Expectation value of an operator to the power of n

Simplifying will give you the desired result.Hi all"How do I prove that <A^n>=<A>^n? It seems intuitive but how do I rigorously prove it?" The user provides their attempt at a proof, but it is only rigorous enough when the wavefunction is an eigenstate of A with eigenvalue λ. However, a counterexample is given for n=2 to show that the variance is not necessarily zero. The user is advised to use the Binomial Theorem to derive a general expression for <A^n>-<A>^n.
  • #1
patric44
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Homework Statement
prove that : <A^n>=<A>^n
Relevant Equations
<A^n>=<A>^n
hi all
how do I prove that
$$
<A^{n}>=<A>^{n}
$$
It seems intuitive but how do I rigorously prove it, My attempt was like , the LHS can be written as:
$$
\bra{\Psi}\hat{A}.\hat{A}.\hat{A}...\ket{\Psi}=\lambda^{n} \bra{\Psi}\ket{\Psi}=\lambda^{n}\delta_{ii}=\lambda^{n}
$$
and the RHS equal:
$$
<A>^{n}=[\bra{\Psi}A\ket{\Psi}]^{n}=\lambda^{n}[\bra{\Psi}\ket{\Psi}]^{n}=\lambda^{n}[\delta_{ii}]^{n}=\lambda^{n}
$$
Is my proof rigurus enough or there are other formal proof for that
 
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  • #2
A counter example. For ground state of a partricle in a box [-a,a],
<x>=0 but [tex] <x^2> \ \ >\ \ <x>^2=0 [/tex]

Your proof seems to be all right only when ##\Psi## is an eigenstate of A with eigenvalue ##\lambda##.
 
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  • #3
patric44 said:
Homework Statement: prove that : <A^n>=<A>^n
Note that for ##n = 2##, the variance is not necessarily zero:
$$\sigma^2(A) =\langle A^2 \rangle - \langle A \rangle^2 \ne 0$$In general, you can derive an expression for ##\langle A^n \rangle - \langle A \rangle^n## by starting with:
$$\langle[A - \langle A \rangle]^n \rangle$$And expanding using the Binomial Theorem.
 
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