Coordinate and Momentum Representation. Commuting Observables and Simultaneous Measurements. January 30, 2012

Transcription

1 Coordinate and Momentum Representation. Commuting Observables and Simultaneous Measurements. January 30,

2 Coordinate and Momentum Representations Let us consider an eigenvalue problem for a Hermitian operator Â:  n = A n n and n m = δ nm In the basis of its own eigenstates the operator  acts as a multiplication operator. In the above example the operator  has a discrete spectrum and the eigenstates can be normalized. However, in many cases we have to consider operators with continuous spectrum, e.g. r and ˆp, when the eigenstates cannot be normalized and we have to use delta-function normalization:  α = α α and α α = δ(α α ) The basis { α } defines α-representation. 2

3 The operators in α-representation are continuous matrices, i.e. they act as integral operators. Also α Ψ = Ψ(α) can be viewed as a wave function in α-representation. The eigenvalue problem for operator  in α-representation reads: α  α α Ψ dα = A α Ψ 3

4 Consider coordinate operator ˆr. The eigenvalue problem can be expressed as: ˆr r = r r and r r = δ(r r ) Thus r r = δ(r r ) = ψ r (r ) is an eigenfunction of r. Note. The vector r is not strictly a ket-vector because ψ r (r ) is not normalizable. However the bra-vector r (linear functional) is well defined, and r Ψ = Ψ(r) represents a conventional wave function, which can be viewed as a state-vector in coordinate representation. Moreover, we can deal with a physical δ-function, i.e. with very sharp function with small but finite width defined by physical factors, e.g. resolution of a measuring device etc. An example of such a function is: (2π) 3/2 a 3 exp( r 2 /a 2 ). 4

5 The Schrödinger equation ĤΨ(r) = EΨ(r), where Ĥ = 2 2m 2 + V (r) is an eigenvalue problem in the coordinate representation. How this equation will be modified if instead of coordinate we use another continuous dynamical variable such as momentum p? To transform the eigenvalue problem to the momentum representation we use: where r = p p r dp, (1) p r = r p = ψ p(r) = (2π ) 3/2 exp( ipr/ ). Here we used our knowledge of the momentum operator eigenfunction ψ p (r) ( De Broglie wave). 5

6 Ψ = r r Ψ dr = p p r r Ψ drdp = p p Ψ dp Here p Ψ ϕ(p) = p r r Ψ dr = 1 (2π) 3/2 e ipr/ ψ(r)dr is the wave function in the momentum representation. ˆr Ψ = Now we use that r r r Ψ dr = r p r = i p p r p r p r r Ψ drdp and r p r r Ψ dr = i p p r r Ψ dr = i p p Ψ 6

7 ˆr Ψ = r r r Ψ dr = p (i p ) p Ψ dp, i.e. in momentum representation ˆr = i p Similarly ˆp Ψ = r ( i ) r Ψ dr = p p p Ψ dp In momentum representation any function f(ˆp) will be a multiplication operator f(p) and any function f(ˆr) = f(i p ). Thus Schrödinger equation in momentum representation can be expressed as: p 2 ϕ(p) + V (i )ϕ(p) = Eϕ(p) (2) 2m 7

8 Commuting Observables and Simultaneous Measurements Since Hermitian operators represent physical observables it is natural to use their eigenstates as a basis in Hilbert Space. But, sometimes this basis cannot be completely specified because of the repeated eigenvalues. Then we have to find another compatible observable to eliminate this ambiguity and to label the basis using the eigenvalues of both operators. In other words, we have to find a complete set of observables characterizing the system. Example: Consider a 1D particle in free space: d 2 Ĥ = 2 2mdx 2 8

9 The eigenvalues E of this Hamiltonian are two-fold degenerate and, therefore, the ket-vectors E do not completely specify the basis. To overcome this difficulty we choose such E that are also eigenstates of the momentum operator ˆp x = i d/dx: x E, + =(2π ) 1/2 exp(ip x x/ ) x E, =(2π ) 1/2 exp( ip x x/ ), where p x = 2mE. In this example: The eigenvalues of Ĥ are degenerate because of the parity symmetry x x and momentum conservation. We choose another operator, ˆp x, which commutes with but does not possess the parity symmetry. Ĥ 9

10 Now we want to answer a question: Under what conditions two observables  and ˆB possess a complete set of common eigenvectors? These two observable are called compatible or simultaneously measurable. The eigenvectors AB of the complete set are characterized by two sharp values of the observables  and ˆB. Necessary Condition. If the complete set AB exists the operators must commute. We require:  AB = A AB ˆB AB = B AB Operating on the second equation by Â, on the first equation b ˆB and subtracting the first equation from the second one, we obtain: 10

11 ( ˆB ˆBÂ) AB = (BA AB) AB = 0 for every ket-vector of the complete set AB. Therefore [Â, ˆB] = ( ˆB ˆBÂ) = 0 Sufficient Condition. Let us suppose that [A, B] = 0 and consider a repeated eigenvalue A of  such that:  A i = A A i i = 1,..., r. Lets operate on each of these equations with commutation condition: ˆB and use the ˆB A i =  ˆB A i = A ˆB A i i = 1,..., r. 11

12 or Â( ˆB A i ) = A( ˆB A i ) (3) Eq. (3) means that ˆB A i is also an eigenstate of Â. Therefore ˆB A i must be a linear combination of r-eigenvectors A i : ˆB A i = r j=1 A j ˆB A i A j (4) Eq. (4) shows that A i is not necessarily an eigenstate of ˆB. However this eigenstate can be easily constructed by changing the basis in the r-dimensional subspace spanned by A i. We construct a new basis: A k B = i S ki A i (5) such that ˆB A k B = B A kb (6) 12

13 Substituting Eq. (5) into Eq. (6) and applying A j to both sides we obtain: A j ˆB A i S ki = B S kj (7) i i This is a linear homogeneous system of equations which has a non-trivial solution if and only if its determinant is zero det( A j ˆB A i Bδ ji ) = 0 (8) The value of B and the unitary matrix S ki can be found the same way as previously. This concludes the procedure of finding a complete set of observables. 13