Quantum Physics ecture 8 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
2 ψ x 2 + 2m 2 Eigenvalue & eigenfunction ( E U )ψ = 0 2 2m An eigenfunction equation. eigen meaning proper or characteristic. In general there is more than one solution ψ n (eigenfunction) Each solution corresponds to a specific value of energy i.e. eigenfunctions ψ n with corresponding eigenvalues E n - a re-statement of energy quantisation! 2 x 2 + U Hamiltonian Ĥ ψ = Eψ The eigenfunctions ψ n are a complete orthogonal set - c.f. Fourier
particle in a box re-visited 2 ψ x 2 + 2m 2 ( E U )ψ = 0 U described by square well potential with infinitely high potential barriers : (1) U = 0 for 0 < x < (2) U= (and so ψ = 0) for x 0 & x Objective: find ψ for 0 < x < 0 x Region (1) SSSE becomes d 2 ψ dx 2 + 2m 2 Eψ = 0 SHM type equation: Oscillatory solutions e ikx
particle in a box solution of SSSE d 2 ψ dx 2 + 2m 2 Eψ = 0 Recall free particle wavefunction Without time dependence. C is a constant Ψ = Cexp i p x = C exp( ikx) 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: Ψ = Cexp( ikx) + Bexp( ikx) where C and B may be complex Expand this using exp(iθ) = cosθ + isinθ
particle in a box solution of SSSE Ψ = C cos kx + isin kx 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 = so ( ) + B cos( kx) + isin( kx) = C cos kx + isin kx k = nπ ( ) ( ) + B( cos kx isin kx) = ( C + B)cos kx + i( C B)sin kx Ψ = 2iC sin kx = Asin kx n = 1,2,3...
particle in a box energy levels k = nπ n = 1,2,3... p = nπ also E = p2 2m E n = n2 π 2 2 2m 2 = n2 h 2 8m 2 E 3 E 2 quantised energy, as before! E 1
particle in a box wavefunctions Same function form as standing waves of earlier model Properties: ψ n = Asin p x = Asin (1) ψ is single-valued and continuous (2) likewise dψ/dx (with the exception that dψ/dx is non-continuous at x = 0, x = ) 2mE n x = Asin nπ x ( ) - an artificial feature of (ideal) infinitely high potential barriers!
particle in a box wavefunctions (3) ψ normalisable? ψ n = Asin( nπ x ) + 2 2 ψ n dx = ψ n dx = 1 = A 2 sin 2 nπ x = A 2 2 0 0 ( ) dx dx cos 2nπ x 0 0 ( ) dx ( ) = A 2 2 x sin 2nπ x 2nπ 0 sin 2 θ = 1 1 cos2θ 2 ( ) = A 2 2 = 1 A = 2 ψ n = 2 sin ( nπ x )
Probability density ψ n = 2 sin ( nπ x ) ψ n 2 = 2 ( nπ sin2 x) From plot, for n = 1, ψ 2 = 2/ (max value) at x = /2 but for n = 2, ψ 2 = 0 at x = /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..
Expectation value of position x = + xψ 2 dx = 2 x sin 2 ( nπ x)dx = 2 x 2 4 x sin 2nπ x 2( 2nπ ) 0 ( ) ( ) cos 2nπ x 8( 2nπ ) 2 0 = 2 ( 2 ) 4 = 2 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
Finite Square Potential Well U U = U o U = U o - + 0 x U = 0 0 region I region II region III x Comparison with infinite case (left): (1) potential is finite at edges of well (2) three regions: I (x < 0), II (0 x ) and III (x > ) (3) specify that regions I and III extend to infinite distance (4) consider only case of E U o Classical view: infinite and finite well cases are equivalent Quantum view: E n values differ & particle in the walls
Apply SSSE to region I Since U = U o E, SSSE becomes U = U o U = U o - + d 2 ψ dx + 2m ( 2 2 E U o )ψ = 0 U = 0 0 region I region II region III where quantity (E - U o ) is negative, expect different solution! Re-write as: d 2 ψ dx 2 a2 ψ = 0 a = 2m ( U E o ) where Solutions are real exponentials: ψ I = Ae ax + Be ax x For ψ I to be finite as x approaches - : must have B = 0 As x is -ve, hence exponential decay Finite ψ I means that particle exists in the wall! ψ I = Ae ax
Other regions? U = U o U = U o - + U = 0 0 region I region II region III x In region I, the wavefunction is exponential decay (green), likewise in region III (use same arguments, think of boundary x = 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..
Region II solution For infinite wall case had ψ = Asin p x + Bcos p x 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 ψ = F sin p x + G cos p x In that case, the main effect of 2 nd 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 λ 1 > 2) (effect of adding cos term!) and hence energies are slightly lower!
Illustrate the complete wavefunction ψ plot shows: longer wavelength, or lower energies ψ 2 plot shows: particle can exist in the walls