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Quantum Physics Lecture 7 Steady state Schroedinger Equation (SSSE):
eigenvalue & eigenfunction
“particle in a box” re-visited
Wavefunctions and energy states normalisation probability density
Expectation value of position
Finite potential well
Eigenvalue & eigenfunction
An eigenfunction equation. “eigen” meaning “proper” or “characteristic”.
In general there is more than one solution ψn (eigenfunction) Each solution corresponds to a specific value of energy i.e. eigenfunctions ψn with corresponding eigenvalues En
- a re-statement of energy quantisation!
The eigenfunctions ψn are a complete orthogonal set - c.f. Fourier
∂ 2ψ∂x2
+ 2m
2 E −U( )ψ = 0
−2
2m∂2
∂x2+U
⎛⎝⎜
⎞⎠⎟ψ = Eψ
Hamiltonian − H
“particle in a box” re-visited
described by “square well” potential with infinitely high potential barriers : (1) U = 0 for 0 < x < L (2) U= ∞ (and so ψ = 0) for x ≤ 0 & x ≥ L
Objective: find ψ for 0 < x < L
Region (1) SSSE becomes
‘SHM’ type equation: Oscillatory solutions eikx
L U
x0 L
∞ ∞
d 2ψdx2
+ 2m
2 Eψ = 0
∂ 2ψ∂x2
+ 2m
2 E −U( )ψ = 0
“particle in a box” solution of SSSE
Recall free particle wavefunction Without time dependence. C is a constant
This is a free particle travelling in +ve x-direction Wave travelling in -ve x-direction also possible More general solution adds both (making a standing wave!)
General solution:
where C and B may be complex
Expand this using exp(iθ) = cosθ + isinθ
d 2ψdx2
+ 2m
2 Eψ = 0
Ψ = Cexp i p
x
⎛⎝⎜
⎞⎠⎟= C exp ikx( )
Ψ = Cexp ikx( ) + Bexp −ikx( )
“particle in a box” solution of SSSE
Now apply boundary conditions at box walls…
At x = 0 must have ψ = 0 So ψ (0) = C + B = 0 and hence C = - B
Putting into equation for ψ gives A is another constant
Also ψ = 0 at x = L so k =
nπL
n = 1,2,3....
Ψ = C cos kx + isin kx( ) + B cos −kx( ) + isin −kx( )( )= C cos kx + isin kx( ) + B cos kx − isin kx( )= C + B( )cos kx + i C − B( )sin kx
Ψ = 2iC sin kx = Asin kx
“particle in a box” energy levels
quantised energy, as before! E1
E3
E2
∴ p
L = nπ also E = p2
2m
∴En =n2π 2
2
2mL2= n2h2
8mL2
k =
nπL
n = 1,2,3....
“particle in a box” wavefunctions
Same function form as “standing waves” of earlier model
Properties:
(1) ψ is single-valued and continuous
(2) likewise dψ/dx
(with the exception that dψ/dx is non-continuous at x = 0, x = L)
- an artificial ‘feature’ of (ideal) infinitely high potential barriers!
ψn = Asin p
x⎛
⎝⎜
⎞⎠⎟ = Asin 2mEn
x
⎛
⎝⎜⎜
⎞
⎠⎟⎟ = Asin nπ
Lx( )
“particle in a box” wavefunctions (3) ψ normalisable? ψ n = Asin nπ L x( )
ψ n = 2L sin
nπL x( )
ψn
2dx
−∞
+∞∫ = ψn
2dx
0
L∫ =1
= A2 sin2 nπL
x( )dx0
L∫= A2
2dx
0
L∫ − cos 2nπL
x( )dx0
L∫⎡⎣⎢
⎤⎦⎥
= A2
2x − L
2nπsin 2nπ
Lx( )⎡
⎣⎢⎤⎦⎥0
L
= A2
2L =1
∴ A = 2L
sin2θ = 1
21− cos2θ( )
Probability density
From plot, for n = 1, |ψ|2 = 2/L (max value) at x = L/2 but for n = 2, |ψ|2 = 0 at x = L/2
depends dramatically on the quantum number!
(This is also in contrast to classical view of equal probability everywhere throughout the box!)
But consider expectation value of position……..
ψ n = 2L sin
nπL x( )
ψn2= 2 Lsin
2 nπL x( )
Expectation value of position
Average position is the middle of the box, irrespective of n!
(Same as classical…) N.B. Classical view (uniform probability across box) is equivalent to very large n
�
x = xψ 2dx−∞
+∞∫ = 2L x sin20
L∫ nπL x( )dx
= 2Lx 2
4−x sin 2nπ L x( )2 2nπ L( ) −
cos 2nπ L x( )8 2nπ L( )2
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥ 0
L
= 2LL24( ) = L2
Finite Square Potential Well
Comparison with infinite case (left): (1) potential is finite at edges of well (2) three regions: I (x < 0), II (0 ≤ x ≤ L) and III (x > L) (3) specify that regions I and III extend to infinite distance (4) consider only case of E ≤ Uo Classical view: infinite and finite well cases are equivalent Quantum view: En values differ & “particle in the walls”
U
x0 L
∞ ∞
U = 0x
0 L
-∞ +∞U = Uo U = Uo
region I region II region III
Apply SSSE to region I
where quantity (E - Uo) is negative, expect different solution! Re-write as:
Solutions are real exponentials:
For ψI to be finite as x approaches -∞ : must have B = 0 As x is -ve, hence exponential decay Finite ψI means that particle exists in the wall!
U = 0x
0 L
-∞ +∞U = Uo U = Uo
region I region II region III
Since U = Uo ≥ E, SSSE becomes
ψ I = Aeax + Be− ax
ψ I = Aeax
d 2ψdx2
+ 2m
2 E −Uo( )ψ = 0
d 2ψdx2
− a2ψ = 0 a =
2m Uo − E( )
where
Other regions?
In region I, the wavefunction is exponential decay (green), likewise in region III
(use same arguments, think of boundary x = L as x´ = 0)
Region II is different because U = 0 so (E - U) is positive, expect complex ‘wave’ solutions, as previous infinite well case? Yes, but with important difference…..
U = 0x
0 L
-∞ +∞U = Uo U = Uo
region I region II region III
Region II solution For infinite wall case had
where B = 0 because ψ = 0 at x = 0; this is not a requirement in finite case (as ψ is non-zero in regions I and III). Therefore, keep general solution as:
But expect G is small
In that case, the main effect of 2nd term is at the walls When apply boundary conditions, ensuring that ψ and dψ/dx both match across wall (more complicated maths), find that the wavelengths are slightly longer (eg λ1 > 2L) (effect of adding cos term!) and hence energies are slightly lower !
ψ = Asin p
x⎛
⎝⎜
⎞⎠⎟+ Bcos p
x⎛
⎝⎜
⎞⎠⎟
ψ = F sin p
x⎛
⎝⎜
⎞⎠⎟+G cos p
x⎛
⎝⎜
⎞⎠⎟
Illustrate the complete wavefunction ψ plot shows: longer wavelength, or lower energies
|ψ |2 plot shows: particle can exist in the walls
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