25 1.7 Spin and angular momentum electron mass.The energy of a magnetic moment in an external field B is given by-LB. so we can write the Hamiltonian for the interaction between the spin of an electron and a magnetic field as H=-ugB.6, (1.76) where denotes the Bohr magneton,=are the three Pauli matrices. a=(90)4=(9。）=(09) (1.77) We see that electrons with spin up and down(measured along the axis parallel to the field) m Hamiltonian (1.76)causes the spin to precess around the direction of the magnetic field. We investigate this precession in detail in Exercise 7. 1.7.2 Two spins Let us now consider a system of wo spin-particles,and investigate the total spin of the two-particle system.A natural basis to express the two-particle spin state in is 1个↑，I个J),I4t),and). (1.78) We define the operator of total spin asS=+S2)the sum of the spin of the two separate particles.The total spin along the z-axis of the four states in (1.78)can now easily be determined.We see that all four states are eigenstates of SoL with respective eigenvalues m of 1.0.0.and-1.If we put this result in the context of the discussion in Section 1.7,it seems a bit confusing.The set of eigenvalues we found suggests that the total object (the two electron)behaves asa spin-1 particle.Its total spinsi one,and the spin along the z-axis can take values ranging from-s to s.i.e.from-I to 1.The confusing thing is that there obviously exist two different states with m.=0.and this does not fit the picture:a spin-1 particle has only one state with m=0. To find out what is going on.we must examine the four eigenstates more closely and calculate ther .The first state.)yields the alues 2h2 d h for respectively and This means that the two-particl eige e state indeed beh aves a a spin-1 particle with m=1,or in other words,using the notation Is,m),we find that )=11.1).We then apply the lowering operator Sto.=S)-+S(2).-to this state. which yields the state)+)),where the factor 1/v2 is included for normaliza- tion.We see that we find an m0 state which is a superposition of the two states we ussed ab ve. To identify whichm=0state we have found,we evaluate its eigenvalue of S and find that+)=11.0).Applying the lowering operator once again. we can verify that the state)corresponds to .-1).as expected. We thus have found three spin states of a two-electron system which behave like the states of a single spin-I particle.This means that we still have a fourth state which seems not tofit the picture.This fourth state(it must be orthogonal to the three discussed above)
25 1.7 Spin and angular momentum electron mass. The energy of a magnetic moment in an external field B is given by −μ · B, so we can write the Hamiltonian for the interaction between the spin of an electron and a magnetic field as Hˆ = −μBB · σˆ, (1.76) where μB denotes the Bohr magneton, μB = |e|h¯/2me, and σˆ are the three Pauli matrices, σˆx = 0 1 1 0 , σˆy = 0 −i i 0 , and σˆz = 1 0 0 −1 . (1.77) We see that electrons with spin up and down (measured along the axis parallel to the field) acquire an energy difference of 2μBB. This splitting is usually called the Zeeman splitting of the electron. If the magnetic field and the quantization axis are not aligned, then the Hamiltonian (1.76) causes the spin to precess around the direction of the magnetic field. We investigate this precession in detail in Exercise 7. 1.7.2 Two spins Let us now consider a system of two spin- 1 2 particles, and investigate the total spin of the two-particle system. A natural basis to express the two-particle spin state in is |↑↑, |↑↓, |↓↑, and |↓↓. (1.78) We define the operator of total spin as Sˆtot = Sˆ(1) + Sˆ(2), the sum of the spin of the two separate particles. The total spin along the z-axis of the four states in (1.78) can now easily be determined. We see that all four states are eigenstates of Sˆtot,z with respective eigenvalues mz of 1, 0, 0, and −1. If we put this result in the context of the discussion in Section 1.7, it seems a bit confusing. The set of eigenvalues we found suggests that the total object (the two electrons) behaves as a spin-1 particle. Its total spin s is one, and the spin along the z-axis can take values ranging from −s to s, i.e. from −1 to 1. The confusing thing is that there obviously exist two different states with mz = 0, and this does not fit the picture: a spin-1 particle has only one state with mz = 0. To find out what is going on, we must examine the four eigenstates more closely and calculate their Sˆ2 tot and Sˆtot,z. The first state, |↑↑, yields the eigenvalues 2h¯ 2 and h¯ for respectively Sˆ2 tot and Sˆtot,z. This means that the two-particle state |↑↑ indeed behaves as a spin-1 particle with mz = 1, or in other words, using the notation |s, mz, we find that |↑↑ = |1, 1. We then apply the lowering operator Sˆtot,− = Sˆ(1),− + Sˆ(2),− to this state, which yields the state √ 1 2 {|↑↓ + |↓↑}, where the factor 1/ √2 is included for normalization. We see that we find an mz = 0 state which is a superposition of the two states we discussed above. To identify which mz = 0 state we have found, we evaluate its eigenvalue of Sˆ2 tot, and find that √ 1 2 {|↑↓+|↓↑} = |1, 0. Applying the lowering operator once again, we can verify that the state |↓↓ corresponds to |1, −1, as expected. We thus have found three spin states of a two-electron system which behave like the states of a single spin-1 particle. This means that we still have a fourth state which seems not to fit the picture. This fourth state (it must be orthogonal to the three discussed above)
6 Elementary quantum mechanics reads))).To see what kind of state this is,we again apply the operators and Land find that)=10.0).This completes the picture of the possible two-electron spin states:there are three states with s=1,or in other words the s=I state forms a spin triplet,and there is a single s=0 state,a spin singlet. What we have just considered is the simple on-trivial mul spin problen there exists When combining more than two spins,or spins higher thans=.the number of possible total spin states grows rapidly.The problem of figuring out the correct total spin states however.stays of the same complexity and one could treat it in a similar way as we did above.Fortunately,for many combinations of spins this has already been done,and the results can simply be looked up when needed. In Exercises2 and 9 we investigate a two-spin system in an external magnetic field where we also allow for a spin-spin coupling.We use the simplest model for this coupling, sometimes called the Heisenberg model,given by the Hamiltonian Apin=S)S A classical picture which could justify the form of this Hamiltonian is to say that the two 裙magnetie dipole m Section17hih山ccorinto amics,leads toa dipole-dipole e.A fa ed 1p8 interaction is diagonal in the basis of the singlet and triplets,and that it effectively splits the states differing in s by an energy h2J. 1.8 Two-level system:The qubit The simplest non-trivial quantum system has only two levels,and therefore its Hilbert space is two-dimensional.There are many realizations of two-level systems that either its two lowermost states are relevant.An electron with spin-confined in a certain orbital state is a two-level system as well,the two possible states corresponding to spin-up and spin-down. To describe a two-level system,we label the two basis states )and )By definition of the basis,any arbitrary state of the system can be written with two complex numbers. and B.as 1)=a0)+11) (1.79 where la+182=1. Control question.Do you see how the constraint la2+82=1 follows from the normalization condition(1.4)and(1.17)? When neither nor Bequals zero.we say that the system is in a superposition of )and )rather than in just one of these basis states.If we should try to determine in which ba state this two-level system is,there is a chance of finding )and a chance 2of finding 1):a system in a superposition is able to give both outcomes
26 Elementary quantum mechanics reads √ 1 2 {|↑↓ − |↓↑}. To see what kind of state this is, we again apply the operators Sˆ2 tot and Sˆtot,z, and find that √ 1 2 {|↑↓ − |↓↑} = |0, 0. This completes the picture of the possible two-electron spin states: there are three states with s = 1, or in other words the s = 1 state forms a spin triplet, and there is a single s = 0 state, a spin singlet. What we have just considered is the simplest non-trivial multi-spin problem there exists. When combining more than two spins, or spins higher than s = 1 2 , the number of possible total spin states grows rapidly. The problem of figuring out the correct total spin states, however, stays of the same complexity and one could treat it in a similar way as we did above. Fortunately, for many combinations of spins this has already been done, and the results can simply be looked up when needed. In Exercises 2 and 9 we investigate a two-spin system in an external magnetic field, where we also allow for a spin–spin coupling. We use the simplest model for this coupling, sometimes called the Heisenberg model, given by the Hamiltonian Hˆ spin-spin = J Sˆ(1) · Sˆ(2). A classical picture which could justify the form of this Hamiltonian is to say that the two spins both carry a magnetic dipole moment (see Section 1.7.1) which then, according to classical electrodynamics, leads to a dipole–dipole coupling. A favored antiparallel spin configuration would correspond to a positive J. We indeed find in Exercise 2 that this interaction is diagonal in the basis of the singlet and triplets, and that it effectively splits the states differing in s by an energy h¯ 2J. 1.8 Two-level system: The qubit The simplest non-trivial quantum system has only two levels, and therefore its Hilbert space is two-dimensional. There are many realizations of two-level systems that either occur naturally or are artificially made. For instance, a particle confined in a potential well with two separate minima of almost equal depth can be seen as a two-level system if only its two lowermost states are relevant. An electron with spin- 1 2 confined in a certain orbital state is a two-level system as well, the two possible states corresponding to spin-up and spin-down. To describe a two-level system, we label the two basis states |0 and |1. By definition of the basis, any arbitrary state of the system can be written with two complex numbers, α and β, as |ψ = α|0 + β|1, (1.79) where |α| 2 + |β| 2 = 1. Control question. Do you see how the constraint |α| 2 + |β| 2 = 1 follows from the normalization condition (1.4) and (1.17)? When neither α nor β equals zero, we say that the system is in a superposition of |0 and |1 rather than in just one of these basis states. If we should try to determine in which basis state this two-level system is, there is a chance |α| 2 of finding |0, and a chance |β| 2 of finding |1: a system in a superposition is able to give both outcomes
27 1.8 Two-level system:The qubit The notion of superposition forms the basis of the field of quantum information,which proposes to use quantum mechanical systems for all kinds of information processing task. The main idea of this field is to use two-level systems as information-carrving units.so- called quantum bits or qubits.This is both in contrast and similarity to the"classical"bits of a regulat computer Classical bits can be in either of two possible states. in a strir f bits,and the a osle ues ltqnb e 'Apeluns 'pasn sitq jo Jaqumu at ol feuonodod st paois uoneuojut be in the state 0)or )but in addition.in contrast to a classical bit,it can also be in any superposition of the two. A quick way to appreciate the advantage of using quantum information is to discuss the amount of inform ation that can be stored in N qubits.Any state of a qubit can be written as in(1.79)and therefore is haracterized by two comple bers,a and B.or equivalently by four real numbers.These four numbers are not all independent.(1)The normalization condition forms one constraint.fixing a relation between the four numbers. (ii)It turns out that the overall phase factor of a state is irrelevant (the states)and) are equivalent),giving the same expectation values for any observable.This leaves just t pendent real ers to characterize any state qubit.Let us now try to enco ye n in Nelass =64 bits.For the information stored in our qubit we therefore need 2Nelass bits. So far there is no dramatic gain:one could even use two "analogue bits"-bits which n take y real value be een 0 and 1-to store the s nt of information.The advantage becomes clear upor increasing the number of qubits.Fo a set of N qubits,the Hilbert space has dimension 2 and an arbitrary superposition of 2 basis states is thus characterized by 2N complex or 2.2N real numbers. Control question.What is a natural choice of basis states?Think of the states of N classical bits. Again,the normalization ondition and irrelevance of the overall phase factor redu ce this number by two,to 2.(2-1)real numbers.This means we need 2Nelass.(2N-1) classical bits to represent the information stored in N qubits.Thus,the information storage capacity of qubits grows exponentially with N.in contrast to the linear growth for classical dca be e give triking e sic which can be stored ife-time of music-500 times as m -require just nine extra qubits.Let us assume that all useful information accumulated by humankind is stored in~10 books.Well,we could encode all this information in just 43 qubits. In recent years several quantum algorithms have been developed which make use of the advantages of qubits over classical bits.They concem factoring integers,simulating quan- tum systems and searching databases.Using these algorithms,some these tasks could be performed exponentially faster on a quantum computer than on a classical one.Thes discoveries triggered a great interest in the experimental physics of two-level systems,and many physical qubit implementations have been proposed and are being investigated. Let us now focus on the quantum mechanics of a single qubit.The operator of any physical quantity in the two ensional Hilbert space nned by the basis ()0)c
27 1.8 Two-level system: The qubit The notion of superposition forms the basis of the field of quantum information, which proposes to use quantum mechanical systems for all kinds of information processing task. The main idea of this field is to use two-level systems as information-carrying units, socalled quantum bits or qubits. This is both in contrast and similarity to the “classical” bits of a regular computer. Classical bits can be in either of two possible states, “0” or “1.” This suffices to represent any kind of information in a string of bits, and the amount of information stored is proportional to the number of bits used. Similarly, a qubit can also be in the state |0 or |1, but in addition, in contrast to a classical bit, it can also be in any superposition of the two. A quick way to appreciate the advantage of using quantum information is to discuss the amount of information that can be stored in N qubits. Any state of a qubit can be written as in (1.79) and therefore is characterized by two complex numbers, α and β, or equivalently by four real numbers. These four numbers are not all independent. (i) The normalization condition forms one constraint, fixing a relation between the four numbers. (ii) It turns out that the overall phase factor of a state is irrelevant (the states |ψ and eiϕ|ψ are equivalent), giving the same expectation values for any observable. This leaves just two independent real numbers to characterize any state of the qubit. Let us now try to encode in classical bits the information stored in a single qubit. We assume that for any practical purpose an accuracy of 10−19 suffices, which means that any real number can be encoded in Nclass = 64 bits. For the information stored in our qubit we therefore need 2Nclass bits. So far there is no dramatic gain: one could even use two “analogue bits” – bits which can take any real value between 0 and 1 – to store the same amount of information. The advantage becomes clear upon increasing the number of qubits. For a set of N qubits, the Hilbert space has dimension 2N, and an arbitrary superposition of 2N basis states is thus characterized by 2N complex or 2 · 2N real numbers. Control question. What is a natural choice of basis states? Think of the states of N classical bits. Again, the normalization condition and irrelevance of the overall phase factor reduce this number by two, to 2 · (2N − 1) real numbers. This means we need 2Nclass · (2N − 1) classical bits to represent the information stored in N qubits. Thus, the information storage capacity of qubits grows exponentially with N, in contrast to the linear growth for classical information. To give some striking examples: all the music which can be stored in a 80 GB iPod can be written in 33 qubits. A whole life-time of music – 500 times as much – requires just nine extra qubits. Let us assume that all useful information accumulated by humankind is stored in ∼ 108 books. Well, we could encode all this information in just 43 qubits. In recent years several quantum algorithms have been developed which make use of the advantages of qubits over classical bits. They concern factoring integers, simulating quantum systems and searching databases. Using these algorithms, some of these tasks could be performed exponentially faster on a quantum computer than on a classical one. These discoveries triggered a great interest in the experimental physics of two-level systems, and many physical qubit implementations have been proposed and are being investigated. Let us now focus on the quantum mechanics of a single qubit. The operator of any physical quantity in the two-dimensional Hilbert space spanned by the basis {|1, |0} can
28 Elementary quantum mechanics be represented by a 2 x 2 matrix.Explicitly,any Hermitian operator can be written as (中二）=1+++ (1.80 where I is the unit matrix,and the matrices are the Pauli matrices(1.77).The Pauli matri- povide convent basis in the space of mid Solving the eigenvalue equation )=El)for any Hamiltonian now reduces to a 2x 2 linear algebra problem in the basis of and 0). ()()=(8) (1.81) To simplify,we skip in the Hamiltonian the term proportional to 1.which means that we count energy starting from the mean of the two possible energy eigenvalues.Therefore, e is the energy difference (1A)-(0A0),and this defines the diagonal elements of the Hamiltonian in(1.81).Apart from that,there are generally also non-diagonal elements sof this Hamiltonian let us consider a qubit made from two electron states 1)and 0),which are localized near the minima of a double well potential (see Fig.1.2).The energy e is then the difference of the corresponding energy levels,while t accounts for tunneling through the potential barrier separating the states. Control question.Another way to write the Hamiltonian in (1.81)would be (1X11-[0XO])+t1X0+*10X11.Do you see the equivalence? Control question.A spin-electron in a magnetic field can be regarded as a qubit with Hamiltonian H=guaB.G.Let us choose 0)and 1)to be the eigenstates of 6:with eigenvalues+1.What are e and t in terms of B if we write the Hamiltonian in this basis?What are they if )and are the eigenstates of? If the elementgoes tozero.we see that the eigenstates of the Hamiltonian coincide with the basis states 11)and 10).Then.the state ()=11)has energy s/2 and the state (0)=10)has energy -s/2.Generally,when t0.this is not the case.With finite t.the 1) 10 Example of therealization.The qubit is formed from two,and a double well potential.The energy difference(1)-()is denoted tunneling between the two states is enabled by t and t
28 Elementary quantum mechanics be represented by a 2 × 2 matrix. Explicitly, any Hermitian operator can be written as q + z x − iy x + iy q − z = q1 + xσˆx + yσˆy + zσˆz, (1.80) where 1 is the unit matrix, and the matrices σˆ are the Pauli matrices (1.77). The Pauli matrices therefore provide a convenient basis in the space of 2 × 2 matrices and are frequently used in this context. Solving the eigenvalue equation Hˆ |ψ = E|ψ for any Hamiltonian now reduces to a 2 × 2 linear algebra problem in the basis of |1 and |0, 1 2 ε t t ∗ −1 2 ε α β = E α β . (1.81) To simplify, we skip in the Hamiltonian the term proportional to 1, which means that we count energy starting from the mean of the two possible energy eigenvalues. Therefore, ε is the energy difference 1|Hˆ |1−0|Hˆ |0, and this defines the diagonal elements of the Hamiltonian in (1.81). Apart from that, there are generally also non-diagonal elements t and t ∗ present. To illustrate one of the possible physical realizations of this Hamiltonian, let us consider a qubit made from two electron states |1 and |0, which are localized near the minima of a double well potential (see Fig. 1.2). The energy ε is then the difference of the corresponding energy levels, while t accounts for tunneling through the potential barrier separating the states. Control question. Another way to write the Hamiltonian in (1.81) would be Hˆ = ε 2 {|11| − |00|} + t |10| + t ∗ |01|. Do you see the equivalence? Control question. A spin- 1 2 electron in a magnetic field can be regarded as a qubit with Hamiltonian Hˆ = 1 2 gμBB · σˆ. Let us choose |0 and |1 to be the eigenstates of σˆz with eigenvalues ±1. What are ε and t in terms of B if we write the Hamiltonian in this basis? What are they if |0 and |1 are the eigenstates of σˆx? If the element t goes to zero, we see that the eigenstates of the Hamiltonian coincide with the basis states |1 and |0. Then, the state 1 0 = |1 has energy ε/2 and the state 0 1 = |0 has energy −ε/2. Generally, when t = 0, this is not the case. With finite t, the Fig. 1.2 Example of the realization of a qubit. The qubit is formed from two electron states, |0and|1, close to the minima of a double well potential. The energy difference1|Hˆ|1−0|Hˆ|0is denoted by ε, and tunneling between the two states is enabled byt andt ∗.
29 19 Harmonicoscllator eigenvalue equation gives the two eigenstates of A. )= +）片 (1.82) /2E+E±e with (1.83) If we write=,the same solutions allow for a compact presentation, 1 /e0/P(cos0±1) 由=2主cos而（sin9) (1.84 with the angle 6 being defined as tan6 =2tl/e. =(5- (1.85) making the description of the qubit very simple.The dynamics of a qubit are also important. they are illustrated by Exercise 7. 1.9 Harmonic oscillator Another simple system which plays a very important role in quantum mechanics is the prThe simplest mechanical verin of the humonie oco encounter t many times in the t of this D00 most notab y a single particle confined in a parabolic potential.If we restrict ourselves for simplicity to one dimension.we can write according to(1.5)the Hamiltonian describing the particle as +受 (1.86 2m where characterizes the strength of the confinement andp=-ihax. Since the potential energy becomes infinite forxt,the particle cannot move through the ntire e.It thus must be in a bound state,and have a discrete en Let us try to find the of ad the.The trick is to introduce two new operators,which at first sight look a bit strange, =贾(+)=贾(-) (1.87) However,if we express the Hamiltonian(1.86)in terms of these new operators,we see that it becomes very compact, i=na仿+，i=aa (1.88)
29 1.9 Harmonic oscillator eigenvalue equation gives the two eigenstates of Hˆ , |± = 1 2E2 ± + E±ε 1 2 ε + E± t ∗ , (1.82) with E± = ±Equbit = ±! ε2 4 + |t| 2. (1.83) If we write t = |t|eiφ, the same solutions allow for a compact presentation, |± = 1 √2(1 ± cos θ) eiφ/2(cos θ ± 1) e−iφ/2 sin θ , (1.84) with the angle θ being defined as tan θ = 2|t|/ε. We label the energy eigenstates of a qubit as |+ and |−, to distinguish from the basis states |0 and |1. In the energy basis the Hamiltonian of (1.81) is conveniently diagonal, Hˆ = Equbit 0 0 −Equbit , (1.85) making the description of the qubit very simple. The dynamics of a qubit are also important, they are illustrated by Exercise 7. 1.9 Harmonic oscillator Another simple system which plays a very important role in quantum mechanics is the harmonic oscillator: we encounter it many times in the rest of this book, most notably in Chapters 7 and 8. The simplest mechanical version of the harmonic oscillator consists of a single particle confined in a parabolic potential. If we restrict ourselves for simplicity to one dimension, we can write according to (1.5) the Hamiltonian describing the particle as Hˆ = pˆ2 2m + mω2 2 xˆ 2, (1.86) where ω characterizes the strength of the confinement and pˆ = −ih¯∂x. Since the potential energy becomes infinite for x → ±∞, the particle cannot move through the entire space. It thus must be in a bound state, and have a discrete energy spectrum. Let us try to find the eigenenergies of Hˆ and the corresponding eigenstates. The trick is to introduce two new operators, which at first sight look a bit strange, aˆ = "mω 2h¯ xˆ + ipˆ mω and aˆ† = "mω 2h¯ xˆ − ipˆ mω . (1.87) However, if we express the Hamiltonian (1.86) in terms of these new operators, we see that it becomes very compact, Hˆ = h¯ω(nˆ + 1 2 ), nˆ ≡ ˆa†aˆ. (1.88)