- #1
Petar Mali
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Of course every prove of this type have some mistake.
If nobody will see I will post solution!
[tex]\hat{A}[/tex], [tex]\hat{B}[/tex] - linear, hermitian operator which commutator is
[tex][\hat{A},\hat{B}]=i\hbar\hat{I}[/tex]
[tex]\hat{I}[/tex] - unit operator
Eigen problems of operators are:
[tex]\hat{A}|\psi \rangle=a|\psi \rangle[/tex]
[tex]\hat{B}|\psi \rangle=b|\psi \rangle[/tex]
[tex]|\psi \rangle[/tex] - normalized eigen vector
[tex]\langle \psi|\psi\rangle=1[/tex]
Look now
[tex][\hat{A},\hat{B}]=i\hbar\hat{I}[/tex]
Take a sandwich of right and left side with vector [tex]|\psi\rangle[/tex]
[tex]\langle \psi|[\hat{A},\hat{B}]|\psi\rangle=i\hbar\langle \psi|\hat{I}|\psi\rangle[/tex]
[tex]\frac{1}{i\hbar}\langle \psi|(\hat{A}\hat{B}-\hat{B}\hat{A})|\psi \rangle=1[/tex]
[tex]\frac{a}{i\hbar}(\langle \psi|\hat{B}|\psi\rangle-\langle \psi|\hat{B}|\psi\rangle)=1[/tex]
[tex]0=1[/tex]
Where is mistake? :D
If nobody will see I will post solution!
[tex]\hat{A}[/tex], [tex]\hat{B}[/tex] - linear, hermitian operator which commutator is
[tex][\hat{A},\hat{B}]=i\hbar\hat{I}[/tex]
[tex]\hat{I}[/tex] - unit operator
Eigen problems of operators are:
[tex]\hat{A}|\psi \rangle=a|\psi \rangle[/tex]
[tex]\hat{B}|\psi \rangle=b|\psi \rangle[/tex]
[tex]|\psi \rangle[/tex] - normalized eigen vector
[tex]\langle \psi|\psi\rangle=1[/tex]
Look now
[tex][\hat{A},\hat{B}]=i\hbar\hat{I}[/tex]
Take a sandwich of right and left side with vector [tex]|\psi\rangle[/tex]
[tex]\langle \psi|[\hat{A},\hat{B}]|\psi\rangle=i\hbar\langle \psi|\hat{I}|\psi\rangle[/tex]
[tex]\frac{1}{i\hbar}\langle \psi|(\hat{A}\hat{B}-\hat{B}\hat{A})|\psi \rangle=1[/tex]
[tex]\frac{a}{i\hbar}(\langle \psi|\hat{B}|\psi\rangle-\langle \psi|\hat{B}|\psi\rangle)=1[/tex]
[tex]0=1[/tex]
Where is mistake? :D