Frederick Kramer

2022-07-10

$\hslash \approx 0$ and the spread of QM wave function

Is there a direct mathematical method to show that if a quantum wave funtion is initially sharply localized, then it will stay sharply localized if $\hslash \approx 0$? In that case the Ehrenfest theorem implies the transition from quantum mechanics to classical mechanics.

Of course, we are dealing with the propagation of a wave function, but let's not mess with path integrals. Thus, the structure of the general solution of Schrödinger equation should imply the result - if possible.

Is there a direct mathematical method to show that if a quantum wave funtion is initially sharply localized, then it will stay sharply localized if $\hslash \approx 0$? In that case the Ehrenfest theorem implies the transition from quantum mechanics to classical mechanics.

Of course, we are dealing with the propagation of a wave function, but let's not mess with path integrals. Thus, the structure of the general solution of Schrödinger equation should imply the result - if possible.

Mekjulleymg

Beginner2022-07-11Added 14 answers

The propagator for a one-dimensional free particle is, for example,

$K({x}^{\prime}-x,{t}^{\prime}-t)=\sqrt{\frac{m}{2\pi \mathrm{i}\hslash ({t}^{\prime}-t)}}\mathrm{exp}(-\frac{m({x}^{\prime}-x{)}^{2}}{2\pi \mathrm{i}\hslash ({t}^{\prime}-t)}).$

Meaning that $\psi ({x}^{\prime},{t}^{\prime})=\int K({x}^{\prime}-x,{t}^{\prime}-t)\phantom{\rule{thinmathspace}{0ex}}\psi (x,t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$. Here, $m$ is the particle mass.

In plain(er) English, time evolution of a free particle comes down to a convolution with a Gaussian kernel with width proportional to $\hslash $. Mathematically, you can take $\hslash \to 0$, the Gaussian function becomes a $\delta $ function and the wave function does not spread any more. Physically that means that the spread is negligible if $\hslash $ is very small compared to $\frac{m{L}^{2}}{T}$, where $L$ and $T$ are the physically relevant length and time scales, respectively.

If you want to take the classical limit properly, you need decoherence, as discussed in other answers.

$K({x}^{\prime}-x,{t}^{\prime}-t)=\sqrt{\frac{m}{2\pi \mathrm{i}\hslash ({t}^{\prime}-t)}}\mathrm{exp}(-\frac{m({x}^{\prime}-x{)}^{2}}{2\pi \mathrm{i}\hslash ({t}^{\prime}-t)}).$

Meaning that $\psi ({x}^{\prime},{t}^{\prime})=\int K({x}^{\prime}-x,{t}^{\prime}-t)\phantom{\rule{thinmathspace}{0ex}}\psi (x,t)\phantom{\rule{thinmathspace}{0ex}}\mathrm{d}x$. Here, $m$ is the particle mass.

In plain(er) English, time evolution of a free particle comes down to a convolution with a Gaussian kernel with width proportional to $\hslash $. Mathematically, you can take $\hslash \to 0$, the Gaussian function becomes a $\delta $ function and the wave function does not spread any more. Physically that means that the spread is negligible if $\hslash $ is very small compared to $\frac{m{L}^{2}}{T}$, where $L$ and $T$ are the physically relevant length and time scales, respectively.

If you want to take the classical limit properly, you need decoherence, as discussed in other answers.

ttyme411gl

Beginner2022-07-12Added 6 answers

The problem with trying to understand the spread of the wavefunction in the classical limit by taking $\hslash \approx 0$ or taking the limit $\hslash \to 0$ is that in reality $\hslash \ne 0$. Taking the real non-zero value of $\hslash $there are cases of perfectly ordinary objects for which Ehrenfest's theorem doesn't imply anything like classical behaviour. Rather the wavefunction of that object spreads out a lot over time. The classical limit is actually a result of decoherence and information being copied out of 'classical' objects into the environment.

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