21 1.7 Spin and angular momentum (yearly)revolution around the sun,and,besides that,some angular momentum due to its (daily)revolution around its own axis.Although this picture might help to visualize the concept of spin,in reality it is not accurate.Elementary particles,such as electrons,are structureless point particles,and it is impossible to assign to them a spatial mass distribu- tion revolving around an axis.Let us thus present a more abstract but more adequate way to consider spin You might recall from your courses on classical mechanics that it is possible to derive the familiar conservation laws(of momentum,energy,etc.)from the symmetries of space and time.For example,the properties of a system are not expected to depend on where we choose the origin of the coordinate system.in other words,we assume that space is homog nent in more detail.We e free particle.and we assume that its state is given by the wave func tion(r).A shift of the coordinate system by a vector a changes the wave function (r)(r)=(r+a).where the prime indicates the new coordinate system.Since this is nothing but a change of basis,we can look for the"translation operator"Ta which shifts the wave function ver a,i.e.'(r)=Tav(r).To find Ta.we expand v(r+a)in a Taylor series r+a=+a异m+(品)o+ (1.64) =v(r)=e-(r). and see that the perator is given bywhere the defined in terms of its Taylor expansion.An infinitesimally small translation a is effected by the momentum operator itself,i.e.Ta)=1-(a).p. So what does this have to do with conservation laws?Since we assumed the Hamiltonian of the system to be invariant under translations T,we find that=A must hold a.This implies that [A=0.the translation operator and the Hamiltonian .As s complete set of eigenstates. or,to reason one step further,the eigenstates of the translation operator are stationary states.Now we can finally draw our conclusion.What do we know about the eigenstates of Ta?The translation operator consists only of products of momentum operators.T=1-Ea-p- (a-)+ its eigenstates are therefore the momentum eigenstates.As explained above.these states must be stationary:moment is thus a con erved quantity Another interesting symmetry to investigate is the isotropy of space (space looks the same in all directions).Let us proceed along similar lines as above,and try to find the operators which correspond to conserved quantities for an isotropic Hamiltonian.The(con- served)momentum operator turned out to be the operator for infinitesimal translations,so let us now look for the ope r of infinitesimal ro ns.In three e-dimensional space,an rotation can be de ed into rotations about the three orthogonal axes.We defin rotation operator R()as the operator which rotates a wave function over an angle6 about the a-axis(where a is x,y,orz).We are thus interested in the operatorsJa defined by e=R(0).which yields=)= 21 1.7 Spin and angular momentum (yearly) revolution around the sun, and, besides that, some angular momentum due to its (daily) revolution around its own axis. Although this picture might help to visualize the concept of spin, in reality it is not accurate. Elementary particles, such as electrons, are structureless point particles, and it is impossible to assign to them a spatial mass distribu￾tion revolving around an axis. Let us thus present a more abstract but more adequate way to consider spin. You might recall from your courses on classical mechanics that it is possible to derive the familiar conservation laws (of momentum, energy, etc.) from the symmetries of space and time. For example, the properties of a system are not expected to depend on where we choose the origin of the coordinate system, in other words, we assume that space is homogeneous. Let us investigate the implications of this statement in more detail. We consider a single free particle, and we assume that its state is given by the wave func￾tion ψ(r). A shift of the coordinate system by a vector a changes the wave function ψ(r) → ψ (r) = ψ(r + a), where the prime indicates the new coordinate system. Since this is nothing but a change of basis, we can look for the “translation operator” Tˆa which shifts the wave function over a, i.e. ψ (r) = Tˆaψ(r). To find Tˆa, we expand ψ(r + a) in a Taylor series, ψ(r + a) = ψ(r) + a · ∂ ∂r ψ(r) + 1 2  a · ∂ ∂r 2 ψ(r) + ... = ea·∂rψ(r) = e − i h¯ a·pˆ ψ(r), (1.64) and see that the translation operator is given by Tˆa = e − i h¯ a·pˆ , where the exponent is defined in terms of its Taylor expansion. An infinitesimally small translation δa is effected by the momentum operator itself, i.e. Tˆ(δa) = 1 − i h¯ (δa) · pˆ. So what does this have to do with conservation laws? Since we assumed the Hamiltonian of the system to be invariant under translations Tˆa, we find that Tˆ † aHˆ Tˆa = Hˆ must hold for any a. This implies that [Hˆ , Tˆa] = 0, the translation operator and the Hamiltonian commute. As a result, Hˆ and Tˆa share a complete set of eigenstates, or, to reason one step further, the eigenstates of the translation operator are stationary states. Now we can finally draw our conclusion. What do we know about the eigenstates of Tˆa? The translation operator consists only of products of momentum operators, Tˆa = 1− i h¯ a · pˆ − 1 2h¯ 2 (a · pˆ) 2 + ... , its eigenstates are therefore the momentum eigenstates. As explained above, these states must be stationary: momentum is thus a conserved quantity. Another interesting symmetry to investigate is the isotropy of space (space looks the same in all directions). Let us proceed along similar lines as above, and try to find the operators which correspond to conserved quantities for an isotropic Hamiltonian. The (con￾served) momentum operator turned out to be the operator for infinitesimal translations, so let us now look for the operator of infinitesimal rotations. In three-dimensional space, any rotation can be decomposed into rotations about the three orthogonal axes. We define the rotation operator Rˆ α(θ) as the operator which rotates a wave function over an angle θ about the α-axis (where α is x, y, or z). We are thus interested in the operators Jˆα defined by e − i h¯ θJˆα = Rˆ α(θ), which yields Jˆα = ih¯∂θRˆ α(θ)|θ=0