Formalism of Quantum Mechanics

Transcription

1 The theory of quantum mechanics is formulated by defining a set of rules or postulates. These postulates cannot be derived from the laws of classical physics. The rules define the following: 1. How to describe a system (the wave function).. How to describe the time evolution of the wave function (the Schrodinger equation). 3. How to describe observable quantities (operators). 4. How to describe the outcomes of a measurement of observable quantities; how to describe the system after the measurement.

2 The wave function The state of a physical system is represented by a wave function which contains all the information that can be known about the system. The wave function is in general complex (it has real and imaginary parts). In configuration (coordinate) space the wave function of a particle is a function of the position of the particle at a given time:! (x, y,z,t) For N particles, this can be generalized to! (x 1, x,...x N, y 1, y...y N, z 1, z...z N,t)

3 Normalization The wave functions! and c! represent the same state, where c is a complex number. Hence we can always multiply the wave function by an arbitrary complex number without changing it. Consider the normalization integral I =! (x,t) dx =! * (x,t)! (x,t) dx If I is a finite number, then the wave function is square integrable. If we multiply the wave function by a constant c = 1 then the wave function is said to be normalized (to unity): I! (x,t) dx = 1

4 The Born interpretation: Probabilities The absolute square of the normalized wave function! : P(x,t)dx =! (x,t) dx is the probability of the particle being in the region dx around position x at time t. P(x,t) is called the probability density curve Hence the normalization requirement is a statement that the probability of finding the particle somewhere is 1.! (x,t) dx = P(x,t)dx = 1

5 The Born interpretation: Probabilities The probability of finding a particle in a region a < x < b is then b! (x,t) dx a This is the area under the probability density curve between x=a and x=b:

6 Example: Find the value of the constant C of the normalized wave function! (x) = Ce x / L, x # 0! (x) = 0, x < 0 where L = 1nm. Plot the wave function and the probability density curves and find the probability of finding the particle in the region x! 1nm! (x,t) dx = 1 % # c e $x / L dx = $ c L 0 # c = L = e$x / L % 0 = c L = 1 P(x) =! (x) x /1nm = e

7 Example: Find the value of the constant C of the normalized wave function! (x) = Ce x / L, x # 0! (x) = 0, x < 0 where L = 1nm. Plot the wave function and the probability density curves and find the probability of finding the particle in the region x! 1nm P(x) =! (x) x /1nm = e P(x! 1nm) = # # 1nm P(x)dx $ c e % x /L dx = % c L 1nm $ P(x) = = 13.5% e%x / L 1nm = c L e%

8 Superposition principle If the wave functions! 1 and! represent two possible states of the system, then any linear combination! = c c also represents a possible state of the system. This is the superposition principle. This allows for superpositions of a particle wave function in two different locations. (Recall double slit experiment). Note:! = c c = c c = ( c * 1 * 1 + c * * ) ( c c ) + c 1 * c 1 * + c 1 c * 1 *

9 Momentum space wave functions Given a wave function in coordinate space representing the position of a particle, the corresponding momentum wave function is the Fourier integral of the the position wave function:!( p,t) = #i 1! e (! ) % ( xp) $ (x,t) dx 1/ If the position wave function is normalized, the momentum wave function is also normalized to unity:!(p,t) dp = 1

10 Momentum space wave functions We can interpret the square of the quantity P( p,t) =!( p,t) dp of the normalized wave function as the probability at time t of the momentum of a particle being in the volume dp around p Hence the normalization requirement is a statement that the probability of finding the particle with some value of momentum is 1.!(p,t) dx = P( p,t)dp = 1

11 Momentum space wave functions The probability of finding a particle with momentum a < p < b is then b!(p,t) dp a This is the area under the momentum probability density curve between p=a and p=b:!( p,t) p

12 Given a wave function representing the momentum of a particle, the corresponding position space wave function is the inverse Fourier integral of the momentum wave function:! (x,t) = i 1! e (! ) # ( px) $( p,t)dp 1/ If the momentum wave function is normalized, the position wave function is also normalized to unity. Note that the position and momentum wave functions represent the same state of the system. All information about the state can be obtained from either wave function.

13 Time-dependent Schrödinger equation The wave function of a particle undergoing a force F(x) is the solution to the Schrödinger equation: i!!! (x,t) = #!t m! (x,t) + U(x) (x,t)!x U(x) is the potential energy associated with the force: F =! U x

14 Time-dependent Schrödinger equation: Separation of variables i!!! (x,t) = #!t m! (x,t) + U(x) (x,t)!x Since U(x) does not depend on time, solutions can be written in separable form as a part that is only position dependent and a part that is only time dependent:! (x,t) = (x)#(t) Inserting this into the above equation, we get i!!(x) t #(t) = $#(t)! m!(x) + U(x)!(x)#(t) x

15 Time-dependent Schrödinger equation: Separation of variables! (x,t) = (x)#(t) i!!(x) t #(t) = $#(t)! m!(x) + U(x)!(x)#(t) x Dividing by!(x,t), i!!(t) t!(t) = # 1 $(x)! m $(x) + U(x) x Left hand side (LHS) is a function of t alone Right hand side (RHS) is a function of x alone LHS=RHS only if LHS = E and RHS = E (E is a constant)

16 Time-dependent Schrödinger equation: Separation of variables! (x,t) = (x)#(t) i!!(t) t!(t) = # 1 $(x)! m $(x) + U(x) x LHS=RHS only if LHS = E and RHS = E (E is a constant) i!!(t) 1 #(x) d dt!(t) = E! m d #(x) + U(x) = E dx

17 Time-dependent Schrödinger equation: Separation of variables! (x,t) = (x)#(t) Solutions for the time-dependent equation: i! d dt!(t) = E!(t)!(t) = e i#t, # = E! Check: d dt!(t) = ie i! e! Et # i! d i dt!(t) = Ee! Et = E!(t)

18 Time-dependent Schrödinger equation: Separation of variables! (x,t) = (x)#(t) Time-independent Schrödinger equation:!! m d (x) + U(x)(x) = E(x) dx This equation is not always easy to solve analytically, but can be solved numerically on a computer. However we can analytically solve some special cases.