- #1
kq6up
- 368
- 13
I am a QM beginner so go easy on me. I have just noticed something. Let $$\hat{O}$$ be an hermitian operator. Then $$\left( \hat { O } \right) ^{ \dagger }\neq \hat { O } $$ when it is by itself. For example $$\left( \hat { p } \right) ^{ \dagger }=i\hbar \frac { \partial }{ \partial x } \neq \hat { p } =-i\hbar \frac { \partial }{ \partial x }$$.
In context with an eigenfunction $$ \left( \hat { O } \Psi \right) ^{ \dagger }=\Psi ^{ \ast }\hat { O } ^{ \dagger }\neq \Psi ^{ \ast }\hat { O } $$. However, I guess it still doesn't work unless I give it a $$\Psi$$ to chew on the right. I guess this might finally work: $$ \left( \hat { O } \Psi \right) ^{ \dagger }\Psi =\Psi ^{ \ast }\hat { O } ^{ \dagger }\Psi =\Psi ^{ \ast }\hat { O } \Psi $$.
However if I test this with an actualy example: $$\hat { O } =\hat { p } \quad and\quad \Psi ={ e }^{ ikx }\quad then\quad { e }^{ -ikx }(-i\hbar \frac { \partial }{ \partial x } ){ e }^{ ikx }\neq { e }^{ -ikx }(i\hbar \frac { \partial }{ \partial x } ){ e }^{ ikx }$$. Then that does not work as well.
Could someone help clarify what is going on here? In what sense are these operators hermitian. I guess not in the same sense that a matrix is hermitian.
Thanks,
Chris Maness
In context with an eigenfunction $$ \left( \hat { O } \Psi \right) ^{ \dagger }=\Psi ^{ \ast }\hat { O } ^{ \dagger }\neq \Psi ^{ \ast }\hat { O } $$. However, I guess it still doesn't work unless I give it a $$\Psi$$ to chew on the right. I guess this might finally work: $$ \left( \hat { O } \Psi \right) ^{ \dagger }\Psi =\Psi ^{ \ast }\hat { O } ^{ \dagger }\Psi =\Psi ^{ \ast }\hat { O } \Psi $$.
However if I test this with an actualy example: $$\hat { O } =\hat { p } \quad and\quad \Psi ={ e }^{ ikx }\quad then\quad { e }^{ -ikx }(-i\hbar \frac { \partial }{ \partial x } ){ e }^{ ikx }\neq { e }^{ -ikx }(i\hbar \frac { \partial }{ \partial x } ){ e }^{ ikx }$$. Then that does not work as well.
Could someone help clarify what is going on here? In what sense are these operators hermitian. I guess not in the same sense that a matrix is hermitian.
Thanks,
Chris Maness