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With single slit electron diffraction, prove Heisenberg's uncertainty s principle.

Mumbai university > FE > SEM 2 > Applied Physics II

Marks: 5M

Year: Dec 2015

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Heisenberg's uncertainty principle

  • The uncertainty principle discovered by Heisenberg reveals a major difference between classical and quantum mechanics.

  • Whereas in classical mechanics the observables are functions on phase space and therefore constitute a commutative algebra, in quantum mechanics the observables are operators, and their algebra is non-commutative.

  • Heisenberg's principle, stated as a mathematical theorem below, asserts that two observables of a quantum system cannot be measured simultaneously with absolute accuracy unless they commute as operators.

  • There is an intrinsic uncertainty in their simultaneous measurement, one that is not due simply to experimental errors.

  • Let us suppose that A is an observable of a given quantum system.

  • Thus, A is a (densely defined) self-adjoin operator on a Hilbert space H. Let ψ € H with |ψ| = 1 represent a state of the system. Then ${A}_ψ = (ψ, Aψ)$ is the expected value of the observable A in the state *.

  • The dispersion of A in the state ψ is given by the square root of its variance; that is

    $Δ_ψA = ((A - (A)_ψ I)^2 )^{1/2} = || Aψ - (A)_ψ ψ|| $

  • Note that if ψ is an eigenvector of A with eigenvalue λ, then ${A}_ψ = λ $ and $△_ψ A = 0$.

Heisenberg's uncertainty principle

  • If A and B are observables of a quantum system, then for every state 11/ common to both operators we have

    $Δ_ψ A Δ_ψ B ≥ \frac{1}{2} |[A,B]_ψ| $

Proof

  • Subtracting a multiple of the identity from A and another such multiple from B does not change their commentator, and it does not affect their variances.

  • Hence we may suppose that ${A}_ψ = {B}_ψ = 0$.

  • Now, we have, using the self-adroitness of both operators,

    $|[A,B]_ψ| = |ψ, (AB - BA)ψ|$

    $|[A,B]_ψ| = |(Aψ, Bψ) - (Bψ, Aψ)|$

    $|[A,B]_ψ| = 2|Im (Aψ , Bψ)|$

  • But the Cauchy—Schwarz inequality tells us that

    $|Im (Aψ , Bψ)| ≤ ||Aψ|| ||Bψ||$

  • Combining (2.2) with (2.3) yields (2.1), as desired.

  • A simple consequence of this result is the fact that in a quantum particle system, the position and momentum of a particle cannot be measured simultaneously with absolute certainty.

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