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

1 Answer 1

i think the answer to the question is that the relation of nth raw moment of the operator $A$ ($<\hat{A^n}>$)

i.e $<\hat{A^n}> = Cn + D$,

does not hold in general (not even for Gaussian random variables sometimes assumed as being simple). The rest follows from that.

Now another question is what would be the physical or mathematical meaning of a negative variance. This possibly could involve dealing directly in a complex space (e.g complex probability), however this is under discussion (or convention).

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