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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?

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    $\begingroup$ You would have to construct a limiting process for each case. e.g. the case for momentum would be a different limiting process than for position. And yes, the factor of length will change the wavefunction's units and magnitude to compensate. $\endgroup$ Commented Apr 19, 2023 at 18:14
  • $\begingroup$ Technically it doesn't follow all that trivially. Physicists like to handwave these things into existence, but in mathematical reality you have to study the solution theory of quantum mechanical equations very carefully. To give you an idea of what is really involved in this, you can take a look at en.wikipedia.org/wiki/Spectrum_(functional_analysis). The complexity of the phenomenon shouldn't discourage you to learn more about the handwaving methods of physics. It's just to show that "proving" that they make sense is not easy. $\endgroup$ Commented Apr 19, 2023 at 18:34
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    $\begingroup$ Intuitively, the small $dx$ comes from the fact that the kets $|x\rangle$ have infinite norm. Rather than being orthonormal, they satisfy:$$\langle y|x\rangle=\delta(x-y)$$ By dimensional analysis you can see that the discrete sum needs to be corrected by the small $dx$ to let the norms diverge. Naturally, this is a heuristic, but can be made rigorous by considering say a periodic box and letting its volume go to infinity. $\endgroup$
    – LPZ
    Commented Apr 19, 2023 at 19:15
  • $\begingroup$ see physics.stackexchange.com/a/359982/50583 $\endgroup$
    – ACuriousMind
    Commented Apr 19, 2023 at 19:40

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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}

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    $\begingroup$ Yes, that's the usual physical handwaving. This post math.stackexchange.com/questions/4086667/… shows why it doesn't really work. To a physicist it doesn't matter most of the time that it doesn't work, to a mathematician it does. The proper mathematical theory to do this "correctly" has been around for eighty plus years now, I believe. That physicists are not using it much (if at all) is practical: it doesn't actually address the real problems with continuous states that pop up in QFT where they matter. $\endgroup$ Commented Apr 19, 2023 at 20:34
  • $\begingroup$ Although I agree that this argumentation is handwaving, I don't see how the question you cited shows that it doesn't work. The question is about some normalization issue which is resolved in the answer to the question. In my answer, normalization is also taken into account accurately. I think the caveats are more subtle (and I'd like to learn them some day) $\endgroup$
    – E. Anikin
    Commented Apr 19, 2023 at 20:58
  • $\begingroup$ I agree, the problems with continuous spectra of linear operators are extremely subtle. Like I said, they also don't matter much in physics. Your answer is, IMHO, perfectly fine for most physicists most of the time. It gets the job done. I just wanted to point out that the proper mathematical techniques that were developed in functional analysis are very different from the physics approach. They are also vastly more powerful. If you want to see a beautiful hard-core application of functional analysis to a physics problem, look at Cedric Villani's work on Landau damping. $\endgroup$ Commented Apr 19, 2023 at 21:03
  • $\begingroup$ Thank you! I'll take a look at Villani. $\endgroup$
    – E. Anikin
    Commented Apr 19, 2023 at 21:06

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