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3502_11_octo05_LG

# 3502_11_octo05_LG - Chem 3502/5502 Physical Chemistry...

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Chem 3502/5502 Physical Chemistry II (Quantum Mechanics) 3 Credits Fall Semester 2009 Laura Gagliardi Lecture 11, October 5, 2009 Solved Homework We are asked to find < T > for the first harmonic oscillator wave function ! 0 x ( ) = k μ " h # \$ % % & ( ( 1/ 4 e ) k μ x 2 / 2 h So, for < T > n =0 we have T n = 0 = k μ ! h " # \$ % & 1/ 4 e ( k μ x 2 /2 h ( h 2 2 μ d 2 dx 2 " # \$ % & () ) * k μ ! h " # \$ % & 1/ 4 e ( k μ x 2 /2 h dx We need to evaluate the second derivative. It is d 2 dx 2 e ! k μ x 2 /2 h = d dx d dx e ! k μ x 2 / 2 h " # \$ % & ( ) = d dx ! x k μ h e ! k μ x 2 / 2 h & * ( ) + = ! k μ h e ! k μ x 2 /2 h + x 2 k μ h 2 e ! k μ x 2 / 2 h Thus, T n = 0 = ! h 2 2 μ k μ " h # \$ % & ( 1/ 2 k μ h 2 # \$ & x 2 e ! k μ x 2 / h !) ) * dx ! k μ h # \$ % & ( e ! k μ x 2 / h !) ) * dx + , - - - . / 0 0 0 We've already evaluated the two integrals in previous work. Using the appropriate formulae provides

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11-2 T n = 0 = ! h 2 2 μ k μ " h # \$ % & ( 1/ 2 k μ h 2 # \$ & h 2 k μ # \$ % & ( " h k μ # \$ % & ( 1 /2 ! k μ h # \$ % & ( " h k μ # \$ % & ( 1/ 2 ) * + + + + , - . . . . = ! h 2 2 μ k μ " h # \$ % & ( 1/ 2 ! k μ 2 h # \$ % & ( " h k μ # \$ % & ( 1/ 2 ) * + , - . = h 4 k μ Recall from our last homework that x 2 n = 0 = h 2 k μ ! " # \$ % & The expectation value of the potential energy < V > is simply ( k /2)< x 2 >, or 1 2 k x 2 n = 0 = k 2 h 2 k μ ! " # \$ % & = h 4 k μ This completes the proof that < T > = < V > for the n = 0 state of the QMHO. It is similarly straightforward, if increasingly tedious, to prove this for the first excited state, and indeed for any state. Angular Momentum Angular momentum is a vector quantity, defined as the cross product of the position vector and the momentum vector. In cartesian coordinates, it is most easily expressed as the determinant L = i j k x y z p x p y p z (11-1) where x , y , and z are the components of the position vector (i.e., the coefficients multiplying the unit vectors i , j , and k , respectively) and p x , p y , and p z are the
11-3 components of the momentum vector. A 3 x 3 determinant may be evaluated by Cramer's rule (no relation to your enthusiastic instructor, as far as I know...), which states that the determinant is equal to the sum of the three down-right wraparound multiplications minus the sum of the three up-right wraparound multiplications. That is i j k x y z p x p y p z = yp z i + zp x j + xp y k ! yp x k ! zp y i ! xp z j = yp z ! zp y ( ) i + zp x ! xp z ( ) j + xp y ! yp x ( ) k (11-2) Thus, the components of L , namely, L x , L y , and L z , are the terms in parentheses preceding the corresponding unit vectors. In the absence of a torque on a system, angular momentum is a conserved quantity, just as linear momentum is conserved in the absence of a force on a system. The magnitude of the angular momentum is (as for any vector quantity) | L | 2 (typically also written L 2 , since it is just a number). From eq. 11-2, that implies L 2 = L x 2 + L y 2 + L z 2 (11-3) In a quantum mechanical system, the discussion thus far continues to apply, except that the momentum components themselves are the operators p q = ! i h " " q (11-4) where q is either x , y , or z . Let us now consider the commutation properties for any two components of the angular momentum. We'll take L x , and L y as an example. For an arbitrary function f we have

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11-4 L x , L y [ ] f = ! i h ( ) y " " z ! z " " y # \$ % & ! i h ( ) z " f " x ! x " f " z # \$ & ! ! i h ( ) z " " x ! x " " z # \$ & ! i h ( ) y " f " z ! z " f " y # \$ % & ( ) * * * + , - - - = ! h 2 y " f " x + yz " 2 f " x " z ! xy " 2 f " z 2 ! z 2 " 2 f " x " y + xz " 2 f " y " z ! yz " 2 f " x " z + z 2 " 2
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