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# 物理代写|量子力学代写Quantum mechanics代考|PHYS4141 Compatible observables and complete vector space

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## 物理代写|量子力学代写Quantum mechanics代考|Compatible observables and complete vector space

Given a physical system, there is a maximal number of compatible observables. These form a set of operators $\left{A_1, A_2, \cdots A_N\right}$ that commute with each other
$$\left[A_i, A_j\right]=0 .$$
In principle the number $N$ corresponds to the number of degrees of freedom of a system in its formulation in Classical Mechanics. This corresponds to the number of canonical position-momentum pairs, plus spin, charge, color, flavor, etc. degrees of freedom, if any. Thus, for a system of $k$ spinless neutral particles in $d$ dimensions there are $N=k \times d$ degrees of freedom that correspond to compatible positions or momenta needed to describe the system. The operators $\left{A_1, A_2, \cdots A_N\right}$ could be chosen as the $N$ positions or the $N$ momenta or some other $N$ compatible operators constructed from them, such as energy, angular momentum, etc.. As explained below, for each choice there is a corresponding complete set of eigenstates that define a basis for the same physical system. A physical state may be written as a linear combination of basis vectors for any one choice of basis corresponding to the eigenstates of compatible observables. Furthermore any basis vector may be expanded in terms of the vectors of some other basis since they all describe the same Hilbert space.

Let us consider the matrix elements of two compatible observables $A_1$ and $A_2$ in the basis that diagonalizes $\left(A_1\right){i j}=\alpha{1 i} \delta_{i j}$. The matrix elements of the zero commutator give
$$0==\left(\alpha_{1 i}-\alpha_{1 j}\right)\left(A_2\right)_{i j}$$

## 物理代写|量子力学代写Quantum mechanics代考|Incompatible observables

Let us now consider an observable $B$ that is not compatible with the set $\left{A_1, A_2, \cdots A_N\right}$, that is $\left[B, A_i\right] \neq 0$. Then we may derive
$$\left(\alpha_{k i_k}-\alpha_{k j_k}\right)(B)_{i j}==0 \text { (?) }$$
The right hand side may or may not be zero; this depends on the states $i, j$ and operators $A, B$. Consider the subset of states for which the result is non-zero. In general when $i \neq j$ the eigenvalues of the $A_k$ ‘s are different, and the nonzero result on the right implies that $B$ cannot be diagonal. Therefore, there cannot exist a basis in which incompatible observables would be simultaneously diagonal. If there were such a basis, they would commute by virtue of being diagonal matrices, and this contradicts the assumption.

Let us consider two incompatible observables and their respective sets of eigenstates
$$A\left|\alpha_i\right\rangle=\alpha_i\left|\alpha_i\right\rangle, \quad B\left|\beta_i>=\beta_i\right| \beta_i>.$$
Since each set of eigenstates is complete and orthonormal, and they span the same vector space, they must satisfy
$$\begin{array}{ccc} \sum_k\left|\alpha_k><\alpha_k\right| & =\mathbf{1}= & \sum_k\left|\beta_k><\beta_k\right| \ <\alpha_i \mid \alpha_j> & =\delta_{i j}= & <\beta_i \mid \beta_j> \end{array} .$$

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## 物理代写|量子力学代写Quantum mechanics代考|Compatible observables and complete vector space

$$\left[A_i, A_j\right]=0 .$$

$$0==\left(\alpha_{1 i}-\alpha_{1 j}\right)\left(A_2\right){i j}$$

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