--%>

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 : 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 : Benefits of soapy detergents over the

    What are the benefits of soapy detergents over the soap less detergents? Briefly state the benefits?

  • Q : What are Ethers and its types? Ethers

    Ethers are the compounds with general formula or CnH2n+

  • Q : Formula of diesel Write a short note on

    Write a short note on the formula of diesel, petrol and also CNG?

  • Q : Depression in the freezing point When

    When 0.01 mole of sugar is dissolved in 100g of a solvent, the depression in freezing point is 0.40o. When 0.03 mole of glucose is dissolved in 50g of the same solvent, depression in the freezing point will be:(a) 0.60o  (b) 0.80o

  • 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 : Problem on colligative properties

    Choose the right answer from following. The magnitude of colligative properties in all colloidal dispersions is : (a) Lowerthan solution (b)Higher than solution(c) Both (d) None

  • Q : Problem on Molar solution Can someone

    Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent

  • Q : Vapour pressure of volatile substance

    Provide solution of this question. According to Raoult's law the relative lowering of vapour pressure of a solution of volatile substance is equal to: (a) Mole fraction of the solvent (b) Mole fraction of the solute (c) Weight percentage of a solute (d) Weight perc

  • Q : Solutions The relative lowering of

    The relative lowering of vapour pressure of 0.2 molal solution in which solvent is benzene