Stationary)perturbation)theory)nsmn1.uh.edu/cratti/PHY6316-Spring_2018_files/QM_Lecture_11_Cla… ·...
Transcript of Stationary)perturbation)theory)nsmn1.uh.edu/cratti/PHY6316-Spring_2018_files/QM_Lecture_11_Cla… ·...
ì Stationary perturbation theory Lecture notes 11 (based on CT, Sec4on 11)
Introduction Ø In the case of harmonic oscillator and hydrogen atom, their
Hamiltonians are simple enough for their eigenvalue equa4ons to be solved exactly
Ø This happens only in a small number of problems
Ø There exist approxima4on methods which enable us to find analy4cally approximate solu4ons of the eigenvalue equa4on in certain cases
Ø Sta4onary perturba4on theory is widely used in quantum physics
Ø One isolates the main effects and, when they have been understood, one tries to explain the finer details
Description of the method Ø We can use perturba4on theory when the Hamiltonian of the
system being studied can be wriHen as H=H0+W, where we know the eigenstates and eigenvalues of H0 and W is small compared to H0 (its matrix elements are much smaller than those of H0)
Ø H0 is 4me-‐independent and it is called the “unperturbed Hamiltonian”
Ø W is called “the perturba4on”
Ø If W is not 4me-‐dependent, we say that it is a “sta4onary perturba4on”
Ø We need to find the modifica4ons in the eigenvalues and eigenstates due to the perturba4on
Description of the method Ø We assume that W is propor4onal to a real, dimensionless
parameter λ:
Ø Here, is an operator whose matrix elements are comparable to those of H0 in magnitude
Ø Perturba4on theory consists of expanding the eigenvalues and eigenstates of H in powers of λ, keeping only a finite number of terms
Description of the method Ø We assume the eigenvalues and eigenvectors of H0 to be known
Ø We assume that the unperturbed energies form a discrete spectrum:
the index i labels the various eigenvectors associated with the energy level in the case of degeneracy
Ø The set of vectors forms an orthonormal basis of the state space:
Ø We can rewrite : the eigenvalues E(λ) of H(λ) generally depend on λ
Description of the method Example of varia4on of the eigenvalues E(λ) of H(λ) with respect to λ
Ø For λ=0 we obtain the spectrum of H0
Ø The eigenvalues and are doubly degenerate
Ø The applica4on of the perturba4on removes the degeneracy in but not the one in
Ø An addi4onal two-‐fold degeneracy appears at λ=λ1
Description of the method Ø We are looking for the eigenstates |ψ(λ)> and the eigenvalues
E(λ) of the Hermi4an operator H(λ):
Ø We assume that they can be expanded in powers of λ:
Ø We replace the expression for H(λ):
Ø We want this equa4on to be sa4sfied for λ small but arbitrary: we must equate the coefficients of successive powers of λ
Description of the method Ø Thus we have:
Ø 0th-‐order terms
Ø 1st-‐ order terms
Ø 2nd-‐order terms
Ø qth-‐order terms
Description of the method Ø We will neglect terms of order higher than 2 in λ
Ø We require that |ψ(λ)> is normalized, and we fix the phase by requiring that <0|ψ(λ)> is real
Ø To 0th-‐order, this implies that |0> must be normalized: <0|0>=1
Ø To 1st-‐order, we can write
which is =1 if the λ term is zero. But, since we chose the phase such that <0|1> is real, we must have <0,1>=<1,0>=0
A similar argument, for 2nd-‐order in λ yields <0|2>=<2|0>=-‐1/2<1|1>
Description of the method Ø In general we have:
Ø Therefore, confining ourselves to 2nd-‐order in λ, we have found the perturba4on equa4ons and we have to supplement them with the above condi4ons
Description of the method Ø This equa4on expresses the fact that |0> is an eigenvector of H0
with eigenvalue ε0
Ø This was to be expected
Ø We therefore choose an eigenvalue of H0
Ø There can exist one or many energies E(λ) of H(λ) which approach when λà0
Ø The number of eigenstates corresponding to these E(λ) cannot vary discon4nuously when λ varies around 0. Therefore, this equals the degeneracy of
Perturbation of a non-‐degenerate level
Ø Consider a non-‐degenerate level of the unperturbed Hamiltonian H0
Ø The eigenvector associated with it is unique within a constant factor
Ø We want to determine the change in the energy and vector by adding a perturba4on W to the Hamiltonian
Ø We choose , which means that |0> is propor4onal to
Ø We choose
Perturbation of a non-‐degenerate level
Ø We call En(λ) the eigenvalue of H(λ) which approaches as λà0
Ø We assume that λ is small enough that this eigenvalue remains non-‐degenerate
Ø We now calculate the first terms of the expansion of En(λ) and |ψn(λ)> in powers of λ
Perturbation of a non-‐degenerate level
Ø First order correc+ons
Ø The equa4ons we need are:
and
Ø We project the first equa4on onto
Ø The first term is zero since is eigenvector of H0 with eigenvalue ε0
Perturbation of a non-‐degenerate level
Ø The equa4on then becomes
Ø Therefore, to first order in the perturba4on we have:
Ø The first-‐order correc4on to a non-‐degenerate energy is simply equal to the mean value of the perturba4on term W in the unperturbed state
Perturbation of a non-‐degenerate level
Ø We must now find the correc4on to the eigenvector
Ø We project the equa4on onto all vectors of the basis other than
Ø
Ø The last term is zero due to the orthogonality of the eigenvectors
Ø Besides, we can let H0 act on :
Perturbation of a non-‐degenerate level
Ø Therefore, the coefficients of the expansion of |1> on the unperturbed basis states is:
Ø The last coefficient, , is zero because and <1|0>=0 due to the boundary condi4ons
Ø Therefore:
Ø To 1st-‐order in perturba4on theory we have:
Perturbation of a non-‐degenerate level
Ø Second order correc+ons
We need the following equa4ons:
And <0|2>=<2|0>=-‐1/2<1|1>
Ø We project the first one onto
Ø Again the first term is zero, as well as for the orthogonality of the states
Perturbation of a non-‐degenerate level
Ø We get:
namely, replacing the expression for vector |1>:
Ø Therefore, the energy En(λ) can be wriHen, to second order, as:
Perturbation of a non-‐degenerate level
Ø If we limit the energy expansion to first order in λ, we can have an idea of the error involved by evalua4ng the second order term
Ø Consider the expression for ε2. It contains a sum of terms whose numerators are posi4ve or zero.
Ø We call ΔΕ the absolute value of the difference between and that of the closest level. We have:
Ø This gives an upper limit for |ε2|
Perturbation of a non-‐degenerate level
Ø We can rewrite it as
Ø Recalling the closure rela4on:
we get:
Perturbation of a non-‐degenerate level
Ø Mul4plying both sides by λ2 we obtain an upper limit for the second order term in the expansion of En(λ) as:
where ΔW is the root-‐mean-‐square devia4on of the perturba4on W in the unperturbed state
Ø This indicates the order of magnitude of the error commiHed by taking only the first-‐order correc4on into account
Perturbation of a degenerate state Ø We now take a level which is gn-‐fold degenerate
Ø The corresponding eigensubspace of H0 is
Ø In this case, the choice is not enough to determine the vector |0>
Ø Under the ac4on of W, the level generally gives rise to several dis4nct sublevels
Ø Their number fn is between 1 and gn
Ø The total number of orthogonal eigenvectors of H associated with the fn sublevels is always equal to gn: if fn<gn, some of these sublevels are degenerate
Perturbation of a degenerate state Ø We limit ourselves to 1st-‐order in λ for the energies and to 0th-‐
order for the eigenvectors
Ø We need to determine ε1 and |0>
Ø We project the equa4on
onto the gn basis vectors
Ø Since the are eigenvectors of H0 with eigenvalue we obtain the gn rela4ons
Perturbation of a degenerate state Ø We use the closure rela4on
Ø The vector |0> is orthogonal to all the basis vectors for which p≠n
Ø Therefore the sum over p reduces to a single term:
Ø We have gn2 numbers : we arrange them in a gnxgn matrix of raw index i and column index i’
Perturbation of a degenerate state Ø This square matrix is cut out of the matrix which
represents in the basis
Ø The column vector of elements is an eigenvector of with the eigenvalue ε1
Ø Effec4vely, this corresponds to solving the vector equa4on
Perturbation of a degenerate state
To calculate the eigenvalues (to first order) and the eigenvectors (to zeroeth order) of the Hamiltonian corresponding to a degenerate unperturbed state , diagonalize the matrix which represents the perturba4on W inside the eigensubspace associated with
/
Perturbation of a degenerate state Ø We call ε1j the dis4nct roots of the characteris4c equa4on of
Ø These eigenvalues are all real and the sum of their degrees of degeneracy is equal to gn
Ø Each eigenvalue introduces a different energy correc4on
Ø The degenerate level splits, to first order in λ, into fn(1) dis4nct sublevels whose energies can be wriHen as
Perturbation of a degenerate state Ø If fn(1)=gn we say that, to first order, the perturba4on completely
removes the degeneracy of
Ø If fn(1)<gn we say that the degeneracy, to first order, is only par4ally removed
Ø We now choose an eigenvalue ε1j of
Ø If it is non-‐degenerate, the eigenvector |0> is uniquely determined
Ø Then there exists a single, non-‐degenerate eigenvalue E(λ) of H(λ) which is equal to
Variational method Ø Consider a physical system whose Hamiltonian H is 4me-‐
independent
Ø We assume that the en4re spectrum of H is discrete and non-‐degenerate:
Ø We know H but not its eigenvalues and eigenstates
Ø The varia4onal method is useful when we do not know how to diagonalize H exactly
Variational method Ø We choose an arbitrary ket |ψ> of the state space of the system
Ø For sure we have:
where E0 is the ground state energy and the equality holds only when |ψ> is the eigenstate of H with eigenvalue E0
Ø To prove the above inequality, we expand |ψ> on the basis of eigenstates of H:
Variational method Ø We then have:
Ø And of course:
which proves the inequality
Ø This is a basis for a method of approximate determina4on of E0
Ø We choose a family of kets |ψ(α)> which depend on a certain number of parameters α
Ø We minimize <H>(α) with respect to the parameters α
Ø This is a good approxima4on of the ground state E0
Variational method Ø More generally, the mean value of the Hamiltonian H is sta4onary
in the neighborhood of its discrete eigenvalues
Ø Consider as a func4onal of |ψ>
Ø Calculate its increment δ<H> when |ψ> becomes |ψ>+|δψ>
Ø To do so, we write: and differen4ate both sides:
Variational method Ø Since <H> is a number, we can write:
Ø The mean value <H> is sta4onary if
which means that
Ø We set: and rewrite the above as
Variational method Ø The last rela4on must be sa4sfied by any infinitesimal ket |δψ>
Ø In par4cular, if we choose , where δλ is an infinitely small real number, we get:
Ø The norm of the ket must be zero, therefore the ket itself is zero
Ø This means that:
Ø <H> is sta4onary if and only if the state vector |ψ> to which it corresponds is an eigenstate of H and the sta4onary values of <H> are the eigenvalues of the Hamiltonian
Variational method Ø The varia4onal method can be generalized and applied to the
approximate determina4on of the eigenvalues of H
Ø If the func4on <H>(α) obtained from the trial kets |ψ(α)> has several extrema, they give the approximate values of some of its energies En
Variational method Ø We will now apply this method to the one-‐dimensional harmonic
oscillator
Ø We will consider the Hamiltonian:
and solve its eigenvalue equa4on approximately by varia4onal calcula4ons
Ø The Hamiltonian is even under parity transforma4ons, therefore its ground state is necessarily represented by an even wavefunc4on
Variational method Ø We consider even trial func4ons
Ø For example, we take the one-‐parameter family
Ø The square of the norm of the ket is equal to:
Variational method Ø We find:
Ø so that:
Variational method Ø The deriva4ve of the above func4on goes to zero when:
and we then have:
Ø We find that the minimum value of <H>(α) is exactly equal to the energy of the ground state of the harmonic oscillator
Ø This is due to the simplicity of the problem we are studying: the wavefunc4on of the ground state happens to be one of the func4ons of the trial family
Ø The varia4onal method in this case gives the exact solu4on
Variational method Ø To calculate the first excited state we have to choose
wavefunc4ons which are orthogonal to the ground state
Ø We choose the trial family of odd func4ons:
Ø In this case:
and
Variational method Ø The above results yield:
Ø This func4on, for the same value of α0 as above, presents a minimum equal to:
Ø Again we find exactly the energy E1 and the associated eigenstate because the trial family includes the correct wavefunc4on
Variational method Ø So far, the families we chose always included the exact
wavefunc4on
Ø We now try a totally different type of wavefunc4on:
Ø A simple calcula4on yields:
and:
Variational method Ø The minimum of this func4on is obtained at:
and is equal to:
Ø This minimum is 4mes the exact ground state energy
Ø The error we commit with this method is:
Variational method Ø We found a minimum of H which is rela4vely close to the real one
Ø However, the corresponding approximate state is quite far from the true eigenstate
Ø The wavefunc4on decreases too rapidly for small values of x and way too slowly at large x
Ø We have to be very careful when physical proper4es other than the energy are calculated using the varia4onal method
Ø For example we find that
which is not very far from the actual one,
Variational method Ø However, the expecta4on value of X4 is infinite in the trial
func4on and finite for the real wavefunc4on
Ø It is impossible to evaluate the error in a varia4onal calcula4on, if we do not know the exact solu4on of the problem
Ø This method can be very flexible, and usually gives a good approxima4on for the energy
Ø It is par4cularly valuable when physical arguments give us an idea of the qualita4ve or semi-‐quan4ta4ve form of the solu4ons