1. Ch 4. Using Quantum Mechanics on Simple Systems
- Discussion of constrained and not constrained particle motion
ex) free particle, In 2-D or 3-D boxes, In vice versa
Continuous energy spectrum of Q.M free particle
- Discrete energy spectrum and preferred position of Q.M particles in the box (Quantized energy levels)
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4.1. The free particle
Free particle : no forces
Classical 1-dimension, no forces :
Solution : x = x0 + v0t
x0 ,v0 : initial condition, constants of integration
Explicit value : must be known initial condition
How about the free particle in Q.M?
Time-independent Schrödinger Equation in 1-dimension is
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constant V(x) : can choose the reference V(x)=0 (absolute potential reference doesn’t exist)
→ reduced to
Use these notations
Solution is given by
Obtain Ψ(x,t) : multiply each e-i(E/ℏ)t or equivalently e-iωt (E = ℏω)
Eigenvalue : not quantized(all energy allowed : k is continuous variable)
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Plane wave cannot be localized.
→ cannot speak about the position of particle.
Then, what about probability of finding a particle?
→ also cannot calculate (wave function cannot be normalized in interval -∞ < x < ∞)
However, if x is ‘restricted’ to the interval –L ≤ x ≤ L then
P(x) : independent of x → no information about position
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What about the momentum of particle?
ψ+(x) : state of momentum + ℏk(positive direction)
ψ-(x) : state of momentum –ℏk(negative direction)
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4.2 The particle in a One-dimensional box
1-dimensional box : particle in the range 0<x<a only
impenetrable : infinite potential
V(x) = 0 for 0 < x < a
= ∞ for x ≥ a , x ≤ 0
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Schrödinger Equation is changed by
If ψ(x) ≠ 0 outside the box, then value of ψ’’(x) becomes infinite because value of V(x) is infinity outside the box.
However, 2nd derivative exists and well-behaved
→ ψ(x) must be 0 outside the box
boundary condition : ψ(0) = ψ(a) =0
Inside the box : same as the free particle
We can write the solution by the sin and cos.
MS310 Quantum Physical Chemistry

8. From the consideration of boundary conditions

9. Normalization
Energy of the particle
‘Quantization’ arises by the boundary condition

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Particle is ‘quantized’, n : quantum number
Ground state : n=1
However, energy of n=1 is not zero : zero point energy(ZPE)
particle in a box : ‘stationary’ wave(not a traveling wave)
Also, n increase → # of node increase → wave vector k increase because
Finally, what about a classical limit?
→ same as result of C.M(same probability in everywhere)
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graph of ψn(x) and ψn*(x)ψn(x)
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Graph of ψn2(x) / [ψ12(x)]max
n increase : large energy
Lower resolution : cannot precise measure → near to C.M
Result of Q.M ‘approach’ to the C.M when classical limit
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4.3 Two- and Three- dimensional boxes
Boundary condition : similar to 1-dimensional box
V(x, y, z) = 0 for 0 < x < a, 0 < y < b, 0 < z < c
= ∞ otherwise
Inside the box, Schrödinger Equation is given by
Solving by separation of variable
And equation is changed by
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Divide both side by X(x)Y(y)Z(z)
E : independent to coordinate → E = Ex + Ey + Ez and original equation(PDE) reduced to three ODEs.
Solution of each equation is already given.
And energy is given by
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Normalization
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If total energy is sum of independent terms
→ wave function is product of corresponding functions
Solution has a three quantum numbers : nx, ny, nz
→ more than one state may have a same energy
: energy level is degenerate and # of state is degeneracy
ex) if a=b=c, energy of (2,1,1), (1,2,1), and (1,1,2) is same.
in this case, state (2,1,1), (1,2,1) and (1,1,2) is degenerate and degeneracy of the level is 3.
2-dimensional box problem : similar to 3-dimensional problem
(end-of-chapter problem)
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4. Using the postulate to understand the particle in the box and
vice versa
Postulate 1 : The state of a quantum mechanical system is completely specified by a wave function Ψ(x,t). The probability that a particle will be found at time t in a spatial interval of width dx centered at x0 given by Ψ*(x0,t)Ψ(x0,t)dx.
We see the postulates of Q.M using the particle in a box.
Ex) 4.2
ψ(x) = c sin (πx/a) + d sin (2πx/a)
a. Is ψ(x) an acceptable wave function of particle in a box?
b. Is ψ(x) an eigenfunction of the total energy operator Ĥ?
c. Is ψ(x) normalized?
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Sol)
a. Yes.
ψ(x) = c sin (πx/a) + d sin (2πx/a) satisfies the boundary condition, ψ(0) = ψ(a) = 0 and well-behaved function.
Therefore, ψ(x) is acceptable wave function.
b. No.
Result of Ĥψ(x) is not ψ(x) multiplied by constants. Therefore, ψ(x) is not a eigenfunction of the total energy operator.
c. No
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Third integral becomes zero because of orthogonality.
Therefore, ψ(x) is not normalized. However, the function
is normalized when |c|2+|d|2=1
Superposition state depends on time. Why?
Therefore, this state doesn’t describe the stationary state.
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Then, what about a probability of particle in the interval?
Ex) 4.3
probability of ground-state particle in the central third?
Sol) ground state :
Probability of finding a particle in central third is 60.9%.
However, we cannot obtain this result by one individual measurement. We can only predict the result of large number of experiment(60.9%).
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We can understand the two postulates together.
Postulate 3: In any single measurement of the observable that corresponds to the operator Â, the only values that will ever be measured are the eigenvalues of that operator.
Postulate 4 : If the system is in a state described by the wave function Ψ(x,t), and the value of the observable a is measured once each on many identically prepared systems, the average value(also called expectation value) of all of those measurement is given by
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1) wave function is an eigenfunction of operator.
→ all measurement gives same value and it is average value
Ex) ground state of particle in a box
2) wave function is not an eigenfunction of operator.
→ each measurement gives different value
Ex) normalized superposition state
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Last two integrals are zero by orthogonality and final result is
where
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However, result of individual experiment is only E1 or E2 by the postulate 3. How can represent the result?
→ by the postulate 4, result of the large number of individual experiment, probability of E1 is |c|2 and probability of E2 is |d|2, and the ‘average value’ of energy <E> = |c|2E1| + |d|2E2.
More generally, we can think about this case
Ψ(x) = cΨ1(x) + dΨ2(x) + 0(Ψ3(x) + Ψ4(x) + …)
All coefficient except Ψ1(x) and Ψ2(x) is zero. Therefore, no other energy is measured except the E1 and E2.
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Now, consider the momentum.
We know and can calculate the average value of
Momentum of nth state.
In the Q.M, momentum of particle : cannot be zero
(energy E = p2 / 2m cannot be zero in Q.M)
→ average of two superposition state is zero!
We can rewrite the wave function by complex form.
(use the )
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In the case of momentum, two probability of positive momentum and negative momentum is same. Therefore, the average value seems to zero.
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Ex) 4.4
Particle in the ground state.
a. Is wave function the eigenfunction of position operator?
b. calculate the average value of the position <x>.
Sol)
a. position operator :
Therefore, wave function is not an eigenfunction of position operator.
b. expectation value is calculated by postulate 4.
Average value of particle is half, the expected position.
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Summary
The motion of particle which is not constrained shows continuous energy spectrum however, the particle in a box has a discrete energy spectrum.
The state of a quantum mechanical system is completely specified by a wave function Ψ(x,t). The probability that a particle will be found at time t in a spatial interval of width dx centered at x0 given by Ψ*(x0,t)Ψ(x0,t)dx.
In any single measurement of the observable that corresponds to the operator Â, the only values that will ever be measured are the eigenvalues of that operator.
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If the system is in a state described by the wave function Ψ(x,t), and the value of the observable a is measured once each on many identically prepared systems, the average value(also called expectation value) of all of those measurement is given by
MS310 Quantum Physical Chemistry