Postulates and Theorems of Quantum Mechanics Literally, a postulate is something taen as self-evident or assumed without proof as a basis for reasoning. It is simply is Postulate 1: State of a physical system is fully described by a wave function: Ψ ( r1, r,...; t) Variables (r 1, r - xyz coord. particle 1,, t = time) Anything not in wave function cannot be nown in QM. Shorthand represent state by quantum numbers, Ψlm,... or observables Ψ ab,... Alternatively lmn,,,... > - vector bracet (Heisenberg) The wave function must be single-valued, continuous (including its first and second derivative), and quadratically integrable. A function obeying all three conditions is well-behaved. (a) multivalued (b) discontinuous (c) not quadratically integrable Postulate : Any quantity that is physically observable can be represented by a Hermitian operator. A Hermitian operator is a linear operator that satisfies the condition ˆ ˆ ψ 1F ψ d τ = ψ ( F ψ 1) d τ, for any pair of functions ψ 1 and ψ which represent physical states of the system. Note: ψ -- complex conjugate ψ, change i -i wherever it occurs. 1
Postulate 3: The allowable results of an observation of the quantity represented by ˆF are any of the eigenvalues f i of ˆF, where F ˆ ψ = fψ i i i In other words, if ψ 1 is an eigenfunction of F with eigenvalue f i, then a measurement of F is certain to yield the value of f i Postulate 4: The average or expectation value, <F> of any observable F, which corresponds to an operator ˆF, is given by: + ˆ F F ψ Fˆ d < >= The above equation that ψ is normalized ψ τ + ψψ Postulate 5: A quantum mechanical operator corresponding to a physical quantity is constructed by writing down the classical expression in terms of the variables and converting that expression to an operator by means of the following rules: Classical Variable Q. M. operator Expression for operator Operation x ˆx x Multiply by x p x p ˆ x h i x Tae derivative w/r/t x and multiply by h i t ˆt t Multiply by t E Ê h Tae derivative i t w/r/t t and multiply h by i Postulate 6: The wave function Ψ ( x, t) is a solution to the time-dependent Schrödinger equation, ˆ i ψ ( x, t) H ( x, t ) ψ ( x, t ) =, where Ĥ is the Hamiltonian operator. t The classical Hamiltonian is the operator for the total energy, the sum of inetic p and potential energy of the system. H = + U( x, t) m = 1
From Postulate 5 we define the quantum mechanical Hamiltonian, ˆ H = + U( x, t) m x Time-Independent Schrödinger equation Hˆ ψ = Eψ m + (,, ) = ψ U x y z ψ Eψ Where x y z = + + Laplacian Time-dependent Schrödinger equation ψ ψ + U( x, y, z, t) ψ = i m t If the potential U is not a function of t, then we can separate the variables in this iet / equation, as ψ ( xyzt,,, ) = ψ ( xyze,, ) Physical meaning of wave functions (Max Born) 196 Born suggested that ψ ψ should be regarded as probability density for finding the particle for given location in space, The wave function is a complex quantity and its absolute value is: ψ ψ = f + ig and ψ = f ig (Note: to get ψ we replace i with i wherever it occurs) ψ ψ = ( f ig)( f + ig) = f i g = f + g = ψ The quantity ψ ψ = f + g is real and non- negative, as a probability density must be. b a x b ψ dx Pr( ) = one-particle., one-dim syst. a 3
The probability for finding the particle somewhere on the x axis must be 1. Hence, + ψ = 1 dx For one-particle in a 3-dim system:. When ψ satisfies this equation is said to be normalized + + + The normalization requirement is often written: x y z t dxdydz ψ (,,, ) = 1 ψ d τ = 1 and is a shorthand notation that stands for the definite integral over the full ranges of all the spatial coordinates of the system. In reality ψ is not to be thought as a physical wave. Instead it is an abstract mathematical entity that gives information about the state of the system. A more accurate and better name for it is state function, because it describes in full the state of the system at any time. More on operators Reminder: An operator is a rule that transforms a given function into another function. If the operator  transforms f (x) into g (x), then we can write  f ( x ) = g ( x ) Provided that f (x) is differentiable then d/dx gives us f We define the sum and the difference of two operators: (A+B) ˆ ˆ f ( x) A ˆf( x) + B ˆf( x) (A-B) ˆ ˆ f ( x) A ˆf ( x) - B ˆf( x) The product of two operators is defined as: AB ˆ ˆ f ( x) A[B ˆ ˆ f ( x)] In other words, we first operate on f(x) with the operator on the right of the operator product, and then we tae the resulting function and operate on it with the ' operator on the left of the operator product. 3D ˆf ( x) = 3[D ˆf ( x)] = 3 f ( x) We define as commutator [A,B] ˆ ˆ of the operators A and B the operator AB ˆ ˆ - BA ˆ ˆ Therefore, [A,B] ˆ ˆ = AB ˆ ˆ - BA ˆ ˆ If AB ˆ ˆ - BA=0 ˆ ˆ, then [A,B] ˆ ˆ = 0 Operators A and B commute! Other wise they do not commute. 4
In Quantum Mechanics all operators are linear.  is said to be linear if and only if it has the two following properties: A[ ˆ f ( x) + g(x)] = A ˆf ( x) + A ˆg( x) A[c ˆ f ( x)] = ca ˆf ( x) Hermitian Operators (after Charles Hermite) The quantum mechanical average value <A> of the physical quantity A must be a real number. To tae the complex conjugate of a number we replace i by i where it occurs. A real number does not contain i, so a real number equals its complex conjugate: z = z. Therefore, <A> = <A>. We have, A A ˆ < >= Ψ Ψ and < A > = ( A ˆ ) (A ˆ Ψ Ψ = Ψ Ψ) Therefore, A ˆ (A ˆ Ψ Ψ = Ψ Ψ) This equation must hold for all possible state functions of Ψ, that is, for all functions that are continuous, single-valued and quadratically integrable. In the above argument we utilized (fg) = f g, where f and g are complex quantitities. Prove the relation Therefore, a linear operator that obeys the above equation for all well-behaved functions is a Hermitian operator. If  is a Hermitian operator, it follows that ˆ ˆ f Ag = g(a f) necessarily eigenfunctions of any operator) and the integrals are definite integrals over all space., where f and g are arbitrary well-behaved functions (not Eigenvalues of Hermitian Operators Theorem 1: The eigenvalues of a Hermitian operator are real. Consider that f and g are the same function and this function is an eigenfunction of  with eigenvalues b. With f = g and  f = bf then the above last relation becomes, f bf = f ( bf ) Using (bf) = b f and taing the constants outside of the integrals, we get 5
b f f = b f f ( b b ) f = 0 The quantity f is never negative. The only way the definite integral f = 0 would be equal to zero it could happen only and only if f is zero everywhere. However f = 0 is not allowed as an eigenfunction. Therefore b = b. Only a real number is equal to its complex conjugate, so the eignevalues b are real. Theorem : Two eigenfunctions of a Hermitian operator that correspond to different eigenvalues are orthogonal. Orthogonal mean that ψψ i j= 0 when i j Let  f = bf and  g = cg The Hermitian property f Ag ˆ = g(a ˆf) becomes g τ = ( ) τ = τ = τ f g = 0. The theorem is proved! c f d g bf d gb f d b gf d Since a Hermitian operator has real eigenvalues: If c b then, ( c b) f g = 0 Example: Given that  is a linear operator and that Âf1 = bf1 and Âf = bf prove that cf 1 1+ cf, where c1 and c are constants, is an eigenfunction of  with eigenvalue b. Example: If  is a linear operator and that Âf1 = bf1 and Âf = bf, and we f1 fd τ define g 1 and g as g1 f1 and g f + f1 where, verify f1 fd 1 τ that g 1 and g are orthogonal. If c = b then orthogonality need not necessarily hold. degeneracy Let the set of functions g 1, g, g 3, be the eigenfunctions of a Hermitian operator. Since these functions are (or can be chosen to be) orthogonal, we have 6
gg j = 0 where j. We shall always normalize eigenfunctions of gg j j = 1 operators, then. The last two expressions can be written as a single equation, gg j = δ j (Kronecer delta) δ j 1 when j = and δ j 0 when j A set of functions that are orthogonal and normalized is called orthonormal set. Theorem 3: The set of eigenfunctions of a Hermitian operator that represents a physical quantity is a complete set. If F is a well-behaved function and the set g 1, g, g 3, is the set of eigenfunctions of the operator  that corresponds to the physical property A, then F = cg. F has been expanded in terms of the set of g s. How do we find the coefficients c in the above expansion? Attention: 1) gf= cgg ) j j δ j j δ j j δ gf = cgg = cgg = c gg = cδ j j j i j j Recall, 1 when = and 0 when, and c j = cjδ jj = cj Therefore, c j = g jf Changing j to and we get ( δτ ) ( gfδτ) j are constants. ( δτ ) F gf g = where the definite integrals = expansion of F in terms of a nown complete set of F gf g functions g 1, g, g 3, 7