Physical Chemistry 3: Schrodinger Equation – Flashcards

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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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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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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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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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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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