--%>

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 ) (n22/4)

= 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 : Linde liquefaction process Liquefied

    Liquefied natural gas (LNG) is produced using a Linde liquefaction process from pure methane gas at 3 bar and 280 K (conditions at point 1 in figure below). A three-stage compressor with interceding is used to compress the methane to 100 bar (point 2). The first stage

  • Q : Question based on relative lowering of

    Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre

  • Q : Which solution will have highest

    Which solution will have highest boiling point:(a) 1% solution of glucose in water  (b) 1% solution of sodium chloride in water  (c) 1% solution of zinc sulphate in water  (d) 1% solution of urea in waterAnswer: (b) Na

  • Q : Analytical chemistry 37% weight of HCl

    37% weight of HCl and density is 1.1g/ml. find molarity of HCl

  • Q : Pressure Phase Diagrams The occurrence

    The occurrence of different phases of a one component system can be shown on a pressure temperature. The phases present in a one line system at various temperatures can be conveniently presented on a P- versus-T diagram. An example is pro

  • Q : Neutralisation of phosphorous acids

    Provide solution of this question. To neutralise completely 20 mL of 0.1 M aqueous solution of phosphorous acid (H3 PO3) the volume of 0.1 M aqueous KOH solution required is: (a) 40 mL (b) 20 mL (c) 10 mL (d) 60 mL

  • Q : What are homogenous catalyst? Give few

    When a catalyst mixes homogeneously with the reactants and forms a single phase, the catalyst is said to be homogeneous and this type of catalysis is called homogeneous catalysis. Some more examples of homogeneous catalysis are:    SO2

  • Q : Coordination compounds discuss

    discuss practical uses of coordination compounds

  • Q : Problem on Neutralization What weight

    What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solution? (a) 0.45 g  (b)0.90 g  (c) 1.08 g  (d) 1.26 g      Answer

  • Q : Describe physical adsorption and its

    When the forces of attraction existing between adsorbate and adsorbent are van der Waal's forces, the adsorption is called physical adsorption. This type of adsorption is also known as physisorption or van der Waal's adsorption. Since the forces existing between adsorbent and adsorbate are very w