1 Heisenberg Representation

Transcription

1 1 Heisenberg Representation What we have been ealing with so far is calle the Schröinger representation. In this representation, operators are constants an all the time epenence is carrie by the states. We have Ψ(t) >= U(t) Ψ(0) > where A matrix element of an operator is then U(t) = exp( ī h Ht) < Ψ(t) O Ψ(t) > where O is an operator constructe out of position an momentum operators. To contrast the Schröinger representation with the Heisenberg representation (to be introuce shortly) we will put a subscript on operators in the Schröinger representation, so we have X S, P S, an O S. We may then write our matrix element as < Ψ(t) O S Ψ(t) >=< Ψ(0) U (t)o S U(t) Ψ(0) > The Heisenberg representation uses time epenent operators an constant in time states. We efine the Heisenberg operator by O H (t) = U (t)o S U(t) The two representations are clearly completely equivalent, an it is a matter of convenience which one is use in a given problem. Once we have efine Heisenberg operators, we may stuy their equations of motion an compare to the corresponing classical equations of motion. We will carry through the iscussion for a = 1 system where the x coorinate ranges to +. We start with the classical iscussion. There is a Lagrangian L = m (x t ) V (x), with generalize momentum p = L ẋ where ẋ = x/t. From the Lagrangian we construct a classical Hamiltonian, H c = pẋ L = p m + V (x) From the classical Hamiltonian, we get classical equations of motion, ẋ = H c p = p m ṗ = H c x = V x

2 Quantizing the system involves introucing operators X S, an P S, which satisfy The quantum Hamiltonian is [X S, P S ] = i. an the Heisenberg operators H = P SP S m + V (X S), X H (t) = exp( ī h Ht)X S exp( ī h Ht), an P H (t) = exp( ī h Ht)P S exp( ī h Ht). The quantum or Heisenberg equations of motion are t X H(t) = ī h [H, X H(t)] an t P H(t) = ī h [H, P H(t)]. Note that H being constant is the same in Heisenberg an Schröinger representations, H = P SP S m + V (X S) = P HP H m + V (X H) Now consier the Heisenberg equation for X H. We have t X H(t) = ī h [H, X H(t)] = ī h U (t)[h, X S ]U(t) Now [H, X S ] = [ P SP S m + V (X S), X S ] = [ P SP S m, X S] + [V (X S ), X S ] Since every operator commutes with itself, an we are left with [V (X S ), X S ] = 0, [ P SP S m, X S] = 1 m (P SP S X S X S P S P S ) = P S i m, where we use [X S, P S ] = i. Putting the pieces together, we have t X H(t) = ī h U (t) P S i m U(t) = P H(t) m (1)

3 which is the same as the classical result for x/t. Now let us get the Heisenberg equation of motion for P H. We have t P H(t) = ī h [H, P H(t)] = ī h U [V (X S ), P S ]U, where we use the fact that P S commutes with itself. To get [V (X S ), P S ], we first take a matrix element an write < x [V (X S ), P S ] Ψ >=< x (V (X S )P S P S V (X S )) Ψ > = V (x) i x < x Ψ > i ( xv (x) < x Ψ >) = i ( xv (x)) < x Ψ >= i < x V (X S) Ψ > So we have < x [H, P S ] Ψ >= i < x V (X S) Ψ > Since < x an Ψ > were arbitrary, we can write the operator equation [H, P S ] = V (X S) Returning to the equation for P H /t, we now have t P H(t) = U (t) V (X S) U(t) = V (X H) X H, () which again is the quantum version of the classical equation for p/t. It is generally true in a quantum system that the Heisenberg equations of motion for operators agree with the corresponing classical equations. An important example is Maxwell s equations. These remain true quantum mechanically, with the fiels an vector potential now quantum (fiel) operators. Application to Harmonic Oscillator In this section, we will look at the Heisenberg equations for a harmonic oscillator. The notation in this section will be O(t) for a Heisenberg operator, an just O for a Schröinger operator. In terms of the notation of the previous section we have O S = O, an O H (t) = O(t). Of course we have O(0) = O. The Hamiltonian for the oscillator is H = P P m + mω 0X, (3) where ω 0 is the natural frequency of the oscillator. The equations of motions for the Heisenberg operators are as follows, X(t) t = P (t) m, P (t) t = mω 0X(t)

4 As always, the Heisenberg equations for operators are the same as the classical equations of motion. Taking a secon time erivative, we have X(t) t = 1 P (t) m t = ω 0X(t) This is a ifferential equation for X(t). We can certainly solve it as a linear combination of sin ω 0 t an cos ω 0 t. We can write X(t) = A cos ω 0 t + B sin ω 0 t, where A, B are operators which are inepenent of t. At t = 0, we must have X(0) = X, so we get that A = X. Also from Eq.(3), we have so We finally have Similar steps lea to X(t) t=0 = P t m = ω 0B, B = P. X(t) = X cos ω 0 t + P sin ω 0 t. (4) P (t) = P cos ω 0 t = X sin ω 0 t. (5) Matrix Elements an Energy Levels The expressions for X(t), P (t) look so much like their classical counterparts, it might seem unlikely that they contain information about energy levels an matrix elements. As will be seen, they in fact contain a large amount of such information. Suppose the Hamiltonian has an eigenstate n > with energy E n an another eigenstate n > with energy E n. Let us write out the matrix element of X(t). We have < n X(t) n >=< n exp( iht )X exp( iht ) n >= exp(i(e E n n )t) < n X n >. (6) This is the matrix element of the left han sie of Eq.(4). Using exponential forms for sin an cos we have for the matrix element of the right han sie, 1 ( ) e iω0t (< n X n > i < n P n >) + e iω0t (< n X n > +i < n P n >). (7) Eqs.(6) an (7) must match in etail. Let us start with the case E n > E n. Then comparing time epenent factors on both sies, the coefficient of exp( iω 0 t) must vanish, since there is no term of that form in Eq.(6). This gives < n X n > +i < n P n >= 0. (8)

5 Now matching the coefficient of exp(iω 0 t, we also must have exp( i(e n E n)t) ) = exp(iω 0 t) This result implies that if < n X n > 0, then E n = E n + ω 0. Let us assume such a state exists. Now apply the same argument to the matrix element < n X(t) n >. We fin again, if E n > E n that E n = E n + ω 0, provie only that < n X n > 0. Using this argument repeately, we fin a sequence of levels E n, E n + ω 0, E n + ω 0,..., E n + lω 0, with no upper limit on l, provie only that the matrix element of X between a state an the next one higher up is non-vanishing. We will return to the question of which matrix elements of X are non-vanishing later in this section. Before oing that, we explore the case where E n < E n. Returning to Eqs.(6) an (7), this time the coefficient of exp(iω 0 )t must vanish, giving < n X n > i < n P n >= 0. (9) The coefficient of exp( iω 0 t) must match on both sies, so we get E n = E n ω 0, so we have foun a state with energy one unit of ω 0 lower, provie that < n X n > 0. At first sight it woul seem that this process coul be repeate inefinitely. However, this woul eventually result in the energy eigenvalue going negative. But this is impossible. Consier the expecte value of H in a state Ψ >. Writing this out in Hilbert space notation instea of Dirac notation, we have (Ψ, HΨ) = 1 m (Ψ, P P Ψ) + 1 mω 0(Ψ, XXΨ) = 1 m (P Ψ, P Ψ) + 1 mω 0(XΨ, XΨ). (10) Each term on the right sie of Eq.(10) is certainly positive, regarless of the particular state. The consequence is that H can have no negative eigenvalues. This means that the argument we have been using cannot generate an infinite sequence of levels below a given one, but it is allowe to generate an infinite sequence above a given one. The consequence is that there must be a lowest state, which we enote as 0 >. It is then natural to esignate the sequence of levels upwar from 0 > as n >, where E n = E 0 + nω 0,

6 where n is any positive integer. Finally, let us explore the question of which matrix elements of X an P are nonvanishing. We start with the lowest state, 0 >. By the iscussion just given, both X an P, acting on 0 >, must lea to a state of higher energy. Let us take matrix elements between 0 > an 1 >. This puts us in the case covere by Eq.(8) so we have < 1( X + i P ) 0 >= 0. (11) This relates matrix elements of X an P. To get a more powerful result, we consier matrix elements between 0 > an n > for n > 1. It is easy to see that all such matrix elements of X an P must vanish, since the matrix elements < n X(t) 0 > an < n P (t) 0 > have time epenence exp(inω 0 )t). But from Eqs.(4) an (5), the only allowe frequency for the matrix element of a state higher in energy is ω 0. ( This result hols more generally, namely the matrix elements of X an P are only non-vanishing between a state an the next one up or own in energy. ) Now consier the expression (X + i P ) 0 > From Eq.(11) it has no matrix element to 1 >, an since the matrix elements of both X an P vanish for any higher state, we have that (X + i P ) 0 >= 0 (1) We can apply any operator to Eq.(1) an still get a vanishing result, so we also have Multiplying this out we get { (X i P )(X + i P ) 0 >= 0 (13) XX + } P P ( ) + i [X, P ] 0 >= 0 (14) Supplying a factor mω 0/ an using [X, P ] = i, we finally have or (H ω 0 ) 0 >= 0 (15) E 0 = ω 0 To summarize, we have foun the entire sequence of energy levels of the harmonic oscillator. It is also easy to show that X an P only have matrix elements between n > an n ± 1 >, except for n = 0, where the only non vanishing matrix elements are to 1 >. A little more work of the same sort one above will eliver the actual values of these matrix elements. This exercise shows that Heisenberg methos can be quite powerful in certain cases.