Review of Lecture 5
The Quantum Harmonic Oscillator:'
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Potential:V (x) =
12
mω2x2
The ladder operators:
Raising operator:
a+ =1√
2mω h̄(−ip+mωx)
Lowering operator:
a− =1√
2mω h̄(ip+mωx)
Definition of commutator:[A,B] = AB−BA
Canonical commutation relation:[x, p] = ih̄
The ground (lowest) solution of the time-independent Schrödinger equationfor the harmonic oscillator is:
ψ0(x) =(mω
π h̄
)1/4e−
mω
2h̄ x2
E0 =12
h̄ω
To find all other functions we can use ψn(x) = An(a+)nψ0(x).
The possible energies are:
En =
(n+
12
)h̄ω
Exercise 8Find the first excited state of the harmonic oscillator.
Useful integrals: ˆ∞
0x2ne−x2/a2
dx =√
π(2n)!
n!
(a2
)2n+1
ˆ∞
0x2n+1e−x2/a2
dx =n!2
a2n+2
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You can also get the normalization algebraically using
Then,
Therefore, the normalization constant An is
Other useful formulas:
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The Harmonic Oscillator: The Analytic Method
We now solve the Schrödinger equation for the harmonic oscillator directly:
Change variables for convenience:
Therefore, we look for solutions in the form
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We look for solutions in the form of a power series
The coefficient of each power of ξ must vanish:
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Problem: not all solutions are normalizable
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We have recovered our previous result that was obtained using the algebraic method!
Note: the condition above will terminate either odd or even power series; the other must be zerofrom the start.
How do we generate the wave functions?
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Apart from the overall factor (a0 or a1) the polynomials hn(ξ ) are the so-called Hermite polyno-mials Hn(ξ ). By tradition, the arbitrary multiplicative factor is chosen so that the coefficient of thehighest power of ξ is 2n. With this convention, the normalized stationary states for the harmonicoscillator are
Summary
The Quantum Harmonic Oscillator: The Analytic Method
Our mission: Solve
− h̄2
2md2ψ
dx2 +12
mω2x2
ψ = Eψ (1)
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Step 1: Change variables
ξ ≡√
mω
h̄x
K ≡ 2Eh̄ω
Resulting equation:d2ψ
dξ 2 = (ξ 2−K)ψ (2)
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Step 2: Determine the solution for ξ → ∞ and separate out the resultingfunction e−ξ 2/2.
ψ(ξ ) = h(ξ )e−ξ 2/2
Resulting equation:d2hdξ 2 −2ξ
dhdξ
+(K−1)h = 0 (3)
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Step 3: Look for solutions of this equation (Equation 3) in the form of apower series:
h(ξ ) = a0 +a1ξ +a2ξ2 + . . .=
∞
∑j=0
a jξj
Resulting equation: Recursion formula
a j+2 =(2 j+1−K)
( j+1)( j+2)a j (4)
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Step 4: Need to truncate∞
∑j=0
a jξj somewhere to ensure that all solutions are
normalizable. Thus the power series must terminate, i.e.
an+2 = 0
for some n = jmax.
Resulting equation:K = 2n+1
En =
(n+
12
)h̄ω, n = 0,1,2, . . .
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Step 5: Put it all together and generate the wave functions
ψn(x) =(mω
π h̄
)1/4 1√2nn!
Hn(ξ )e−ξ 2/2
where Hn are Hermite polynomials.
As expected, we get the same result as before using a± operators.
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Note that the probability of finding the particle outside of the classically allowed regions is notzero. The potential energy curve represents the classically-allowed maximum displacement of theoscillator. All the wave functions extend beyond the curve.
Classically, the energy of the oscillator is E = 12Ka2 = 1
2mω2a2, where a is the amplitude.
Only for large n do we see some resemblance to the classical case:
Computer simulation: http://www.falstad.com/mathphysics.html, 1D Quantum mechanics applet,Harmonic oscillator
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The Free Particle: V (x) = 0 everywhere
We introduce
(Same as inside of the infinite square well, where the potential is zero.) For reasons that will beclear later, we will write the solution in exponential form instead of sin or cos:
There are no boundary conditions to restrict the values of the energy here, and the free particle canhave any positive energy. Adding the time-dependent terms exp(−iEt/h̄), the wave function is
A function that depends on x± vt can represent any wave of fixed profile that travels with speed vin the direction of∓x. Note that a point on the waveform, for example a maximum, corresponds toa fixed value of x± vt. Every point on the waveform is moving with the same speed, so the shapedoes not change as it propagates. Hence a function that depends on x− vt is a wave moving to theright, while one that depends on x+ vt is a wave moving to the left.
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We can put these two expressions together by allowing k to be both positive and negative:
These solutions are analogous to the “stationary states” of the infinite square well and harmonicoscillator, but here we have no boundary conditions. Hence k is a continuous variable and theenergies are not quantized. The solutions are propagating waves.
As waves, we can assign a wavelength λ :
According to the de Broglie formula, the waves carry momentum
The speed of the waves can be found directly from the solution by writing the exponent as x− vt:
However, the classical speed of a free particle with energy E is given by E = 12mv2:
Therefore, it appears that the wave function travels at only half the speed of the particle that it issupposed to represent! We will return to this problem later.
The other problem is that the resulting wave function is not normalizable:
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