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