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CSIR NET Chemical Sciences

Physical Chemistry

The commutator [x2,px2] is equal to [x: position operator, px: momentum operator]\text{The comm

Question

$\text{The commutator } [x^2, p_x^2] \text{ is equal to } [x: \text{ position operator, } p_x: \text{ momentum operator}]$

$2x i \hbar$

$2 i \hbar$

$4 i \hbar$

$2 i \hbar (x p_x + p_x x)$

Solution

Correct option is D

If two observables A and B have linear operators

, the commutator is defined by,

The commutator is itself a (composite) operator. Acting the commutator on ψ gives:

If ψ is an eigenfunction with eigenvalues a and b for observables A and B respectively, and if the operators commute:

then the observables A and B can be measured simultaneously with infinite precision, i.e., uncertainties ΔA=0

, ΔB=0simultaneously. ψ is then said to be the simultaneous eigenfunction of A and B. To illustrate this:

It shows that measurement of A and B does not cause any shift of state, i.e., initial and final states are same (no disturbance due to measurement). Suppose we measure A to get value a. We then measure B to get the value b. We measure A again. We still get the same value a. Clearly the state ( ψ) of the system is not destroyed and so we are able to measure A and B simultaneously with infinite precision.
If the operators do not commute: