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# 物理代写|量子力学代写Quantum mechanics代考|PHYS402 Operators

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

Next we discuss operators. Usually in quantum mechanics we use only linear or antilinear operators. A linear operator $L$ and an antilinear operator $A$ are both mappings from the ket space into itself,
$$L: \mathcal{E} \rightarrow \mathcal{E}, \quad A: \mathcal{E} \rightarrow \mathcal{E},$$
but they have different properties when acting on linear combinations of kets. In particular, we have
\begin{aligned} &L\left(c_1\left|\psi_1\right\rangle+c_2\left|\psi_2\right\rangle\right)=c_1 L\left|\psi_1\right\rangle+c_2 L\left|\psi_2\right\rangle, \ &A\left(c_1\left|\psi_1\right\rangle+c_2\left|\psi_2\right\rangle\right)=c_1^* A\left|\psi_1\right\rangle+c_2^* A\left|\psi_2\right\rangle . \end{aligned}

The only antilinear operator of interest in nonrelativistic quantum mechanics is the time-reversal operator, which we will discuss in Notes 22 . For now we ignore antilinear operators, and concentrate exclusively on linear operators.

Linear operators themselves can be multiplied by complex numbers and added, so they form a complex vector space in their own right. Linear operators can also be multiplied with one another; the product $A B$ means, apply $B$ first, then $A$ (to some ket). The multiplication is associative,
$$A(B C)=(A B) C,$$
for any linear operators $A, B, C$, but it is not commutative, that is
$$A B \neq B A$$
in general. The lack of commutativity of two linear operators $A$ and $B$ is measured by their commutator, defined by
$$[A, B]=A B-B A .$$
For reference, we also define the anticommutator,
$${A, B}=A B+B A$$

## 物理代写|量子力学代写Quantum mechanics代考|Rules for Commutators

The commutator [, ] obeys the following properties, which are trivial consequences of the definition (37). In the following, capital letters are linear operators and lower case letters are complex numbers. First, the commutator is linear in both operands,
\begin{aligned} &{\left[c_1 A_1+c_2 A_2, B\right]=c_1\left[A_1, B\right]+c_2\left[A_2, B\right]} \ &{\left[A, c_1 B_1+c_2 B_2\right]=c_1\left[A, B_1\right]+c_2\left[A, B_2\right]} \end{aligned}

it is antisymmetric,
$$[A, B]=-[B, A]$$
and it obeys the Jacobi identity,
$$[A,[B, C]]+[B,[C, A]]+[C,[A, B]]=0 .$$
Any bracket operation [, ] defined on any vector space (not just spaces of linear operators) that satisfies properties (41)-(43) qualifies that vector space as a Lie algebra.

In addition, the commutator satisfies the following properties, sometimes referred to as the derivation property or Leibnitz rule:
\begin{aligned} &{[A B, C]=A[B, C]+[A, C] B} \ &{[A, B C]=B[A, C]+[A, B] C} \end{aligned}
Calculations in quantum mechanics often require one to reduce complicated commutators into simpler ones that are known. Rules (44) are especially useful for this purpose. It is interesting that properties (41)-(44) are also valid for the Poisson bracket in classical mechanics [except that the ordering of the factors in Eq. (44), which must be respected in quantum mechanics, is immaterial in classical mechanics]. See Sec. B.21.

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

$$L: \mathcal{E} \rightarrow \mathcal{E}, \quad A: \mathcal{E} \rightarrow \mathcal{E},$$

$$L\left(c_1\left|\psi_1\right\rangle+c_2\left|\psi_2\right\rangle\right)=c_1 L\left|\psi_1\right\rangle+c_2 L\left|\psi_2\right\rangle, \quad A\left(c_1\left|\psi_1\right\rangle+c_2\left|\psi_2\right\rangle\right)=c_1^* A\left|\psi_1\right\rangle+c_2^* A\left|\psi_2\right\rangle .$$

$$A(B C)=(A B) C,$$

$$A B \neq B A$$

$$[A, B]=A B-B A \text {. }$$

$$A, B=A B+B A$$

## 物理代写|量子力学代写Quantum mechanics代考|Rules for Commutators

$$\left[c_1 A_1+c_2 A_2, B\right]=c_1\left[A_1, B\right]+c_2\left[A_2, B\right] \quad\left[A, c_1 B_1+c_2 B_2\right]=c_1\left[A, B_1\right]+c_2\left[A, B_2\right]$$

$$[A, B]=-[B, A]$$

$$[A,[B, C]]+[B,[C, A]]+[C,[A, B]]=0 .$$

$$[A B, C]=A[B, C]+[A, C] B \quad[A, B C]=B[A, C]+[A, B] C$$

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