PerturbationTheory.nb 1 Quantum Mechanics Perturbation Theory Time-Independent Perturbation ◼ A Toy Model of Qubit ◼ General Ideas of Perturbation Theory Perturbation theory: a set of approximation schemes that allows us to extend our knowledge about an exactly solvable quantum system to its vicinity (in the parameter space). Why do we need perturbation theory? " Exact solutions are rare. ⇒ We have to rely on perturbation theory to go beyond them and make the most of these exact solutions. " Separation of scales: physics often takes place at different energy scales, e.g. … → quarks (GeV) → nucleus (MeV) → atoms (eV) → molecules (100meV) → … ⇒ We can refine our descriptions by adding perturbative corrections progressively. What are the central problems in perturbation theory? " Time-independent perturbation: given the spectrum (eigenstates and eigenenergies) of H0, find the spectrum of H = H0 + λ V, in power series of λ (given that λ is small). -ⅈ H0 t " Time-dependent perturbation: given the (bare) propagator U0(t) = e , find the propa- gator t . U(t) = + exp -* , -t H(t ) , 0 (1) H(t) $ H0 % & V (t), in power series of & (given that & is small). [We will explain the notations in Eq. (1) later.] How is perturbation theory useful? " Conceptually: to establish effective Hamiltonians, to analyze renormalization group flows " " Practically: to calculate scattering amplitudes, response functions, spectral weights " ! Qubit Model and its Exact Solution Let us start with a toy model of a single qubit. Consider H(&) $ H0 % & V, (2) PerturbationTheory.nb 2 z x where H0 $ / and V $ / , s.t. H(&) can be more explicitly written as H(&) $ /z % & /x 1 & (3) 0 . & -1 What are the eigenenergies and eigenstates of H(&)? H(&) 123(&)4 $ E3(&) 123(&)4. (4) " Eigenenergies 2 E3(&) $ 3 1 % & . (5) " Eigenstates 51 % 1 % &2 6 174 % & 184 12%(&)4 $ , 2 51 % &2 % 1 % &2 6 (6) 51 % 1 % &2 6 184 - & 174 12-(&)4 $ . 2 51 % &2 % 1 % &2 6 ! Taylor Expansion Assuming & is small (i.e. & 9 1), Eq. (5) and Eq. (6) can be expanded in power series of & " As a reminder, the Taylor expansion of a function f (&) is given by ; k <& f (0) f (&) $ : &k = k$0 k (7) f >(0) f (3)(0) $ f (0) % f .(0) & % &2 % &3 % ". 2 6 " Applying Eq. (7) to Eq. (5) and Eq. (6), we get &2 &4 E3 $ 3 1 % - % " , (8) 2 8 & &2 3 &3 11 &4 12%(&)4 $ 174 % 184 - 174 - 184 % 174 % ", 2 8 16 128 (9) & &2 3 &3 11 &4 12-(&)4 $ 184 - 174 - 184 % 174 % 184 % ". 2 8 16 128 Goal: obtain these power series without first calculating the exact solution! - This is possible as n n long as we know how to evaluate the derivatives <& E(0) and <& 123(0)4. The perturbation theory is essentially an iterative algorithm to calculate these derivatives order by order, based on our knowledge about H0 and V. PerturbationTheory.nb 3 The perturbation theory is essentially an iterative algorithm to calculate these derivatives order by order, based on our knowledge about H0 and V. ! Non-Degenerate Perturbation Theory ! Problem Setup The starting point is the following Hamiltonian (linearly parameterized by &) H(&) $ H0 % & V. (10) " This implies H(0) $ H0, <& H(0) $ V, (11) 2 3 <& H(0) $ <& H(0) $ " $ 0. " To simplify the notation, we will suppress the argument & if it is evaluated at & $ 0. H $ H0, <& H $ V, (12) 2 3 <& H $ <& H $ " $ 0. Rule: Everything is treated as a function of &, like f (&). But if the dependence on & is not explic- itly spelt out, we assume it to be the function evaluated at & $ 0, i.e. f ? f &$0 $ f (0). Consider the eigen equation H(&) 1n(&)4 $ En(&) 1n(&)4. (13) For each given &, there is a different H(&), and hence a different set of En(&) and 1n(&)4, labeled by n $ 1, 2, 3, ". " En(&) is the nth energy level. It is a real number depending on &. " 1n(&)4 is the nth eigenstate (in correspondence to En(&)). It is a state vector in the Hilbert space that can change with &. Note: The notation 1n(&)4 does not imply that the index n is & depen- dent, it should be understood as 1 (&)4 $ 2 (&) 1 4 n : nm m . (14) m We assume a discrete spectrum without degeneracy, such that the “nth” level/state is uniquely defined. [The case with degeneracy will be discussed latter.] Statement of the problem: suppose we know the eigenenergies and eigenstates at and only at & $ 0, H 1n4 $ En 1n4, (15) and we also know what the perturbation is: V $ <& H, PerturbationTheory.nb 4 Vmn $ @mA V 1n4 $ @mA <& H 1n4, (16) calculate En(&) and 1n(&)4 in power series of & (to any desired order) in terms of En, Vmn and 1n4. ! Hellmann-Feynman Theorems " Applying <& to both sides of H 1n4 $ En 1n4, <& H 1n4 % H 1<& n4 $ <& En 1n4 % En 1<& n4. (17) " Note: 1<& n4 stands for the derivative of the state 1n4 (not the index n) 1<& n4 $ :<& 2nm(&) 1m4 . (18) m &$0 It does not imply that the integer index n can be differentiated. " Overlap with @mA from the left, @mA <& H 1n4 % @mA H 1<& n4 $ <& En @m n4 % En @m <& n4. (19) Using @mA H $ @mA Em, @mA <& H 1n4 % Em @m <& n4 $ <& En @m n4 % En @m <& n4 (20) ! @mA <& H 1n4 $ <& En @m n4 % (En - Em) @m <& n4. " Note that <& H $ V and @m n4 $ Bmn, Vmn $ @mA V 1n4 $ <& En Bmn % (En - Em) @m <& n4. (21) This establishes a relation between the matrix element Vmn (of the perturbation) and the deriva- tives <& En and 1<& n4. " When m $ n, Eq. (21) implies <& En $ Vnn. (22) This is the first Hellmann-Feynman theorem. " When m C n, Eq. (21) implies Vmn @m <& n4 $ , En - Em (23) Vmn @<& m n4 $ . Em - En This is the second Hellmann-Feynman theorem. Tip: the energy denominator is always given by the energy of the state that is being differentiated minus the energy of the other state. The Hellmann-Feynman theorems tell us how the derivative of the energy <& En or the state 1<& n4 and @<& mA on the left-hand-side is related to something on the right-hand-side which does not contain <&. ! This effectively reduces the order of <& by one. ! Applying them itera- tively, we will be able to calculate the derivatives to any order for both energies and states, which are all we need to construct the power series of the perturbative expansion. PerturbationTheory.nb 5 The Hellmann-Feynman theorems tell us how the derivative of the energy <& En or the state 1<& n4 and @<& mA on the left-hand-side is related to something on the right-hand-side which does not contain <&. ! This effectively reduces the order of <& by one. ! Applying them itera- tively, we will be able to calculate the derivatives to any order for both energies and states, which are all we need to construct the power series of the perturbative expansion. ! Energy Corrections According to the Taylor expansion, ; <k & En k 1 2 2 En(&) $ : & $ En % <& En & % <& En & % ". (24) = k$0 k 2 " We already know from Eq. (22) that <& En $ Vnn $ @nA <& H 1n4. (25) " We continue to evaluate 2 <& En $ <& @nA <& H 1n4 ( ) 2 26 $ @<& nA <& H 1n4 % @nA <& H 1n4 % @nA <& H 1<& n4. 2 Note that <& H $ 0 according to the setup in Eq. (12). 2 <& En $ @<& nA <& H 1n4 % @nA <& H 1<& n4 $ :(@<& n m4 @mA <& H 1n4 % @nA <& H 1m4 @m <& n4) m Vnm Vmn (27) $ : Vmn % Vnm - - m En Em En Em Vnm Vmn $ 2 : . - m En Em But, wait a moment " The energy denominator diverges when m $ n, what is wrong? - Note that Eq. (23) only holds for m C n, so we must be careful. Let us restart from the 2nd line of Eq. (27), 2 <& En $ :(@<& n m4 @mA <& H 1n4 % @nA <& H 1m4 @m <& n4) m $ : (@<& n m4 @mA <& H 1n4 % @nA <& H 1m4 @m <& n4) % mCn (@<& n n4 @nA <& H 1n4 % @nA <& H 1n4 @n <& n4) (28) Vnm Vmn $ 2 : % Vnn(@<& n n4 % @n <& n4) - mCn En Em Vnm Vmn $ 2 : % Vnn <& @n n4. - mCn En Em Given that @n n4 $ 1, taking <& on both sides, <& @n n4 $ <& 1 $ 0. So PerturbationTheory.nb 6 2 Vnm Vmn <& En $ 2 : . (29) - mCn En Em So to the 2nd order in &, the perturbative correction to the energy is given by Vnm Vmn 2 En(&) $ En % Vnn & % : & % ". (30) - mCn En Em ◻ Comment on Gauge Fixing In fact, @n <& n4 is the connection of the vector bundle which can always be set to zero by gauge fixing along the path of &. To see this, we start with @n n4 $ 1 ! @<& n n4 % @n <& n4 $ <& @n n4 $ <& 1 $ 0 (31) ! Re @n <& n4 $ 0. So @n <& n4 can only be purely imaginary. But we are free to perform the gauge transformation 1n(&)4 # e* D(&) 1n(&)4, (32) under which, @n <& n4 # @n <& n4 % * <& D. (33) We can always choose <& D to transform @n <& n4 to zero. So in addition to Eq. (23), we can fur- ther require @n <& n4 $ @<& n n4 $ 0. (34) ! State Corrections According to the Taylor expansion, ; k E<& nF k 1 2 2 1n(&)4 $ : & $ 1n4 % 1<& n4 & % E<& nF & % ". (35) = k$0 k 2 By Eq. (23), we know Vmn 1<& n4 $ : 1m4 @m <& n4 $ : 1m4 . (36) - mCn mCn En Em Let us continue to calculate the next order derivative [Please bear with me ...] @ A <& 1 4 2 m H n E<& nF $ <& : 1m4 - mCn En Em @mA <& H 1n4 @<& mA <& H 1n4 $ : 1<& m4 % 1m4 % - - mCn En Em En Em - (37) PerturbationTheory.nb 7 @mA <& H 1<& n4 @mA <& H 1n4 1m4 - 1m4 (<& En - <& Em) 2 En - Em (En - Em) Vlm Vmn Vml Vln $ : : 1l4 % : 1m4 % - - - - mCn lCm Em El En Em lCm Em El En Em Vml Vln Vmn :1m4 - 1m4 (Vnn - Vmm) - - 2 lCn En Em En El (En - Em) Push l $ m or l $ n terms out of the summation, so as to combine the first three summations under GlCm,n (sum over l excluding both m and n), 2 Vlm Vmn Vml Vln Vml Vln E<& nF $ : : 1l4 % 1m4 % 1m4 % - - - - - - mCn lCm,n Em El En Em Em El En Em En Em En El Vnm Vmn Vmn Vnn 1n4 % 1m4 % Em - En En - Em Em - En En - Em Vmm Vmn Vmn Vnn Vmn Vmm 1m4 - 1m4 % 1m4 2 2 (38) En - Em En - Em (En - Em) (En - Em) Vlm Vmn Vml Vln Vml Vln $ : : 1l4 % 1m4 % 1m4 - - - - - - - mCn lCm,n Em El En Em Em El En Em En Em En El Vnm Vmn Vmn Vnn Vmn Vmm 1n4 - 2 1m4 % 2 1m4 2 2 2 (En - Em) (En - Em) (En - Em) The double summation GmCn GlCm,n means to sum over l and m, under the constraint that l, m, n are mutually exclusive.