Elementary quantum mechanics From the way we just wrote it,we can interpret it as a Lorentzian peak centered around x=0with an infinitely small width but with a fixed area under the peak.Using this delta function.we finally writ 2π D=】 (1.62 were we sum over all possible final states f).thereby making Ithe total decay rate from the have reproduced Fermi's far ous golden We see that the rate (1.62)is zero except when Er =Ei.that is,the initial and final state have the same energy.Indeed.since we made sure that all time-dependence in the Hamiltonian is so slow that the system always evolves adiabatically,we expect energy con- servation to forbid any transition accompanied by a change in energy.For adiscrete energ spectrum,two ene E:and E ill never be in practice.and n ths ons.We do however expect transitions for a continuous spectrum:th sum over final states is to be converted to an integral,and the delta function selects from the continuous distribution of energy levels those providing energy conservation.In Chapter9 we illustrate this in detail by computing the spontaneous emission rates from excited atoms Higher-order terms from the expansion(1.54)can be used to evaluate higher-order tran sition rates.For ins ance,the second-orde rterm include es the possibility to make atr from the initial to the final state via another state.This intermediate state is called virtud since its energy matches neither the initial nor final energy.this forbidding its true real ization.These higher-order rates become particularly important when the perturbation A does not have matrix elements directly between the initial and final states.If the pertur bation.howeve does have elements between the initial and a virtual state and be this virtual and the final state,transi ons can take place.and are second-order with a rate proportional to the second power of H.The work-out of this rate by time-dependent per turbation theory (see Exercise 1)proves that the rate is given by Fermi's golden rule with an effective matrix element A)→∑创府n Ei-Ev (1.63) Here,the summation is over all possible virtual states.The energy mismatch between the initial and virtual states enters the denominators in this expression. 1.7 Spin and angular momentum From our introductory courses on quantum mechanics,we know that elementary particles (electrons,neutrons.etc.)all have a special degree of freedom called spin.An intuitive picture,which is often given when the ncept of spin is introduced,is that the spin of particle can be regarded as the angular momentum of the particle due to the fact that i is spinning around its own axis.For example,an electron in an atomic orbital could be compared to the earth orbiting the sun:it possesses orbital angular momentum due to its
20 Elementary quantum mechanics From the way we just wrote it, we can interpret it as a Lorentzian peak centered around x = 0 with an infinitely small width but with a fixed area under the peak. Using this delta function, we finally write i = f 2π h¯ | f |Hˆ |i |2δ(Ei − Ef), (1.62) were we sum over all possible final states |f, thereby making i the total decay rate from the initial state |i. We have reproduced Fermi’s famous golden rule. We see that the rate (1.62) is zero except when Ef = Ei, that is, the initial and final state have the same energy. Indeed, since we made sure that all time-dependence in the Hamiltonian is so slow that the system always evolves adiabatically, we expect energy conservation to forbid any transition accompanied by a change in energy. For a discrete energy spectrum, two energies Ei and Ef will never be exactly aligned in practice, and in this case there are no transitions. We do however expect transitions for a continuous spectrum: the sum over final states is to be converted to an integral, and the delta function selects from the continuous distribution of energy levels those providing energy conservation. In Chapter 9 we illustrate this in detail by computing the spontaneous emission rates from excited atoms. Higher-order terms from the expansion (1.54) can be used to evaluate higher-order transition rates. For instance, the second-order term includes the possibility to make a transition from the initial to the final state via another state. This intermediate state is called virtual since its energy matches neither the initial nor final energy, this forbidding its true realization. These higher-order rates become particularly important when the perturbation Hˆ does not have matrix elements directly between the initial and final states. If the perturbation, however, does have elements between the initial and a virtual state and between this virtual and the final state, transitions can take place, and are second-order with a rate proportional to the second power of Hˆ . The work-out of this rate by time-dependent perturbation theory (see Exercise 1) proves that the rate is given by Fermi’s golden rule with an effective matrix element f |Hˆ |i → v f |Hˆ |v v|Hˆ |i Ei − Ev . (1.63) Here, the summation is over all possible virtual states. The energy mismatch between the initial and virtual states enters the denominators in this expression. 1.7 Spin and angular momentum From our introductory courses on quantum mechanics, we know that elementary particles (electrons, neutrons, etc.) all have a special degree of freedom called spin. An intuitive picture, which is often given when the concept of spin is introduced, is that the spin of a particle can be regarded as the angular momentum of the particle due to the fact that it is spinning around its own axis. For example, an electron in an atomic orbital could be compared to the earth orbiting the sun: it possesses orbital angular momentum due to its
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 distribution 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 function 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 (conserved) 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
22 Elementary quantum mechanics Let us now keep the discussion as general as possible,and look for properties of the J without assuming anything specific about the t e of rotations.A useful starting point would be to derive the investigate the case where R()is a simple rotation of the coordinate system about one of the Cartesian axes.You might wonder what other less simple types of rotation there are, but a bit of patience is required,we find this out below.For now,we simply write down the three matrices which rotate a Cartesian coordinate system.For example,for rotations about the z-axis we find )-(g) (1.65) and similar matrices for the other two axes.We calculate j=ihdeR()from these matrices,and find that [心，j]=i.心.j月=j,ndd.j]=, (1.66 We use these commutation relations as definitions for the operators J.The observ ables associated with these operators,which are conserved quantities,give the angular momentum of the system along the three orthogonal axes. Control question.Can you verify that the three components of the"quantum ana- logue"L=fx 6 of the classical angular momentum indeed satisfy the commutation relations(166)? It turs out that it is possible to reveal the complete matrix structure of the operator on the e commutation relations given above.We start by defining an operator comespondins to the square of the total anuar momentum. that (1.67 the total angular momentum operator commutes with all three orthogonal components of J.This means that )2 and.for example.share a common set of eigenstates.It is not difficult to show (see Exercise 5)that any common eigenstate of )2 and 7:can be written as li.m),such that P,m=0+1,m (1.68) J:lj.m2).m2). where jis either integer or half-integer,and for given j.the number mcan take one of the 2j+1 value +1 j-1, We now derive the matrix structure this basis of eigenstatesofActually. it is easier to work with the operatorsxiy(see Exercise 5).The operator J+is a"raising operator,"it couples states with quantum number mz to states with quantum number m:+1.We can thus write 7lj.m)=y(m)j.m:+1).where we would like to know the matrix element y(m).To find it,we evaluate ,m:_+.m)=0+1)-m:(m+1 (1.69
22 Elementary quantum mechanics Let us now keep the discussion as general as possible, and look for properties of the Jˆ without assuming anything specific about the type of rotations. A useful starting point would be to derive the commutation relations for Jˆ, i.e. find [Jˆα, Jˆβ]. For this purpose, we investigate the case where Rˆ α(θ) is a simple rotation of the coordinate system about one of the Cartesian axes. You might wonder what other less simple types of rotation there are, but a bit of patience is required, we find this out below. For now, we simply write down the three matrices which rotate a Cartesian coordinate system. For example, for rotations about the z-axis we find ⎛ ⎝ x y z ⎞ ⎠ = ⎛ ⎝ cos θ − sin θ 0 sin θ cos θ 0 0 01 ⎞ ⎠ ⎛ ⎝ x y z ⎞ ⎠ , (1.65) and similar matrices for the other two axes. We calculate Jˆ = ih¯∂θRˆ (θ)|θ=0 from these matrices, and find that [Jˆx, Jˆy] = ih¯Jˆz, [Jˆy, Jˆz] = ih¯Jˆx, and [Jˆz, Jˆx] = ih¯Jˆy. (1.66) We use these commutation relations as definitions for the operators Jˆ. The observables associated with these operators, which are conserved quantities, give the angular momentum of the system along the three orthogonal axes. Control question. Can you verify that the three components of the “quantum analogue” Lˆ = rˆ × pˆ of the classical angular momentum indeed satisfy the commutation relations (1.66)? It turns out that it is possible to reveal the complete matrix structure of the operators Jˆ just based on the commutation relations given above. We start by defining an operator Jˆ2 ≡ Jˆ2 x + Jˆ2 y + Jˆ2 z , corresponding to the square of the total angular momentum. We see that [Jˆ2, Jˆx] = [Jˆ2, Jˆy] = [Jˆ2, Jˆz] = 0, (1.67) the total angular momentum operator commutes with all three orthogonal components of Jˆ. This means that Jˆ2 and, for example, Jˆz share a common set of eigenstates. It is not difficult to show (see Exercise 5) that any common eigenstate of Jˆ2 and Jˆz can be written as |j, mz, such that Jˆ2|j, mz = h¯ 2j(j + 1)|j, mz, Jˆz|j, mz = hm¯ z|j, mz, (1.68) where j is either integer or half-integer, and for given j, the number mz can take one of the 2j + 1 values −j, −j + 1, ... , j − 1, j. We now derive the matrix structure of Jˆx and Jˆy in this basis of eigenstates of Jˆz. Actually, it is easier to work with the operators Jˆ± ≡ Jˆx ± iJˆy (see Exercise 5). The operator Jˆ+ is a “raising operator,” it couples states with quantum number mz to states with quantum number mz + 1. We can thus write Jˆ+|j, mz = γ+(mz)|j, mz + 1, where we would like to know the matrix element γ+(mz). To find it, we evaluate j, mz|Jˆ−Jˆ+|j, mz = h¯ 2[j(j + 1) − mz(mz + 1)]. (1.69)
23 1.7 Spin andangular momentum Control question.Do you see how to derive(1.69)? However.per definition we can also write (j.mm)=y+(m2)) +(m),and com parison with (1.69)gives us the desired y+(m)up to a phase factor.Usually the matrix elements are chosen to be real,so we find J4j,m2)=hvjj+1)-mz(m:+1)lj.m2 +1) (1.70 and,in a similar wav. -j,m)=h0+1)-m(m:-1i,m:-1 (1.71) We now return to the issue of the different types of rotation,which we mentioned in the beginning of this section.As you have already checked.the regular orbital angular erators L=x p satisfy the commutation relations,and their matrix struc- urebe as derived above Let us inveti this obita anurm ntum more detail.If we were to write a wave function in spherical coordinates,=rsin cos y =rsin sin and z=rcose,the z-component of L takes the very simple form L=-indo. Control question.Do you know how to transform the operatorx p to spherical coordinates? The eigenstates of can then be easily determined.They are the solutions of ))which of course are)(r.0).Since the ion must be single-valued everyw for are l=hm:where m:must be an integer.This is of course not inconsistent with wha we found above for the angular momentum operator J.However,we then concluded that eigenvalues of J.can be integer or half-integer.The orbital angular momentum only exploits the first option. Is there a problem with half-i values of a omentum?Supp se we have a about the z-axis to this state.We know that the rotation operator readsR)=eWe set 6 to 2 and make use of the fact that)is an eigenstate of so that R(2rlw)=e2)=-1. (1.72) The state acquires a factor-1 after a full rotation about the z-axis.This sounds strange. but in quantum mechanics this actually is not a problem!A wave function is in any case only defined up to a phase factor.so)and)=)are the same physical state since any possible observable is the same for the two states,(A)=(Aly').Therefore, there is no a-priori reason to exclude the possibility of half-in angular momentum As explained a ve,half-integer angul omentur lar momentumL,so there must be another type of"intrinsic"angular momentum.Thi intrinsic angular momentum is called spin,and is represented by the operators S.so that j=1+s. (1.73)
23 1.7 Spin and angular momentum Control question. Do you see how to derive (1.69)? However, per definition we can also write j, mz|Jˆ−Jˆ+|j, mz = γ+(mz) ∗γ+(mz), and comparison with (1.69) gives us the desired γ+(mz) up to a phase factor. Usually the matrix elements are chosen to be real, so we find Jˆ+| j, mz = h¯ j(j + 1) − mz(mz + 1)| j, mz + 1, (1.70) and, in a similar way, Jˆ−| j, mz = h¯ j(j + 1) − mz(mz − 1)| j, mz − 1. (1.71) We now return to the issue of the different types of rotation, which we mentioned in the beginning of this section. As you have already checked, the regular orbital angular momentum operators Lˆ = rˆ × pˆ satisfy the commutation relations, and their matrix structure must thus be as derived above. Let us investigate this orbital angular momentum in more detail. If we were to write a wave function in spherical coordinates, x = rsin θ cos φ, y = rsin θ sin φ and z = r cos θ, the z-component of Lˆ takes the very simple form Lˆz = −ih¯∂φ. Control question. Do you know how to transform the operator rˆ × pˆ to spherical coordinates? The eigenstates of Lˆz can then be easily determined. They are the solutions of −ih¯∂φψ(r, θ, φ) = lzψ(r, θ, φ), which of course are ψ(r, θ, φ) = e i h¯ lzφf(r, θ). Since the wave function must be single-valued everywhere, we find that the allowed values for lz are lz = hm¯ z where mz must be an integer. This is of course not inconsistent with what we found above for the angular momentum operator Jˆz. However, we then concluded that eigenvalues of Jˆz can be integer or half-integer. The orbital angular momentum only exploits the first option. Is there a problem with half-integer values of angular momentum? Suppose we have a state |ψ which is an eigenstate of Jˆz with eigenvalue h¯/2. Let us apply a full 2π-rotation about the z-axis to this state. We know that the rotation operator reads Rˆz(θ) = e i h¯ θJˆz . We set θ to 2π and make use of the fact that |ψ is an eigenstate of Jˆz, so that Rˆz(2π)|ψ = e i h¯ 2π h¯ 2 |ψ = −|ψ. (1.72) The state acquires a factor −1 after a full rotation about the z-axis. This sounds strange, but in quantum mechanics this actually is not a problem! A wave function is in any case only defined up to a phase factor, so |ψ and |ψ = −|ψ are the same physical state since any possible observable is the same for the two states, ψ|Aˆ|ψ=ψ |Aˆ|ψ . Therefore, there is no a-priori reason to exclude the possibility of half-integer angular momentum. As explained above, half-integer angular momentum cannot be related to the orbital angular momentum Lˆ , so there must be another type of “intrinsic” angular momentum. This intrinsic angular momentum is called spin, and is represented by the operators Sˆ, so that Jˆ = Lˆ + Sˆ. (1.73)
Elementary quantum mechanics The spin angular momentum has nothing to do with actual rotations of the coordinates be related to rotations of other aspects of the structure of theave What be?Nothing forbids complex thar just a function of position.It could,for example,be defined in a two-dimensional vector space (r)=((r).(r)),or in an even higher-dimensional vector space.The spin angular woud then relate to this:effected by can be regarded as rotations within this vector space.This means that.if a certain wave function has an N-dimensional vector representation,its spin operators can be written as Nx N matrices.This dimension relates in a simple way to the total spin quantum number s(analogous to j for the total angular momentum):2s+1 =N.A two-dimensional wave 6 action thus a particle with spins Wave functions whichdo posse a three-dimensional with s a vector structure ong to particles with spin 0 It can be shown that all particles with integer spin must be bosons,and particles with half-integer spin (....are fermions. Fortunately.the most common elementary particles.such as electrons.protons and neu trons,.are all fermions with spin是，that is to say the most“simple”fermions allowed by We can write down their 2x 2 spin operators. s=(90)$-(9。）m=(09) 1.74) In the basis we have chosen,ie.the eigenstates ofa particle with a wave functior (r.0 is thus a spin-particle with spin quantum number m=(or spin up).and a state (0.(r))has a spin quantum number m:=-(or spin down).These two values m=,are the only two possible outcomes if one measures the spin of an electron along a fixed axis.A way to do this would be to make use of the fact that particles with different spin quantum number behave differently in the presence of a magnetic field. 1.7.1 Spin in a magneticfield As we explained above,a quantum state can be rotated in two different ways.One can sim- different types of rotation can be performed independently.it isno rprise that there xis Hamiltonians which couple only to the orbital angular momentum of a particle.as well as Hamiltonians which couple only to its spin.An example of the latter is the Hamiltonian describing the coupling between spin and a magnetic field. Since spin is a form of angular momentum.it adds to the magnetic moment of the i=ys. (1.75 where the constant y depends on the type of particle.In the case of free electrons.this constant is y=lel/me,where lel is the absolute value of the electron charge and me is the
24 Elementary quantum mechanics The spin angular momentum has nothing to do with actual rotations of the coordinates, so it must be related to rotations of other aspects of the structure of the wave function. What could this structure be? Nothing forbids a wave function to be more complex than just a function of position. It could, for example, be defined in a two-dimensional vector space (r) = (ψ↑(r), ψ↓(r))T , or in an even higher-dimensional vector space. The spin angular momentum would then relate to this vector structure: rotations effected by e − i h¯ θ·Sˆ can be regarded as rotations within this vector space. This means that, if a certain wave function has an N-dimensional vector representation, its spin operators can be written as N × N matrices. This dimension relates in a simple way to the total spin quantum number s (analogous to j for the total angular momentum): 2s + 1 = N. A two-dimensional wave function thus describes a particle with spin s = 1 2 , a three-dimensional with spin 1, etc. Wave functions which do not possess a vector structure belong to particles with spin 0. It can be shown that all particles with integer spin must be bosons, and particles with half-integer spin ( 1 2 , 3 2 , ... ) are fermions. Fortunately, the most common elementary particles, such as electrons, protons and neutrons, are all fermions with spin 1 2 , that is to say the most “simple” fermions allowed by Nature. For this reason, we now focus on spin −1 2 particles. We can write down their 2 × 2 spin operators, Sˆx = h¯ 2 0 1 1 0 , Sˆy = h¯ 2 0 −i i 0 , and Sˆz = h¯ 2 1 0 0 −1 . (1.74) In the basis we have chosen, i.e. the eigenstates of Sˆz, a particle with a wave function (ψ↑(r), 0)T is thus a spin −1 2 particle with spin quantum number mz = 1 2 (or spin up), and a state (0, ψ↓(r))T has a spin quantum number mz = −1 2 (or spin down). These two values, mz = ±1 2 , are the only two possible outcomes if one measures the spin of an electron along a fixed axis. A way to do this would be to make use of the fact that particles with different spin quantum number behave differently in the presence of a magnetic field. 1.7.1 Spin in a magnetic field As we explained above, a quantum state can be rotated in two different ways. One can simply rotate the coordinates of the wave function (which is achieved by the operator e − i h¯ θ·Lˆ ), or one can rotate the spin of the wave function (which is done by e − i h¯ θ·Sˆ ). Since these different types of rotation can be performed independently, it is no surprise that there exist Hamiltonians which couple only to the orbital angular momentum of a particle, as well as Hamiltonians which couple only to its spin. An example of the latter is the Hamiltonian describing the coupling between spin and a magnetic field. Since spin is a form of angular momentum, it adds to the magnetic moment of the particle. This contribution is proportional to the spin, and thus reads μˆ = γ Sˆ, (1.75) where the constant γ depends on the type of particle. In the case of free electrons, this constant is γ = |e|/me, where |e| is the absolute value of the electron charge and me is the