an angular dependence specified by the spherical harmonic Yl,m(θ,ϕ). (See Section [sharm].) If the electron is also in an eigenstate of S2 and Sz then the quantum numbers s and ms take the values 1/2 and ±1/2, respectively, and the internal state of the electron is specified by the spinors χ±. (See Section [spauli].) Hence, the simultaneous eigenstates of L2, S2, Lz, and Sz can be written in the separable form ψ(1)l,1/2;m,±1/2=Yl,mχ±. Here, it is understood that orbital angular momentum operators act on the spherical harmonic functions, Yl,m, whereas spin angular momentum operators act on the spinors, χ±.
Because the eigenstates ψ(1)l,1/2;m,±1/2 are (presumably) orthonormal, and form a complete set, we can express the eigenstates ψ(2)l,1/2;j,mj as linear combinations of them. For instance,
ψ(2)l,1/2;j,m+1/2=αψ(1)l,1/2;m,1/2+βψ(1)l,1/2;m+1,−1/2, where α and β are, as yet, unknown coefficients. Note that the number of ψ(1) states that can appear on the right-hand side of the previous expression is limited to two by the constraint that mj=m+ms [see Equation ([e11.23])], and the fact that ms can only take the values ±1/2. Assuming that the ψ(2) eigenstates are properly normalized, we have
Now, it follows from Equation ([e11.26]) that
J2ψ(2)l,1/2;j,m+1/2=j(j+1)ℏ2ψ(2)l,1/2;j,m+1/2, where [see Equation ([e11.12])] J2=L2+S2+2LzSz+L+S−+L−S+. Moreover, according to Equations ([e11.28]) and ([e11.29]), we can write ψ(2)l,1/2;j,m+1/2=αYl,mχ++βYl,m+1χ−. Recall, from Equations ([eraise]) and ([elow]), that L+Yl,m=[l(l+1)−m(m+1)]1/2ℏYl,m+1,L−Yl,m=[l(l+1)−m(m−1)]1/2ℏYl,m−1. By analogy, when the spin raising and lowering operators, S±, act on a general spinor, χs,ms, we obtain S+χs,ms=[s(s+1)−ms(ms+1)]1/2ℏχs,ms+1,S−χs,ms=[s(s+1)−ms(ms−1)]1/2ℏχs,ms−1. For the special case of spin one-half spinors (i.e., s=1/2,ms=±1/2), the previous expressions reduce to S+χ+=S−χ−=0, and S±χ∓=ℏχ±.
It follows from Equations ([e11.32]) and ([e11.34])–([e11.39]) that J2Yl,mχ+=[l(l+1)+3/4+m]ℏ2Yl,mχ+=+[l(l+1)−m(m+1)]1/2ℏ2Yl,m+1χ−, and J2Yl,m+1χ−=[l(l+1)+3/4−m−1]ℏ2Yl,m+1χ−=+[l(l+1)−m(m+1)]1/2ℏ2Yl,mχ+. Hence, Equations ([e11.31]) and ([e11.33]) yield
(x−m)α−[l(l+1)−m(m+1)]1/2β=0,−[l(l+1)−m(m+1)]1/2α+(x+m+1)β=0, where x=j(j+1)−l(l+1)−3/4. Equations ([e11.42]) and ([e11.43]) can be solved to give x(x+1)=l(l+1), and
αβ=[(l−m)(l+m+1)]1/2x−m. It follows that x=l or x=−l−1, which corresponds to j=l+1/2 or j=l−1/2, respectively. Once x is specified, Equations ([e11.30]) and ([e11.45]) can be solved for α and β. We obtain
ψ(2)l+1/2,m+1/2=(l+m+12l+1)1/2ψ(1)m,1/2+(l−m2l+1)1/2ψ(1)m+1,−1/2, and
ψ(2)l−1/2,m+1/2=(l−m2l+1)1/2ψ(1)m,1/2−(l+m+12l+1)1/2ψ(1)m+1,−1/2. Here, we have neglected the common subscripts l,1/2 for the sake of clarity: that is, ψ(2)l+1/2,m+1/2≡ψ(2)l,1/2;l+1/2,m+1/2, et cetera. The previous equations can easily be inverted to give the ψ(1) eigenstates in terms of the ψ(2) eigenstates:
ψ(1)m,1/2=(l+m+12l+1)1/2ψ(2)l+1/2,m+1/2+(l−m2l+1)1/2ψ(2)l−1/2,m+1/2,ψ(1)m+1,−1/2=(l−m2l+1)1/2ψ(2)l+1/2,m+1/2−(l+m+12l+1)1/2ψ(2)l−1/2,m+1/2. The information contained in Equations ([e11.47])–([e11.50]) is neatly summarized in Table [t2]. For instance, Equation ([e11.47]) is obtained by reading the first row of this table, whereas Equation ([e11.50]) is obtained by reading the second column. The coefficients in this type of table are generally known as Clebsch-Gordon coefficients .
m,1/2 | m+1,−1/2 | m,ms | |
[0.5ex] l+1/2,m+1/2 | √(l+m+1)/(2l+1) | √(l−m)/(2l+1) | |
[0.5ex] l−1/2,m+1/2 | √(l−m)/(2l+1) | −√(l+m+1)/(2l+1) | |
[0.5ex] j,mj |
As an example, let us consider the l=1 states of a hydrogen atom. The eigenstates of L2, S2, Lz, and Sz, are denoted ψ(1)m,ms. Because m can take the values −1,0,1, whereas ms can take the values ±1/2, there are clearly six such states: that is, ψ(1)1,±1/2, ψ(1)0,±1/2, and ψ(1)−1,±1/2. The eigenstates of L2, S2, J2, and Jz, are denoted ψ(2)j,mj. Because l=1 and s=1/2 can be combined together to form either j=3/2 or j=1/2 (see previously), there are also six such states: that is, ψ(2)3/2,±3/2, ψ(2)3/2,±1/2, and ψ(2)1/2,±1/2. According to Table [t2], the various different eigenstates are interrelated as follows:
ψ(2)3/2,±3/2=ψ(1)±1,±1/2,ψ(2)3/2,1/2=√23ψ(1)0,1/2+√13ψ(1)1,−1/2,ψ(2)1/2,1/2=√13ψ(1)0,1/2−√23ψ(1)1,−1/2,ψ(2)1/2,−1/2=√23ψ(1)−1,1/2−√13ψ(1)0,−1/2,ψ(2)3/2,−1/2=√13ψ(1)−1,1/2+√23ψ(1)0,−1/2, and
ψ(1)±1,±1/2=ψ(2)3/2,±3/2,ψ(1)1,−1/2=√13ψ(2)3/2,1/2−√23ψ(2)1/2,1/2,ψ(1)0,1/2=√23ψ(2)3/2,1/2+√13ψ(2)1/2,1/2,ψ(1)0,−1/2=√23ψ(2)3/2,−1/2−√13ψ(2)1/2,−1/2,ψ(1)−1,1/2=√13ψ(2)3/2,−1/2+√23ψ(2)1/2,−1/2, Thus, if we know that an electron in a hydrogen atom is in an l=1 state characterized by m=0 and ms=1/2 [i.e., the state represented by ψ(1)0,1/2] then, according to Equation ([e11.57]), a measurement of the total angular momentum will yield j=3/2, mj=1/2 with probability 2/3, and j=1/2, mj=1/2 with probability 1/3. Suppose that we make such a measurement, and obtain the result j=3/2, mj=1/2. As a result of the measurement, the electron is thrown into the corresponding eigenstate, ψ(2)3/2,1/2. It thus follows from Equation ([e11.52]) that a subsequent measurement of Lz and Sz will yield m=0, ms=1/2 with probability 2/3, and m=1, ms=−1/2 with probability 1/3.
−1,−1/2 | −1,1/2 | 0,−1/2 | 0,1/2 | 1,−1/2 | 1,1/2 | m,ms | |
[0.5ex] 3/2,−3/2 | 1 | ||||||
[0.5ex] 3/2,−1/2 | √1/3 | √2/3 | |||||
[0.5ex] 1/2,−1/2 | √2/3 | −√1/3 | |||||
[0.5ex] 3/2,1/2 | √2/3 | √1/3 | |||||
[0.5ex] 1/2,1/2 | √1/3 | −√2/3 | |||||
[0.5ex] 3/2,3/2 | 1 | ||||||
j,mj |
The information contained in Equations ([ecgs])–([ecge]) is neatly summed up in Table [t3]. Note that each row and column of this table has unit norm, and also that the different rows and different columns are mutually orthogonal. Of course, this is because the ψ(1) and ψ(2) eigenstates are orthonormal.
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)