Physics 505 Homework No. 1 Solutions S1-1

Transcription

1 Physics 505 Homework No s S- Some Preliminaries Assume A and B are Hermitian operators (a) Show that (AB) B A dx φ ABψ dx (A φ) Bψ dx (B (A φ)) ψ dx (B A φ) ψ End (b) Show that AB [A, B]/2+{A, B}/2 where the anticommutator {A, B} AB +BA Further, show that the anticommutator is Hermitian and the commutator is anti- Hermitian (that is, [A, B] [A, B]) We know that expectation values of Hermitian operators are real What can you say about the expectation value of an anti-hermitian operator? [A, B]/2 + {A, B}/2 (AB BA)/2 + (AB + BA)/2 AB {A, B} (AB) + (BA) B A + A B {A, B}, so the anticommutator is Hermitian [A, B] (AB) (BA) B A A B (AB BA) [A, B], so the commutator is anti-hermitian Now let A be an anti-hermitian operator Then A dx ψ (Aψ) Also A dx (Aψ) ψ dx ψ (Aψ) which means that A is pure imaginary End

2 Physics 505 Homework No s S-2 2 Generalized Uncertainty Principle Let A and B be Hermitian operators Define the uncertainty in A by the square root of the mean square deviation from the mean: A (A A ) 2 A 2 A 2 Show that A B [A, B] 2 Hint: Use the Schwarz Inequality and the results from problem Comment: The commutator of x and p x is i h, so x p x h/2 Define new Hermitian operators A A A and B B B Then by the Schwarz inequality, A 2 B 2 A B 2, or A B A B [A, B ] /2 + {A, B } /2 [A, B ] /2 Since the expectation value of the commutator is imaginary and the anticommutator is real, each makes a positive contribution to the absolute value, and the anticommutator can be dropped without changing the inequality in the last step So, A B [A, B ] /2 [A, B] [A, B ] [ A, B] + [ A, B ] /2 [A, B] /2 A and B are just numbers, so they commute with the operators and the commutators involving them are 0 End

3 Physics 505 Homework No s S-3 3 We saw in lecture that the eigenfunction of the momentum operator with eigenvalue p is f p (x) (/ 2π h) exp(ipx/ h) We are working in one dimension here and we are assuming (almost) the δ-function normalization described in lecture (The extra h has been inserted for convenience below) We also stated that the eigenfunctions of a Hermitian operator form a complete set This means that for any wave function, we should be able to write ψ(x) c p f p (x), p where c p are the expansion coefficients Since p is a continuous variable, the coefficients become a function and we should write: ψ(x) + ϕ(p)e ipx/ h dp 2π h, where ϕ(p) plays the role of the expansion coefficients Either ψ(x) or ϕ(p) is suitable for describing the state of the system If we are using ψ(x), we are using a position space or configuration space description If we are using ϕ(p), it s a momentum space description (a) Show how to determine ϕ(p) from ψ(x) (The above already gave an expression for ψ(x) in terms of ϕ(p)) We take the dot product of f p (x) with ψ(x) f p ψ dx ϕ(p) 2π h e ipx/ h ψ(x) dx e ipx/ h dp ϕ(p ) e ip x/ h 2π h 2π h dp ϕ(p ) dx p)x/ h 2π h ei(p dp ϕ(p )δ(p p) The configuration and momentum space wavefunctions are just Fourier transforms of each other! End

4 Physics 505 Homework No s S-4 (b) What is the meaning of p2 p ϕ(p) 2 dp? It s the probability that a measurement of the momentum produces a result between p and p 2 (Remember, the square of an expansion coefficient is the probability of being in the corresponding state) End (c) What are the position and momentum operators in momentum space? (You can probably make a good guess, but justify your answer!) We know the operators in configuration space, so let s operate in configuration space and transform to momentum space f p xψ h i dx e ipx/ h xψ(x) 2π h dx e ipx/ h x dp ϕ(p ) e ip x/ h 2π h 2π h dp ϕ(p ) dx x p)x/ h 2π h ei(p dp ϕ(p ) dx h p)x/ h i p 2π h ei(p ( dp h ) ) dx p)x/ h i p ϕ(p 2π h ei(p ( dp h ) ) δ(p p) i p ϕ(p p ϕ(p) integration by parts The position operator is ( h/i)( / p) as you might have guessed By doing essentially the same thing but operating with ( h/i)( / x) we would find that the momentum operator is multiplication by p, also as you might have guessed Note that it s still the case that [x, p] i h End

5 Physics 505 Homework No s S-5 4 Projection operators It s often the case that we want to find the component of a function parallel to another function We just take the dot product with the second function, but then we also need to multiply by the second function A handy notation is ψ ψ This projects onto ψ Operating on ϕ, we get ψ ψ ϕ, which is what we want! Remember ψ ϕ is just a number and ψ is the vector Similarly, operating on ϕ we get ϕ ψ ψ, which is the desired expression for the adjoint vector Suppose you have a complete set of orthonormal basis vectors ψ n What is a compact expression for transforming an arbitrary vector ϕ into this basis set? (This is much easier to write down than to ask!) We want to project ϕ onto each basis vector (this gives the expansion coefficients) and then sum the coefficients times the basis vectors: ϕ ψ n ψ n ϕ n End

6 Physics 505 Homework No s S-6 5 We often need to exponentiate a matrix! As an example, let A be the 2 2 matrix A Compute e ita using two different methods: (a) Use the Taylor expansion for the exponential: e ita (ita) n n! n0 Note that A 2 ( ) ) ( So the even terms in the Taylor expansion involve the identity matrix and the odd terms involve A e ita t 2 /2! + t 4 /4! + ia(t t 3 /3! + t 5 /5! ) cos t + i sint cos t i sin t i sin t cos t End (b) Use the spectral decomposition of A: write A 2 k λ k ψ k ψ k with ψ k ψ l δ kl and use e ita 2 e itλ k ψ k ψ k k Hint: ψ k are just two element column vectors that are eigenvectors of the matrix A The first thing we need to do is find the eigenvalues and eigenfunctions The eigenvalue equation for this matrix is ( or ( λ λ ) a λ b a b ) a 0 b This has a solution (that s not all 0s) only if the determinant vanishes Taking the determinant and setting it to zero we get the characteristic equation λ 2 0,

7 Physics 505 Homework No s S-7 with solutions λ ± and eigenfunctions (eigenvectors) belonging to λ ±, So e ita e it /2 /2 + e it /2 /2 /2 /2 /2 /2 ( (e it + e it )/2 (e it e it ) )/2 (e it e it )/2 (e it + e it )/2 cos t i sin t i sin t cos t End ( ) / 2 ±/ 2