Lecture 15 (Nov. 1, 2017)

Transcription

1 Leture Quantum Theor I, Fall Leture 5 (Nov., Charged Partile in a Uniform Magneti Field Last time, we disussed the quantum mehanis of a harged partile moving in a uniform magneti field B = Bẑ. We deomposed the Hamiltonian as with The Hamiltonian H d has spetrum p H = z + H d, (5. m E (d Π x + Π H d = (5. m. ( n = ω n +, (5.3 with the lotron frequen given b ω =. (5.4 m We then wanted to determine the degenera of these energ levels, alled Landau levels. We defined the guiding enter oordinates, R x := x + Π, R := Π x. (5.5 These variables are onjugate, up to a multipliative fator, [R x, R ] = i = ilb, (5.6 and both ommute with the Hamiltonian. As we showed on the homework, this implies that the Hamiltonian is degenerate. Here, l B is the magneti length, whih is the length sale suh that if we ut out a irular disk of radius l B orthogonal to the magneti field, then the flux through the disk is on the order of one flux quantum: h πbl B = e h = Φ 0, (5.7 e where Φ 0 is the flux quantum. The guiding enter oordinates are the oordinates of the enter of the lotron orbit, whih are a onstant of the motion. This statement is equivalent to the statement that the guiding enter oordinates ommute with the Hamiltonian. In the quantum theor, the size of the orbit is set b the onl length sale in the problem, the magneti length. We expet that the degenera is the number of flux quanta passing through the sample, BA Φ, where A is the area of the sample. Now we will alulate the degenera expliitl. We pik a partiular gauge to work in, for onveniene. We hoose A = Bx, A x = 0. (5.8 This is known as Landau gauge. In this gauge, p x ( p x m H d = m + = p x m + ( mω p (5.9 x.

2 Leture Quantum Theor I, Fall In this gauge, [p, H] = 0, so we an label states b their p eigenvalue. For an fixed p, we get a one-dimensional simple harmoni osillator of frequen ω, whih reprodues the spetrum E (d n = ω (n +. (5.0 The energ eigenfuntions take the form ( ψ n,p = e ip/ p φ n x. (5. This is a plane wave multiplied b a shifted SHO wavefuntion. It is now lear that states with different values of p will be degenerate. The degenera question then beomes the question of how man distint values of p exist for a sstem of a given area. Consider a sample of lengths L x and L along the x- and -diretions, respetivel, with periodi boundar onditions along the -diretion. We then must have whih gives us quantized momentum modes along the -diretion, e ipl/ =, (5. p = πm, m Z. (5.3 L ( p Eah of the degenerate wavefuntions ψ (x, has a modulus φ x n,p n and is entered at ( ( πm x =. (5.4 L However, we annot arbitraril shift these SHO groundstate wavefuntions, beause the sample has a finite length L x in the x-diretion. Thus, the -diretion momentum is quantized beause of the finite length in the -diretion, and has a finite number of possible values beause of the finite length in the x-diretion. The number of independent states is then L x g = ( ( π L = L x L π = A h = BA, (5.5 Φ 0 as we expeted. In lassial mehanis, we an prove that there is no diamagnetism. mehanis, everthing is determined b the partition funtion, Z ˆ dxi dp i (π 3 e β ( i In the presene of a magneti field, the Hamiltonian hanges, Z[A] ˆ dxi dp i (π 3 e β ( i In lassial statistial p i +V ({x m i }. (5.6 (p i e A(x i +V ({x m i }. (5.7 However, we an simpl shift the momenta as p i = p i e A(x i, whih then gives us Z[A] = Z[A = 0]. Thus, lassiall, the magneti field does not affet the statistial mehanis. However, we have just seen that in quantum mehanis, there is an effet. This effet is purel quantum mehanial.

3 Leture Quantum Theor I, Fall We might wonder how a two-dimensional problem of motion in the x, -plane turned into a one-dimensional SHO. The eletron is loalized around the guiding enter oordinates. However, we see that the two oordinates do not ommute with one another. Thus, there is onl one real oordinate in the problem, and the other oordinate is its onjugate variable. This is wh we are onl solving a one-dimensional problem at the end of the da. Note that beause R x and R do not ommute, we annot be in a simultaneous eigenstate of both, so the loation of guiding enter itself must be smeared out. 5. Composite Sstems We now onsider omposite sstems, whih are sstems omposed of two (or more subsstems. Suppose we have two sstems A and B, with Hilbert spaes H A and H B, respetivel. We want to obtain the Hilbert spae of the omposite spae A + B, whih is alled the tensor produt of H A and H B. Consider a basis {φ i } of H A and a basis {χ j } of H B. We will define new vetors φ i χ j that live in a new Hilbert spae, denoted H A+B = H A H B. These φ i χ j will provide a basis for the Hilbert spae H A+B, meaning that an state in H A+B an be deomposed as ψ A+B = ψ ij φ i χ j. (5.8 i,j Often times, we will omit the tensor produt smbol for onveniene, when the meaning is lear. Thus, we will write φ i χ j = φ i χ j. We an then write down operators on H A+B and formulate the quantum theor as we have before. There are lasses of operators that at on onl one of the two subsstems, suh as O = O A B, but in general we an have operators that at nontriviall on both subsstems at one. Consider, in partiular, a sstem of two spin- partiles A and B. Some possible states in the omposite Hilbert spae an be written as a single tensor produt, suh as [ ] A B, A B, A ( B + B. (5.9 There are learl an infinite number of suh states. There are also more interesting states (entangled states that annot be written as a single tensor produt, suh as ( A B A B. (5.0 Some of the operators on this Hilbert spae at onl on one of the subsstems, suh as σa z B := σa z (the seond wa of writing this is a notational onveniene. However, there are more general operators that at nontriviall on both subsstems simultaneousl, suh as σa x σ x B := σ A σ B. 5.. Quantum Entanglement Two subsstems A and B of a quantum mehanial sstem an be entangled with one another. For example, onsider again the sstem of two spin- partiles A and B. Some states, suh as A B, A B, (5. have the propert that if we restrit our view to onl one of the subsstems, then the sstem is in a well-defined state of that subsstem. B ontrast, if we onsider a state suh as ( A B ± A B, (5.

4 Leture Quantum Theor I, Fall then we see that neither spin b itself is in a definite quantum state, despite the fat that the omposite sstem is in a definite state. This is an entangled state. Entanglement is a statement about the relation of subsstems of a sstem to one another. It does not make sense to ask if the state of a sstem is entangled. Instead, we an make statements about whether two subsstems are entangled with one another. If we have subsstems A and B of some omposite sstem, then an unentangled state is preisel one that an be fatored as a tensor produt of states of the two subsstems, An entangled state is one that annot be fatorized in this wa. ψ A+B = ψ A ψ B. (5.3

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