Physical Chemistry 3: Schrodinger Equation

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Forces don’t work well with quantum mechanics. You can redo all of the physics in terms of energy!
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p^ is the momentum operator. What does it equal?
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-i(hbar)d^
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So, p^f = ?
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-i(hbar)df/dx
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The most important physical observable is that of (1). The operator associated with the total energy is called the (2) operator.
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(1) total energy E (2) Hamiltonian (H^)
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The eigenvalue equation for the Hamiltonian is (1). It is the time independent Schrodinger equation.
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H^ (psi) = E (psi)
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The steps of facing a problem in elementary quantum mechanics. 1. Define the classical Hamiltonian for the system. 2. Use Postulate II to replace the classical variables, x, px etc., with their appropriate operators. 3. Solve the Schrodinger equation, H^ (psi) = E (psi), which is now a second order differential equation.
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1. Define the classical Hamiltonian for the system. The total energy for a classical system is…. E_cl_ = T + V. (1) What do T and V stand for?
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(1) T is the kinetic energy and V is the potential energy.
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(1) Kinetic energy in the classical sense = 1/2mv^2. What is the T expression in terms of momentum?
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(1) Momentum (p) = mv p/m = v Substitute into equation we know and get kinetic energy = p^2/2m. This is the 1D equation. The 3D equation is T = 1/2m(p^2x + p^2y + p^2z)
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The potential energy is almost always a function of coordinates. What is the V expression?
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V = V(x) for 1D For 3D V = V(x,y,z)
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So the full Hamiltonian can be written as, H_cl_ = V(x) + p^2/2m.
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So with Postulate II, you go from classical mechanics to quantum. That means that whenever we see an x, turn it into x^. Whenever we see p, turn it into p^. So plugging into the Schrodinger equation we get, [(p^)^2/2m + V(x^)] (psi) = E (psi) (p^)^2 = (p^)(p^) = (-i(hbar)d/dx)^2 = -(hbar)^2 d2/dx2
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Now, if you solve the Schrodinger equation for one dimension, you end up with a (1) order differential equation. It looks like this… ([-(hbar)^2/2m] d2/dx2 (psi)) + (V(x) – E)psi = 0
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(1) second order
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So now let’s determine if things functions are eigenfunctions are not. 1) Is x^3 an eigenfunction? (-(hbar)^2/2m d2/dx2) (x^3) =(-(hbar)^2/2m) (6x) 2) Is cos(x) an eigenfunction? (same value) cos(x) = same value (-cos(x)) 3) What is the eigenvalue of this eigenfunction?
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(1) No. x^3 isn’t an eigen function because x^3 isn’t the same as 6x. (2) Yes! cos(x) is an eigenfunction of this Hamiltonian operator (but not of position or momentum) because the functions match after two derivations. (3) (hbar)^2/2m
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So when you solve the Schrodinger equation, you get eigenfunctions of the Hamiltonian. This means that it is a function of these operators, but not of position or momentum.
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Postulate III implies that if psi is an eigenfunction of a particular operator representing a physical observable, then all measurements of that physical property will yield the associated eigenvalue. However, if psi is not an eigenfunction of a particular operator, then all measurements of that physical property will still yield AN eigenvalue, but we can’t predict for certain which one.
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We can, however, given an expectation, or average, value for the measurement. = integral_allspace_ [(psi*norm) (o^) (psi norm) dx]
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Look on scratch sheet for example problem of average value theorem!
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In quantum mechanics certain pairs of variables cannot, even in principle, be simultaneously known to arbitrary precision. Such variables are called (1)
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(1) complimentary
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General statement for Heisenberg Uncertainty: delta(a)delta(B) >_ 1/2 || where [a^,B^] means the commuter of a^ and B^. Commuter is defined as…
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[a^,B^] is a^B^ – B^a^
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= (1) if they commute.
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(1) 0
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At the very best we can only hope to simultaneously know position and momentum such that the product of the uncertainty in each is (hbar)/2.
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When energy is expressed in terms of momentum, it is said to be a (1).
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(1) Hamiltonian
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H^ (psi) = E (psi) Solutions of this equation leads to stationary state solutions, which are analogous to the stationary state or standing wave solutions obtained for a vibrating spring. The psi functions which are possible solutions of the equation are called (1), characteristic functions, or wave functions. These wave functions have the property that when the Hamiltonian operator acts on each of them, a constant (which is the energy) is formed times the original wave function. Thus, corresponding to each of these wave functions is an allowed energy level, and these levels are called the (2) or characteristic values for the energy of the system.
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(1) eigenfunctions (2) eigenvalues. They are formed and result from the action of the operator on the eigenfunction. These energy eigenvalues are often called the allowed energies of the system.
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So, the greater the amplitude of the light wave in a particular region, the greater is the probability that a photon is present in that region. The psi-squared gives a probability density of the electron.
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For an eigenfunction to be acceptable, one requirement is that it must be finite in all regions of space; otherwise the Schrodinger equation would not apply since the electron would be located at a region where psi was infinite and would not be acting as a wave.
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When you normalize an eigenfunction, you make it so that the probability is equal to this function. Remember, if we obtain an eigenfunction psi, if we multiply psi by any number a, the function a(psi) is still an eigen function, because H^ (apsi) = E (apsi). So there is an infinite set of eigenfunctions difffering from each other by numerical factors. We just want the one that is equal to the probability that the electron is in the volume element of dimensions dx dy dz. This function is the normalized one!
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