7636S ADVANCED QUANTUM MECHANICS Solutions 1 Spring 010 1 Warm up a Show that the eigenvalues of a Hermitian operator A are real and that the eigenkets of A corresponding to dierent eigenvalues are orthogonal b Show that if the state ket α a c a a is normalized then the expansion coecients c a must satisfy a c a 1 a i A number c is shown to be real if c c Let us study Hermitian operator A to whom holds A A and an eigenstate a corresponding an eigenvalue a, such that A a a a Based on evalution of the inner product c a A a a a a the eigenvalue a has the expression a c a a Let us now study what is a? a c ( a A a ( a a a a ( a A a a a a A a a a a Now it has been proven that a Hermitian operator A has real eigenvalues ii To prove that the eigenkets of A corresponding to dierent eigenvalues are orthogonal (ie b a 0, we examine the inner product b A a between two eigenstates a and b corresponding dierent eigenvalues a and b, (a b The inner product can be evaluated two dierent ways: d b A a a b a d b A a b A a b b a In the latter, the hermicity of A is applied Now the above two expression are subtracted from each other 0 (a b b a which implies in case of a b that b a 0
b First of all, ( ( α α c a a c a a c a c a a a c a R, a a a,a a then the normalization condition α α ( α α 1 straight implies that 1 α α c a a The previous proof about orthogonality (aii holds also for a degenerate case, then corresponding an eigenvalue, say, b, we have a set of eigenstates b 1, b,, b j, but anyway all of them are orthogonal to some other eigenstate a corresponding eigenvalue a b Prove the Theorem 1 from lecture notes: If both of the basis { a } and { b } are orthonormalized and complete then there exists a unitary operator U so that b 1 U a 1, b U a, b 3 U a 3, (1 (Unitary operator: U U UU 1 The proof has three stages: construction of operator U, proof of property (1 and proof of unitarity Construction procedure is rather easy, we would like to build an operator that projects the basis state a j to basis state b j : U j b j a j To show that the property (1 holds, operator U operates on an arbitrary basis state a k : U a k j b j a j a k j b j δ jk b k where the orthonormality of basis { a } plays a role The unitarity is checked via brute calculation: UU b j a j ( b i a i b j a j a i b i j i j ij i b j a j a i b }{{} i b j b j 1 j δ ij }{{} completeness of { b }
3 Consider the spin operators S x, S y and S z in the { S z ;, S z ; } basis a Write out the operators S x, S y and S z in the { S z ;, S z ; } basis b Compute the commutators [S x, S y ] and [S, S x ] as well as anticommutator {S x, S y } c Let us dene the ladder operators S ± S x ±is y Compute S ± S z ; and S ± S z ; Let us rst summarize the { S z ;, S z ; } basis represented with the help eigenstates of S x and S y operators with proper phase choice (see eg J J Sakurai, Modern Quantum Mechanics, p 8: S z ; 1 S x ; + 1 S x ; S z ; 1 S y ; + 1 S y ; S z ; 1 S x ; 1 S x ; S z ; i S y ; + i S y ; ( Representation of an operator B in the { S z ;, S z ; } basis B S z ; j S z ; j B S z ; k S z ; k B jk S z ; j S z ; k j k j k and in matrix representation we use the following convention with the indecies B jk S z ; j B S z ; k ( ( B B B Sz ; B S z ; S z ; B S z ; S z ; B S z ; S z ; B S z ; B B a Basis representation for operator S z is after above denitions just the calculation of matrix elements B jk : ( Sz ; S S z z S z ; S z ; S z S z ; S z ; S z S z ; S z ; S z S z ; ( Sz ; S z ; S z ; S z ; ( 1 0 S z ; S z ; S z ; S z ; 0 1 To do the same for S x,y we resort to relations ( and nd out that S x S z ; S z; S y S z ; i S z; S x S z ; S z; S y S z ; i S z; which shows that ( Sz ; S S x x S z ; S z ; S x S z ; ( 0 1 S z ; S x S z ; S z ; S x S z ; 1 0 ( Sz ; S S y y S z ; S z ; S y S z ; ( 0 i S z ; S y S z ; S z ; S y S z ; i 0
b As we now have the representations of operators S i in the S z eigenstate basis we can use them to calculate the (anticommutators (( ( ( ( [S x, S y ] S x S y S y S x 0 1 0 i 0 i 0 1 4 1 0 i 0 i 0 1 0 i S z (The general rule goes [S i, S j ] i ɛ ijk S k, where ɛ ijk is the Levi-Civita permutation symbol Then it happens out that Si /4 for all i x, y, z, therefore S Sx +Sy +Sz 3 /4 and it is clear that [S, S x ] 3 [I, S x ]/4 0 When calculating [S x, S y ] one notices that S x S y S y S x which implies that {S x, S y } 0 c Since matricies are handy objects, let us express ladder opertors S ± also in the familiar { S z ;, S z ; } basis: S ± S x ± is y S + ( 0 1 + ( ( 0 1 0 1 1 0 1 0 0 0 or S ( 0 1 1 0 + ( 0 1 1 0 ( 0 0 1 0 S + S z ; S z ; S S z ; S z ; and operations to S z eigenstates result S + S z ; 0 S S z ; S z ; S + S z ; S z ; S S z ; 0 Now, the physical meaning of the ladder operators can be read Operator S + raises the spin component by and if the spin component cannot be raised further, we get null state Similarly, S lowers the spin component by Both these operators are non-hermitian
4 Prove the Theorem from lecture notes: If T is a unitary matrix, then the matrices X and T XT have the same trace and the same eigenvalues i Trace of a matrix X is the sum of its diagonal elements: Tr (X i X ii and as an reminder the matrix multiplication expressed in index notation goes (AB ij k A ikb kj The unitarity of T has then index expression: T T 1 T T 1 T ik T kj δ ij k k T ik T kj δ ij With these in our mind we are ready to prove the trace invariance: Tr ( T XT (T XT ii T ij X jkt ki i i j k X jk T ki T ij k j i }{{} δ kj k X kk Tr (X ii The matrix X has eigenvalues {a 1, a,, a n } and corresponding eigenvectors { a 1, a,, a n } By constructing a new set of vectors such that b j T a j and evaluating T XT b j T XT T a j T X a j a j T a j a j b j, we observe that b j are the eigenvectors of matrix T XT corresponding the same eigenvalues {a 1, a,, a n } Thus matricies X and T XT has the same eigenvalues if T is a unitary matrix
5 The translation operator for a nite (spatial displacement is given by T (l exp ( i p l where p is the momentum operator and l the displacement vector a Evaluate [x i, T (l] b How does the expectation value x of the position operator change under the translation? a As introduced in lectures the eect of a nite spatial displacement by l is T (l x x + l When evaluating commutator one should remember when x i is an operator and when a pure number, to distinguish these two cases I use now notation ˆx i for operator and x i for number Let us introduce arbitrary state α having presentation in conguration space α dx x x α [ˆx i, T (l] α ˆx i T (l α T (l ˆx i α ˆx i dx x + l x α T (l dx x i x x α dx (x i + l i x + l x α dx x i x + l x α dx l i x + l x α l i T (l α Now we say that [ˆx i, T (l] l i T (l, which is an operator identity since it holds for an arbitrary state Furthermore, we know by generalising the result that ˆxT (l T (lˆx lt (l or ˆxT (l T (l(l + ˆx b Before the translation, the expectation value of position operator ˆx for an arbitrary state is x α ˆx α and after the translation α l ˆx α l α T (lˆxt (l α α T (lt (l(l + ˆx α which is not any big surprise α l + ˆx α l + x,