Graduate Discrete measurement operator definition

[jim jones]

Consider the Gaussian position measurement operators $$\hat{A}_y = \int_{-\infty}^{\infty}ae^{\frac{-(x-y)^2}{2c^2}}|x \rangle \langle x|dx$$ where $|x \rangle$ are position eigenstates. I can show that this satisfies the required property of measurement operators: $$\int_{-\infty}^{\infty}A_{y}^{\dagger}A_{y}dy = 1$$ where $1$ is the identity operator. I am interested in defining an analogue of this continuous Gaussian measurement operator for some discrete eigenstates $|m\rangle$. These are discrete eigenstates rather than continuous as above, hence I'm not sure how to define the Gaussian part of $\hat{A}_y$. I've considered the Binomial distribution $$Pr(X = k) :=
\begin{pmatrix}
n \\
k
\end{pmatrix}p^{k}(1-p)^{n-k}$$ for $p=0.5$, since this is a discrete probability density function which resembles the normal distribution as $n \to \infty$, but I'm not sure how to implement this since this distribution is only centered at positive integers. Does anyone have an idea of what would be a reasonable way to define the Gaussian part of this measurement operator for the discrete case?

Thanks or any assistance, let me know if any clarity is needed.

 

[andrewkirk]

Subtract the mean from the binomial random variable $k$ in order to centre the distribution at zero. Then the limit as $n\to\infty$ will be a Gaussian with zero mean.

$$
Pr(X = k) :=
\begin{pmatrix}
n \\
k'
\end{pmatrix}p^{k'}(1-p)^{n-k'}
\quad\textrm{where }
k'=round(k+np)
$$

 

[jim jones]

Thanks for your response. Consider the following idea if you have a chance, let me know what you think. If we consider $X \sim Binom(n,p)$ with change of variable $Z_c = X -np + c$, hence it follows that $E(Z_c) = c$ and $Var(Z_c) = Var(X)$. Hence we have the distribution shifted to a new mean $c$.

We then have the probability distribution $Pr(Z_k|c) := Pr(Z_c = k -np +c) = Pr(X=k)$. If we define an analogue $\hat{A}_c$ to the measurement operator $\hat{A}_y$ above as we have for the continuous case, then we can define $$\hat{A}_c = \sum_{k}\sqrt{Pr(Z_k|c)}|Z_{k} \rangle \langle Z_k|$$
hence we have $$\sum_c A_{c}^{\dagger}A_c = \sum_{c}\sum_{j} P(Z_{j}|c) | Z_j \rangle \langle Z_j | = \sum_{j} \sum_{c} P(Z_{j}|c) |Z_{j} \rangle \langle Z_j| = \sum_j |Z_j \rangle \langle Z_j | = 1$$ What do you think of this idea?