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I would like to think of "state of an object" as, classically, the simultaneous knowledge of its position, momentum, rotational angular momentum, orbital angular momentum, kinetic energy, etc. In quantum mechanics, "state of an object" would be simultaneous knowledge of the probability distributions of all the observables.

But we are typically taught that "state" is a vector, sometimes is represented by a wave function if we are talking about its position and momentum in space sometimes is just an abstract vector if we are talking about spin.

This definition of state does not seem to agree with what I want a state to be. But is the following true?

Suppose a state induces a probability distribution for each observables, be it position, momentum, kinetic energy, spin, etc. There is no other state that induces the same probability distributions of observables.

(if two states differ by a scalar multiple, they are considered the same)

In particular, if two wave functions induces the same probability distributions of position and momentum, are they the same wave function?

If two wave functions $f,g$ have such properties, then their probability amplitude are equal that $\lvert f(x)\rvert^2=\lvert g(x)\rvert^2$, meaning $f(x)=e^{i\theta(x)}g(x)$ for some real-valued function $\theta$. I believe $\theta$ may have to be a constant function if $f,g$ induces the same probability distributions of momentum.

By the same arguments, their Fourier transform are related by $\hat f(k)=e^{i\phi(k)}\hat g(k)$. This should lead to $$\int_{-\infty}^{\infty}e^{i\theta(x)}g(x)e^{-ikx}dx=\int_{-\infty}^{\infty}e^{i\phi(k)}g(x)e^{-ikx}dx.$$Not sure how that is helpful.

Qmechanic
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edm
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1 Answers1

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If the probability distributions of all the observables are known, then yes the state, $|\psi\rangle$, is known, simply because the distribution of the observable $|\psi\rangle\langle\psi|$ will be nonzero for a single case - either the state is $|\psi\rangle$ (eigenvalue $\langle\psi|\psi\rangle$ observed), or it lies in the orthogonal subspace (eigenvalue 0). This is essentially the same as measuring a complete set of commuting observables.

The problem is that (except for the case of the spin-half system) most "observables" - hermitian operators on the Hilbert space - do not have a clear physical interpretation and measurement setting. A wave packet may seem pretty classical but, as you point out, the spatial probability distribution is not all: a normalized wave function may be written as $$\psi(x)=e^{i\theta(x)}\sqrt{p(x)},\:\:\:\theta(x)\in\mathbb{R},\:p(x)\in\mathbb{R}^+,$$ where $p(x)$ is the probability density; so the position statistic tells nothing about the shape of $\theta(x)$. Given also the momentum probability density $\hat p(k)$, we have the momentum wave function as $$ \hat\psi(k)=e^{i\hat\theta(k)}\sqrt{\hat p(k)}= \frac{1}{\sqrt{2\pi}}\int\!dx\,e^{i[\theta(x)-kx]}\sqrt{p(x)}\,, $$ which is just a functional relationship between the unknown functions $\theta(x),\hat\theta(k)$.

Daniel
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