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  1. What is the proof for the fact that the product of two operators is generally not commutative? $$\hat A\hat {\vphantom{A}B}\not=\hat{\vphantom{A}B} \hat A.$$

  2. What is the difference between $\hat A\hat {\vphantom{A}B}$ and $\hat {\vphantom{A}B} \hat A$?

  3. What are commutator and anti commutator of two operator $\hat A$ and $\hat {\vphantom{A}B}$?

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1. proof is in the pudding: there are operators which are not commutative. pick any 2 square matrices and in general they won't commute – Valentin Jun 7 '12 at 19:59

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What are your operators? There is no reason that they should commute in general, because its not in the definition. A linear operator $\hat{A}$ is a mapping from a vector space into itself, ie. $\hat{A}:V\to V$ (actually an operator isn't always defined by this fact, I have seen it defined this way, and I have seen it used just as a synonym for map). A matrix is a representation of an operator in a particular basis of the vector space. For example, consider the differential operator $\frac{d}{dx}$ and the multiplication by $x$ operator, $\hat{x}$, where the vector space $V=$all polynomials over the real numbers. For example, $x^2\in V$, and $\frac{d}{dx}x^2=2x$, $\hat{x}x^2=x^3$. That is how these operators act on that particular polynomial. The commutator of $\hat{A}$ and $\hat{B}$ is $\hat{A}\hat{B}-\hat{B}\hat{A}$. Using now the polynomial $1$ for example, $(\frac{d}{dx}\hat{x}-\hat{x}\frac{d}{dx})1=\frac{d}{dx}(\hat{x}1)-\hat{x}(\frac{d}{dx}1)=\frac{d}{dx}x=1,$ and so $\hat{x}\frac{d}{dx}\ne\frac{d}{dx}\hat{x}$ since the commutator isn't $0$. In fact for any polynomial $p(x)$, $(\frac{d}{dx}\hat{x}-\hat{x}\frac{d}{dx})p(x)=1$, and so the commutator is just the identity. The anticommutator of $\hat{A}$ and $\hat{B}$ is $\hat{A}\hat{B}+\hat{B}\hat{A}$.

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What are your operators? they are comlpex numbers – user29646 Jun 7 '12 at 20:22
If your operators are complex numbers, then they commute – M Turgeon Jun 7 '12 at 21:21
@M Turgeon so if $\hat A \hat B=\hat B\hat A$ Which means $[\hat A,\hat B]=0$ isn't it true!? – user29646 Jun 7 '12 at 22:07

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