# Biggest variance of $h=\sum_i H_i$?

What's the biggest variance of $$h=\sum_i H_i$$ where $$H_i$$ is the hamiltonian act on the ith qubit?

If the n qubits state is separable, i.e., the state is $$\mid\psi_1\rangle\otimes\mid\psi_2\rangle\otimes\cdots\mid\psi_n\rangle$$. Obviously, the biggest variance, $$\max\limits_{\mid\psi\rangle}(\Delta h)^2$$, is $$N(\lambda_M-\lambda_m)^2/4$$, where $$\lambda_M \text{ and } \lambda_m$$ are biggest and smallest eigenvalue of $$H_i$$ respectively. To see this, we can use the general formula for state $$\mid\psi_i\rangle\equiv \cos(\theta)\mid0\rangle+\sin(\theta)e^{i\phi}\mid1\rangle$$. We then calculate the variance of one part: $$\mid\psi_i\rangle$$ and will get the biggest variance is when $$\theta=\frac{\pi}{4}$$, and the variance is $$(\lambda_M-\lambda_m)^2/4$$. At last the total variance of $$N$$ parts can be added as $$N(\lambda_M-\lambda_m)^2/4$$.

But how to prove if the total state $$\mid\psi\rangle$$ is not necessarily separable, the biggest variance of $$h$$ becomes $$(N\lambda_M-N\lambda_m)^2/4$$ instead of $$N(\lambda_M-\lambda_m)^2/4$$?

Here is an approach that requires no specific knowledge about $$|\psi\rangle$$ whatsoever.

In your description you implied that each $$H_i$$ has the same maximum and minimum eigenvalues $$\lambda_m$$ and $$\lambda_M$$ respectively so I will assume this in the derivation. The process of measuring $$\langle h \rangle$$ empirically can be thought of running a series of experiments with the $$i$$-th experiment returning an energy $$E_i$$ in the range $$[N\lambda_m, N\lambda_M]$$. Then we can average all of the observed $$E_i$$ values to compute $$\langle h \rangle$$, and we can think of $$E$$ being a random variable whose probability density (or mass) function is completely determined by $$h$$ and $$|\psi\rangle$$.

Let $$X$$ be a random variable defined as $$E- N\lambda_m$$ and therefore takes values in the range $$[0, N(\lambda_M - \lambda_m)]$$. The variance of $$X$$ is: \begin{align}\tag{1} \text{Var}(X) &= \frac{1}{n}\sum_i X_i^2 - \bar{X}^2 \\ &\leq \frac{1}{n}\sum_i N(\lambda_M - \lambda_m)X_i - \bar{X}^2 \\ &= (N(\lambda_M - \lambda_m) - \bar{X})\bar{X} \end{align} where the inequality just substitutes one $$X_i$$ for the maximum possible value of $$X$$, and $$n$$ is a number of samples that can be assumed to be taken to infinity. Then set the derivative of this expression to zero, $$\partial_\bar{X}\text{Var}(X) = N(\lambda_M - \lambda_m) - 2\bar{X}=0$$ to get $$\tag{2} \text{argmax}_{\bar{X}} \text{Var}(X) = \frac{N(\lambda_M - \lambda_m)}{2}$$ which results in

$$\tag{3} \max \text{Var}(X) = \left(\frac{N(\lambda_M - \lambda_m)}{2}\right)^2$$

which also gives $$\text{Var}(E) = (\Delta h)^2$$ since variance is not affected by the constant shift of $$N\lambda_m$$.

Note that you can apply this same technique to each $$H_i$$ individually to recover the maximum variance for the case of a separable state; the factor of $$N^2$$ disappears compared to the above derivation and only a factor of $$N$$ is reintroduced by summing up the variances of the separate systems.

• Thanks a lot. I've tried several ways including the method of Lagrange multipliers while still cannot solve it. The answer is helpful. May 24 '21 at 7:21
• But why when I calculate $\text{Var}(E)$ instead of $\text{Var}(E - N\lambda_m)$, and use the same technique, I get $(\frac{N\lambda_M}{2})^2$ instead? May 24 '21 at 7:48
• When you try to derive it that way then the second step in Equation (1) is not necessarily well defined. Namely, the variance $\left(N\lambda_M/2\right)^2$ that you calculate corresponds to a distribution of energies with $\bar{E} = N\lambda_M / 2$, but this choice of $\bar{E}$ might not actually fall in $[N\lambda_m, N\lambda_M]$ in which case the resulting variance does not correspond to any achievable distribution. This was the reason for working with a shifted variable $X$ whose expected value (2) is guaranteed to fall in $[0, N(\lambda_M - \lambda_m)]$ May 24 '21 at 18:18
• also you might find the discussions here very useful: stats.stackexchange.com/questions/45588/… May 24 '21 at 18:19