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In quantum mechanics, I read that when operators corresponding to observable commute, then they form a complete set that can define the state of the system.

But in the case of $1$ dimension, we say that, since $\hat{x}$ (position) and $\hat{p}$ (momentum) don't commute, so to define the state of a system, we only need one of them.

Since, $\Delta X \Delta P = \dfrac{h}{4\pi}$, the "uncertainty" in position and momentum is bound by uncertainty relation, knowing uncertainty in one, we can calculate minimum uncertainty in other, but knowing the expectation value of one, will not allow us to calculate the expectation value of other. Then how can only one of them, represent the complete state of a system?

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    $\begingroup$ I don't think it's right to say that to specify the state of a system we only "need" one of them. I think it's more accurate to say that we can only use one of them, since it's not possible for both to be precisely specified simultaneously. $\endgroup$ – Philip Jun 9 at 7:23
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When we say that you can completely specify the state of a system using a complete set of commuting operators, we can mean two related (almost equivalent) things:

  • If a given state is a simultaneous eigenstate of the complete set of commuting operators, then the eigenvalues associated with the state corresponding to each of the operators (in the complete set of commuting operators) completely characterizes the system. For example, in the one dimensional case you mention, if you know that the state is an eigenstate of position operator with eigenvalue $x_0$ then you know the state completely, in particular, it would be $\delta(x-x_0)$ in position basis. You can calculate everything of interest using this. Similarly, if you know that the the state is an eigenstate of momentum operator with eigenvalue $p_0$ then you know the state completely, in particular, it would be $\exp(ip_0x)$ in position basis. You can calculate everything of interest using this as well.

  • Now, if the given state is not a shared eigenstate of a given complete set of commuting operators then you can still completely characterize the given state by listing the inner product of the given state with the eigenstates of the complete set of commuting operators. For example, in your example, since the position operator forms a complete set of commuting operators, for a generic state $\vert \psi \rangle$, you can completely characterize it by listing $\langle x \vert \psi \rangle = \psi(x)$ for all $x$. You can calculate everything of interest using $\psi(x)$ as you know.

The reason I said that the two bullet points are equivalent is that even if a given state is not a shared eigenstate of a given complete set of commuting operators, you can always find another complete set of commuting operators (which might not be physically interesting) and characterize the given state by listing the associated eigenvalues of the operators in this other set. Now, you can find the inner products of the given state with the eigenstates of the original complete set of commuting operators by a change of basis from the basis spanned by the other complete set of commuting operators to the original complete set of commuting operators.

Finally, when we say that a complete set of commuting operators can be used to fully characterize a state, it doesn't mean that knowing the expectation values of each of the operators would determine the state completely (so that you can calculate expectation values of any generic operator). In order to do something like that, you need a tomographically complete set of operators, i.e., a set of operators that form an operator basis for the vector space of operators on a Hilbert space. See a related recent question of mine: How to recognize a tomographically complete set of operators?

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