# Significance of an operator with a negative variance

Suppose we have a quantum state, well described by its time-independent wave function $\Psi$. And we have a well-defined Hermitian (self-adjoint) operator $\hat{A}$. We successfully evaluate the expectation value of the operator $\hat{A}$. Next we derive the general formula for the higher moments of $\hat{A}$ (i.e. the expectation value of $\hat{A^n}$ for $n=2,3,4…$). Finally we scale the operator $\hat{A}$ appropriately, in order to make the result dimensionless and to remove a possible growth factor (of type $C^n$) in the moments. We obtain:

$$<\hat{A^n}> = Cn + D$$

for $n=1,2,3,...$ and where $C$ and $D$ are constants.

Let us now define a new operator $\hat{B}$ as follows:

$$\hat{B} = \hat{A^{n+1}} - \hat{A^n}$$

We can easily verify that the first and second moment of B are given by:

$$< \hat{B} > = C$$

$$< \hat{B^2} > = 0$$

Therefore the variance of operator $\hat{B}$ is negative! In violation of statistical laws.

Should we conclude from this example that the results derived for the moments of $\hat{A}$ must be flawed? Or should we conclude that the new operator $\hat{B}$ is not a proper operator after all and therefore its strange properties are insignificant with respect to questions about the validity of $\hat{A}$?

\begin{align} <\hat{B}> &= <\hat{A^{n+1}}> - <\hat{A^n}> \\ &= C(n+1) + D - Cn - D = C \end{align}

\begin{align} <\hat{B^2}> &= <\hat{A^{2n+2}}> - 2<\hat{A^{2n+1}}> + <\hat{A^{2n}}> \\ &= C(2n+2) + D - 2C(2n+1) - 2D + C(2n) + D \\ &= 0 \end{align}

\begin{align} \text{Variance of }\hat{B} &= <\hat{B^2}> - \left(<\hat{B}>\right)^2 \\ &= -C^2 \end{align}

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How are <B> and <B^2> calculated, can you add these? – Nikos M. Jun 13 '14 at 14:49
<B> = <A^(n+1)> - <A^n> = C*(n+1) + D - C*n - D = C. – M. Wind Jun 13 '14 at 15:09
<B^2> = <A^(2n+2)> - 2*<A^(2n+1)> + <A^(2n)> = C*(2n+2) + D - 2*C*(2n+1) - 2*D + C*(2n) + D = 0. – M. Wind Jun 13 '14 at 15:12
Thus: Variance of B = <B^2> - <B>^2 = -C^2 is negative. – M. Wind Jun 13 '14 at 15:16
i dont understand how this $<\left(\hat{A}\right)^n> = Cn + D$ holds in general? – Nikos M. Jun 13 '14 at 16:19

i think the answer to the question is that the relation of nth raw moment of the operator $A$ ($<\hat{A^n}>$)
i.e $<\hat{A^n}> = Cn + D$,