--%>

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 : Mole fraction of hydrogen Give me

    Give me answer of this question. In a mixture of 1 gm H2 and 8 gm O2 , the mole fraction of hydrogen is: (a) 0.667 (b) 0.5 (c) 0.33 (d) None of these

  • Q : Why acetic has less conductivity than

    Illustrate the reason, why acetic has less conductivity than Hcl?

  • Q : Help 1) Chromium(III) hydroxide is

    1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer. 2) Explain how dissolving t

  • Q : Vapour pressure related question Help

    Help me to solve this question. Which of the following is incorrect: (a) Relative lowering of vapour pressure is independent (b)The vapour pressure is a colligative property (c)Vapour pressure of a solution is lower than the vapour pressure of the solvent (d)The

  • Q : Problem associated to vapour pressure

    Provide solution of this question. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is P0 , the vapour pressure of solution is:(a) 0.10P0 (b) 1.10P0 (c) 0.90P0 (d) 0.99P0

  • Q : Molar mass Select the right answer of

    Select the right answer of the question. Which is heaviest: (a)25 gm of mercury (b)2 moles of water (c)2 moles of carbon dioxide (d)4 gm atoms of oxygen

  • Q : Molality of a glucose solution What

    What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m  (ii) 0.5m  (iii) 0.2m  (iv) 2m

  • Q : What are biodegradable polymers?

      These are polymers that can be broken into small segments by enzyme-catalysed reactions. The required enzymes are produced by microorganism. It is a known fact that the carbon-carbon bonds of chain growth polymers are inert to enzyme-catalysed reactions, and hence they are non biod

  • Q : Problem based on lowering in vapour

    Help me to solve this problem. An aqueous solution of glucose was prepared by dissolving 18 g of glucose in 90 g of water. The relative lowering in vapour pressure is: (a) 0.02 (b)1 (c) 20 (d)180

  • Q : Molar and Volumetric flow rate problem

    Cyclohexane (C6H12) is produced by mixing Benzene and hydrogen. A process including a reactor, separator, and recycle stream is used to produce Cyclohexane. The fresh feed contains 260L/min C6H6 with 950 L/min of H2