--%>

Quantum Mechanical Operators

The quantum mechanical methods, illustrated previously by the Schrödinger equation, are extended by the use of operators.

1740_Quantum Machanics.png 
Or, with h for h/(2∏), as

2160_Quantum Mechanics.png 

For each of the set of functions that satisfies this equation, the quantity ε is the energy of the particle in the state corresponding to that solution function.

This equation, with which the energy corresponding to each allowed state is calculated, is just one of a number of equations that can be set up to calculate properties of quantum mechanical systems. All these expressions can be looked on as operator equations. Equation can be displayed to show this feature by writing it as

393_Quantum Mechanics1.png 

This expression in the square brackets is an example of an operator. This particular operator is one dimensional Hamiltonian operator. It or its two-or three-dimensional counterparts are given the symbol H. with this notation, equation can be written as

H ψ = ε ψ

Earlier  we looked for functions that solved the Schrodinger equation, such as acted on by the operator H, give back a constant times the function. In this equation is the energy corresponding to that function. Functions that satisfy equations such as are known as eigenfunctoins, and the values of the constant, such as ε of equation are eigenvalues.

The energies of a system are identified as the eigenvalues for the Hamiltonian operator. Any other observable quantity has its own operator. The operator approach is therefore quite general. When an operator from an observable quantity operates on the wave function for the system and gives a result which is constant times the wave function, that constant is the value of the observable quantity.

Normalization: wave functions can be imaginary or complex, i.e. they can involve I = √-1. Let us now allow ψ to be such a function. Its complex conjucate, obtained by replacing I wherever it appears by -i, is denoted ψ *. A complex ψ is normalized if

∫ ψ * ψ d 273_Quantum Mechanics8.png= 1 

Example: normalize the wave functions for a particle on a line given as ψ = (const) sin (n∏x/a).

Solution: a wave function in one dimension is normalized if ∫ ψ * ψ dx= 1. Here we require that

13_Quantum Mechanics2.png 

The integral can be simplified by introducing y = n∏x/a, so that

626_Quantum Mechanics3.png 

Now the integration result given in integral tables can be used to obtain

2132_Quantum Mechanics4.png 

= (a/(n∏)) ((n∏)/2)

= a/2


It follows that (const) = (2/a)1/2 and that the normalized wave function is    

ψ = (2/a)1/2 sin n∏x/a

Example: use the normalized wave function expression ψ = (2/a)1/2 sin (n∏x/a) for a particle-on-a-line and the position operator to obtain the expectation value for the position of a particle on a line segment.

Solution: the position operator is the x coordinate and the expectation value is given by equation here we have

2014_Quantum Mechanics5.png 

Substitution of y = n∏x/a converts this to 

1759_Quantum Mechanics6.png 

Use of the integration result from tables of integrals then gives

438_Quantum Mechanics7.png 

= 2a/(n2 ∏2 ) (n2∏2/4)

= a/2


We have come, by this formal procedure, to the result that the average, or expectation, value for the position of a particle on a line segment is at the middle of the segment. This result is apparent from symmetry of the wave functions.

   Related Questions in Chemistry

  • Q : Molar mass of solute The boiling point

    The boiling point of benzene is 353.23 K. If 1.80 gm of a non-volatile solute was dissolved in 90 gm of benzene, the boiling point is increased to 354.11 K. Then the molar mass of the solute is: (a) 5.8g mol-1  (b)

  • Q : Ionization Potential Second ionization

    Second ionization potential of Li, Be and B is in the order (a)Li>Be>B (b)Li>B>Be (c)Be>Li>B (d)B>Be>Li

  • Q : Explain the preparation of phenols. The

    The methods used for the preparation of phenols are given below:    From aryl sulphonic acids

  • Q : Problem on normality Help me to solve

    Help me to solve this problem. 0.5 M of H2AO4 is diluted from 1 lire to 10 litre, normality of resulting solution is : (a)1 N (b) 0.1 N (c)10 N (d)11 N

  • Q : Molecular mass from Raoults law Provide

    Provide solution of this question. Determination of correct molecular mass from Raoult's law is applicable to: (a) An electrolyte in solution (b) A non-electrolyte in a dilute solution (c) A non-electrolyte in a concentrated solution (d) An electrolyte in a liquid so

  • Q : How to establish nomenclature for

    In the common chemistry terminologies, aliphatic halogen derivatives are named as alkyl halides. The words, n-, sec-, tert-, iso-, neo-, and amyl are

  • Q : Schrodinger equation with particle in a

    Three dimensional applications of the Schrodinger equation are introduced by the particle-in-a-box problem.So far only a one-dimensional problem has been solved by application of the Schrodinger equation. Now the allowed energies and the probability functi

  • Q : Atmospheric pressure Give me answer of

    Give me answer of this question. The atmospheric pressure is sum of the: (a) Pressure of the biomolecules (b) Vapour pressure of atmospheric constituents (c) Vapour pressure of chemicals and vapour pressure of volatile (d) Pressure created on to atmospheric molecules

  • Q : Describe Transformation Matrices. Each

    Each symmetry operation can be represented by a transformation matrix.You have seen what happens when a molecule is subjected to the symmetry operation that corresponds to any of the symmetry elements of the point group to which the molecule belongs. The m

  • Q : Molarity of solution Help me to go

    Help me to go through this problem. When 7.1gm Na2SO4 (molecular mass 142) dissolves in 100ml H2O , the molarity of the solution is: (a) 2.0 M (b) 1.0 M (c) 0.5 M (d) 0.05 M