PHY4221Fall05_Notes5

PHY4221Fall05_Notes5 - PHY4221 Quantum Mechanics I Fall...

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PHY4221 Quantum Mechanics I Fall 2004 Rayleigh-Ritz linear variational method Given the eigenvalue problem of the Hamiltonian operator H : H | Φ i = E | Φ i , (1) the Variational Principle states that any normalized state vector | Ψ i for which the expectation value E ([Ψ]) ≡ h Ψ | H | Ψ i of the Hamiltonian operator H , i.e. the average energy, is stationary is an eigenvector of H , and the corresponding energy eigenvalue is the stationary value of E ([Ψ]). That is, if we could actually carry out a variation of | Ψ i , starting with some initial | Ψ Initial i and varying | Ψ i by small steps, namely by rotating | Ψ i by small amounts in the Hilbert space, such that we could end by finding the true minimum of the functional E ([Ψ]), we would attain the ground state eigenvector and eigenenergy exactly. Similarly, a local minimum would give us an excited state eigenvector and eigenenergy. Obviously, this is not feasible even with modern computers. Instead we usually restrict the | Ψ i to be of a specific analytical form and contain a few undetermined parameters, which are fixed at the values making the variation δE = 0. According to the variational principle, the resultant E ([Ψ]) provides an upper bound of the ground state eigenenergy. Proof: E ([Ψ]) ≡ h Ψ | H | Ψ i = X n E n |h Ψ | Φ n i| 2 > E 0 X n |h Ψ | Φ n i| 2 = E 0 . (2) Alternatively, we may apply the Rayleigh-Ritz linear variational method , namely constructing the | Ψ i as a linear combination of a finite subset of a set of orthonormal basis vectors {| Φ n i} : | Ψ i = N X n =1 a n | Φ n i . (3) where the N + 1 parameters a n are now to be considered as the variational parameters. In order to ensure that the | Ψ i is normalized, i.e. N n =1 | a n | 2 = 1, we apply the method of Lagrange multipliers to tackle the variational problem: 1
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1. We first construct a functional of the form L ≡ h Ψ | H | Ψ i - E ( h Ψ | Ψ i - 1) = N X m,n =1 a * m a n h Φ m | H | Φ n i - E ˆ N X n =1 | a n | 2 - 1 ! . (4) 2. If we arbitrarily choose a 1 , a 2 , ... , a N - 1 as independent so that a N is determined from the normalization condition, then we have L ∂a k = 0 , k = 1 , 2 ,...,N - 1 (5) but L /∂a N is not necessarily zero. However, we still have the undetermined multiplier E at our disposal. We can now choose this multiplier so that L /∂a N does vanish too. As a result, we obtain a set of linear homogeneous algebraic equations: N X n =1 ( h Φ m | H | Φ n i - mn ) a n = 0 , m = 1 , 2 ,...,N (6) Note that this set of linear algebraic equations is the same as the set obtained from performing diagonalization in terms of the truncated basis {| Φ n i ,n = 1 , 2 ,...,N } . 3. Solving the set of homogeneous algebraic equations in Eq.(6) yields the variational
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This note was uploaded on 12/10/2011 for the course PHYS 4221 taught by Professor Cflo during the Spring '11 term at CUHK.

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PHY4221Fall05_Notes5 - PHY4221 Quantum Mechanics I Fall...

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