- #1
johng23
- 294
- 1
I am trying to follow a derivation in a book which is written without bra-ket notation, and presumably without the concept of state vectors. I can easily follow it if I may use the fact that [itex]\sum_{n}|\varphi_{n}\rangle\langle\varphi_{n}|[/itex] is the identity operator.
Analogously to the way I would prove that the above expression is the identity operator: I write [itex]\psi=\sum_{n}c_{n}\varphi_{n}[/itex] as the expansion of a wavefunction on the complete basis set [itex]\varphi_{n}[/itex]. If I use the fact that [itex]c_{m}=\int\varphi^{*}_{m}\psi d^{3}\textbf{r}[/itex], I can write [itex]\psi=\sum_{n}\varphi_{n}\int\varphi^{*}_{n}\psi d^{3}\textbf{r}[/itex]. The statement which I would like to prove is identical to this, If I replace [itex]\psi[/itex] on both sides by another expression [itex]\hat{w}\varphi_{n}[/itex]. Thus I would like to extract [itex]\psi[/itex] from the RHS and equate the rest of the expression to identity. How can I do this? The operator [itex]\hat{w}[/itex] has no special relation to the basis functions [itex]\varphi_{n}[/itex].
Analogously to the way I would prove that the above expression is the identity operator: I write [itex]\psi=\sum_{n}c_{n}\varphi_{n}[/itex] as the expansion of a wavefunction on the complete basis set [itex]\varphi_{n}[/itex]. If I use the fact that [itex]c_{m}=\int\varphi^{*}_{m}\psi d^{3}\textbf{r}[/itex], I can write [itex]\psi=\sum_{n}\varphi_{n}\int\varphi^{*}_{n}\psi d^{3}\textbf{r}[/itex]. The statement which I would like to prove is identical to this, If I replace [itex]\psi[/itex] on both sides by another expression [itex]\hat{w}\varphi_{n}[/itex]. Thus I would like to extract [itex]\psi[/itex] from the RHS and equate the rest of the expression to identity. How can I do this? The operator [itex]\hat{w}[/itex] has no special relation to the basis functions [itex]\varphi_{n}[/itex].