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In quantum mechanics of discrete systems, a state is represented by a linear combination of $n$ basis states: $$|\Psi\rangle = \sum_{i=1}^n \psi_{i}|\psi_i\rangle$$

In a basis like position, where $n\rightarrow\infty$, the state is represented by an integral:

$$|\Psi\rangle = \int\text{d}x\, \psi(x)|x\rangle$$

I understand the intuition in replacing an infinite sum with an integral and how Riemann sums converge to integrals generally, but don't understand where the factor of $\text{d}x$ comes from that makes this replacement valid.

How does the continuous definition follow from the discrete definition, and does the factor of length change the state's units?

Qmechanic
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1 Answers1

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In discrete systems, the basis states are usually normalized to unity: $\langle\psi_i|\psi_j\rangle = \delta_{ij}$. In contrast, in the continuous case the states are normalized by Dirac delta function. For example, for the 1D particle \begin{equation} \langle x|x'\rangle = \delta(x-x'), \end{equation} which means that the norm of the state $|x\rangle$ is $\delta(0) = \infty$. This intuitively explains the emergence of infinitely small $dx$ in the integral.

A link between the continual and discrete cases can be made by considering a lattice approximation for the continuum as @LPZ commented. Assume that the particle on a line can occupy only discrete sites $x_k = k\Delta x$. For each site, there exists a corresponding ket vector $|x_k^{discr}\rangle$ normalized by unity: \begin{equation} \langle x_k^{discr} | x_{k'}^{discr}\rangle = \delta_{kk'}, \end{equation} Arbitrary ket vector can be decomposed into a sum \begin{equation} |\psi\rangle = \sum_k c^{discr}_k |x_k^{discr}\rangle, \end{equation}
where $\sum |c^{discr}_k|^2 = 1$. We would like to take the limit $\Delta x \to 0$ and replace the sums with the integral with the prescription \begin{equation} \sum_k \Delta x F(x_k) \to \int dx F(x) \end{equation} However, it is necessary to change the norms of the basis vectors and expansion coefficients for that.

First, the normalization of $|x_k^{discr}\rangle$ differs from the delta-function normalization which should be in the continuous limit. To fix this, let us introduce new rescaled ket vectors $|x_k\rangle = \frac{1}{\sqrt{\Delta x}}|x_k^{discr}\rangle$. Their scalar product is \begin{equation} \langle x_k | x_{k'}\rangle = \frac{\delta_{kk'}}{\Delta x} \to \delta(x_k - x_{k'}) \end{equation} which approaches delta-function at $\Delta x \to 0$.

Also, let us rescale $c^{discr}_k$ as \begin{equation} c^{discr}_k = \sqrt{\Delta x}c(x_k). \end{equation}

With new $|x_k\rangle$ and $c(x_k)$, the decomposition of $|\psi\rangle$ becomes \begin{equation} |\psi\rangle = \sum_k \Delta x \cdot c(x_k) |x\rangle \to \int dx \,c(x) |x\rangle, \end{equation} as usually defined for the continuous case. The coeficients $c(x)$ are also normalized: \begin{equation} \sum \Delta x |c(x_k)|^2 \to \int |c(x)|^2 dx = 1. \end{equation}

E. Anikin
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