8_Variational_Principle

8_Variational_Principle - VIII. The Variational Principle...

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Unformatted text preview: VIII. The Variational Principle Griffiths Chapter 7 1 A. The Variational Principle The variational principle is another approximation method for quantum mechanics. Perturbation theory allowed us to find approximate wave functions and energies for systems that we couldn’t solve the Schrödinger equation for exactly. 2 A. The Variational Principle The variational principle only tells us about the energy of a system, and only of the ground state. Furthermore, we can only find an upper bound on the ground state energy. On the other hand, the variational principle is easy to understand and use, and the ground state energy of a system is often the most important thing we want to know. 3 A. The Variational Principle 1. Statement of the Principle The ground state energy of a system is, according to the variational principle, Egs ≤ ￿ψ |H |ψ ￿ ground state expectation value of H, 〈H〉 See board VIII.A.1 4 A. The Variational Principle 2. An Example Griffiths Example 7.1: Find an upper bound on the ground state energy for the one dimensional harmonic oscillator. ￿2 d 2 1 H=− + mω 2 x2 2m dx2 2 Solution: Egs 1 = ￿ω 2 What luck! Equal to the exact result. Used a Gaussian trial function with free parameter and then M INIMIZED with respect to the free parameter. See board VIII.A.2 5 A. The Variational Principle 3. Another Example Griffiths Example 7.2: Find an upper bound on the ground state energy for the delta function potential. ￿2 d 2 H=− − αδ (x) 2 2m dx Solution: Egs mα 2 ≤− 2 π￿ For exact result, replace π w ith 2. Negative for b ound state Used a Gaussian trial function again with a free parameter a nd then MINIMIZED with respect to the free parameter. See board VIII.A.3 6 A. The Variational Principle 4. A Third Example Griffiths Example 7.3: Find an upper bound on the ground state energy for the one dimensional square well. where Solution: Egs ￿2 d 2 H=− + V (x) 2 2m dx ￿ 0 , if 0 ≤ x ≤ a V (x) = ∞, otherwise 12￿2 For exact result, replace 12 with π2. ≤ 2 2ma Could NOT use a Gaussian trial function here. It “leaks” outside the potential – it is non-zero in the infinity region of the potential See board VIII.A.4 7 B. The Ground State of Helium The helium atom consists of two electrons and a nucleus of two protons (as well as two neutrons which we can ignore in our model). ￿2 e2 2 2 H=− (∇1 + ∇2 ) − 2m 4π￿0 kinetic energy o f electron 1 kinetic e nergy of e lectron 2 ￿ 2 protons in the￿ nucleus 2 2 1 + − r1 r2 |r1 − r2 | Coulomb potential Coulomb potential b et ween electron b et ween electron 1 a nd the nucleus 2 a nd the nucleus Coulomb potential b et ween electrons There is no analytical solution for the time independent Schrödinger equation for the energy levels of the electrons in a helium atom. 8 B. The Ground State of Helium 1. The Simplest Approximate Solution We can ignore the electron-electron Coulomb interaction term altogether ￿ e 2 2 H=− (∇1 + ∇2 ) − 2m 4π￿0 2 2 ￿ 2 2 1 + − r1 r2 |r1 − r2 | ￿ The Schrödinger equation is now separable between r1 and r2. The ground state energy in this approximation is Egs = -109 eV. The experimentally measured ground state energy of helium is -78.975 eV n ot a very good match The ground state wave function in this approximation is just the product of two hydrogen wave functions. 8 −2(r1 +r2 )/a e ψgs (r1 , r2 ) = 3 See board VIII.B.1 πa 9 B. The Ground State of Helium 2. A Better Approximate Solution 8 −2(r1 +r2 )/a e ψgs (r1 , r2 ) = 3 πa We can use the simplest approximation wave function as a trial function in the variational method. This gives us ￿H ￿ = 8E1 + ￿Vee ￿ expectation v alue of the H amiltonian ground state e nergy of the h ydrogen atom expectation value of the e lectron-electron Coulomb i nteraction term ￿Vee ￿ = ￿ e2 4π￿0 ￿￿ 8 π a3 ￿2 ￿ e−4(r1 +r2 )/a dr1 dr2 |r1 − r2 | The ground state energy in this approximation is Egs = -75 eV. The experimentally measured ground state energy of helium is -78.975 eV See board VIII.B.2 Not bad. 10 B. The Ground State of Helium 3. An Even Better Solution We can use the simplest approximation wave function as a trial function in the variational method, but with a different effective nuclear charge – Z instead of 2. The logic is that the trial function ignores electron- electron interactions completely. We now suppose that the presence of another electron partially “screens” the interaction between the first electron and the nucleus. In other words, the first electron doesn’t see a bare nucleus of charge +2, but rather a nucleus surrounded by a cloud of negative charge that is the second electron. 11 B. The Ground State of Helium Therefore we should use a trial wave function of the form Z 3 −Z (r1 +r2 )/a ψgs (r1 , r2 ) = e 3 πa compare with 23 ψgs (r1 , r2 ) = Ide nt al ic ! If we use this wave function, it is more convenient to write our Hamiltonian ￿ e (∇2 + ∇2 ) − 2 2m 1 4π￿0 ￿ ￿2 e2 2 2 H=− (∇ + ∇2 ) − 2m 1 4π￿0 ￿ H=− as 8 −2(r1 +r2 )/a e 3 πa 2 2 2 2 1 + − r1 r2 |r1 − r2 | Z Z + r1 r2 ￿ 12 e2 + 4π￿0 ￿ ￿ (Z − 2) (Z − 2) 1 + + r1 r2 |r1 − r2 | Just adding and subtracting Z/r terms ￿ B. The Ground State of Helium The expectation value of the Hamiltonian will be ￿H￿ = 2Z 2 E1 + 2(Z − 2) ￿ ￿￿ ￿ e 1 + ￿Vee ￿ 4π￿0 r 2 expectation value with respect to hydrogenic wave f unction ψ100 e xcept with c harge Z – see Griffiths e quation 6.55 ￿ 1 ￿ Z r = same as before but w ith Z instead of 2 ￿Vee ￿ = − 5Z E1 4 a The expectation value of the Hamiltonian will be ￿H￿ = [−2Z 2 + (27/4)Z ]E1 13 B. The Ground State of Helium Minimizing with respect to Z gives us the ground state energy in this approximation of Egs = -77.5 eV. The experimentally measured ground state energy of helium is -78.975 eV Within 2% See board VIII.B.3 Griffiths points out that we could continue adding refinements to the trial function in this way, with more adjustable parameters, and get closer to the measured result. But he gets bored and stops. We’ll stop too. 14 C. The Hydrogen Molecule Ion We have looked at atomic hydrogen and atomic helium. Now let’s look at molecular hydrogen, H2, but ionized so that it only has one electron (since that’s easier). We can use the variational principle to not only find a bound on the ground state energy, but also to determine whether such an ionized molecule will be stable at all – as opposed to forming a hydrogen atom and a free proton. 15 C. The Hydrogen Molecule Ion The hydrogen molecule ion is shown schematically in Griffiths figure 7.5. The Hamiltonian for this system is ￿ e 2 H=− ∇− 2m 4π￿0 2 2 16 ￿ 1 1 + r1 r2 ￿ C. The Hydrogen Molecule Ion The Hamiltonian for this system is ￿ e 2 H=− ∇− 2m 4π￿0 2 2 ￿ 1 1 + r1 r2 ￿ We could use the hydrogen atom ground state wave function as our trial function in the variational principle. ψ0 (r) = √ 1 π a3 e−r/a Better yet, we should let the electron have an equal chance of being associated with either proton. 17 C. The Hydrogen Molecule Ion ψ0 (r) = √ 1 π a3 e−r/a Better yet, we should let the electron have an equal chance of being associated with either proton. Let’s use a combination of hydrogen atom ground state wave functions, one associated with each proton. hydrogen atom wave function a ssociated with 1st proton normalization factor hydrogen atom wave function a ssociated with 2nd proton ψ = A[ψ0 (r1 ) + ψ0 (r2 )] LCAO (Linear Combination of Atomic Orbitals) technique 18 C. The Hydrogen Molecule Ion 1. Normalization Normalizing this trial function gives us a value for A. 1 |A| = 2(1 + I ) 2 where the “overlap integral” I is ￿ I = e−R/a 1 + ￿ R a ￿ + 1 3 ￿ R a ￿2 ￿ and R is the distance between the two protons. (a is still the Bohr radius). See board VIII.C.1 19 C. The Hydrogen Molecule Ion 2. Expectation Value Next, we calculate the expectation value of the Hamiltonian. ￿H￿ = E1 − 2|A|2 ￿ ￿￿ ￿ e 1 1 a￿ψ0 (r1 )| |ψ0 (r1 )￿ + a￿ψ0 (r1 )| |ψ0 (r2 )￿ 4π￿0 a r2 r1 ≡ ≡ See board VIII.C.2 2 D: Direct i ntegral X: Exchange i ntegral a￿ a ￿ −2R/a D= − 1+ e R R X= ￿ 1+ See Griffiths problem 7.8 20 R a ￿ e−R/a C. The Hydrogen Molecule Ion Therefore the expectation of the Hamiltonian is ￿ ￿ (D + X ) ￿H￿ = 1 + 2 E1 1+I But this is for a given distance R between the two protons. Since we don’t know what this separation should be, we should add the potential energy for the protonproton separation. Vpp e2 1 2a = = − E1 4π￿0 R R 21 C. The Hydrogen Molecule Ion Combining all of this, Griffiths gives us the upper bound for the total energy of the system, in units of -E1 and expressed as a function of x = R/a. F (x) = −1 + 2 x ￿ (1 − (2/3)x )e + (1 + x)e 1 + (1 + x + (1/3)x2 )e−x 2 22 −x −2x ￿ C. The Hydrogen Molecule Ion We see that indeed ionic molecular hydrogen can form instead of neutral atomic hydrogen plus a free proton because there is a minimum below the binding energy of neutral atomic hydrogen. The upper bound on the ground state energy for ionic molecular hydrogen is -1.8eV. experimental value is -2.8eV The predicted separation of the protons is 2.4a, or 1.3Å. experimental value is 1.06Å 23 The End Quantum Physics 25 2 ...
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This note was uploaded on 09/28/2011 for the course PHYS 334 taught by Professor Resch during the Spring '08 term at Waterloo.

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