- #1
MatthijsRog
- 14
- 1
Hi,
I'm an undergrad, following my very first serious course in QM. We're following Griffith's book, and so far we're staying close to the text in terms of course structure.
Griffiths starts out his book by postulating that each and every state for any system [itex]\Psi[/itex] must be a solution to the Schrödinger equation. He then introduces two things. Firstly the statistical interpretation of this function. Secondly he introduces the notion of operators [itex]\hat{Q}[/itex] corresponding to observables [itex]Q[/itex] so that the integral over [itex]\Psi^* \hat{Q} \Psi[/itex] gives the expectancy value for the observable [itex]Q[/itex]. So far I understand everything.
After working through some example solutions of the Schrödinger equation, Griffiths moves into the formalism of QM: he recasts it into linear algebra. He shows us that the operators [itex]\hat{Q}[/itex] can be thought of as linear, hermitian operators that work upon functions in Hilbert space (and thus on the wave/state functions). He then introduces the notion of a determinate state: a state that is an eigenfunction of an operator. He proves that if a system exists in such a state, then the spread in the expectancy value is nil: any measurement will result in the same outcome. Additionally, since the operator is Hermitian the eigenvalues form a basis for the Hilbert space.
Now my question: is it true that not all of these eigenstates are admissible states to the problem at hand? All states must solve the Schrödinger equation. So while my wave function must be a linear combination of eigenstates, is it true that it is only a linear combination of eigenstates that solve the Schrödinger equation?
Does that mean that the "plan of attack" for quantum mechanical problems becomes the following? Instead of integrating the awful [itex]\Psi^* \hat{Q} \Psi[/itex] we find the eigenstates of [itex]\hat{Q}[/itex], express our wavefunction in this basis and use the coefficients of this projection to determine the expectancy value? Does that mean we still must find [itex]\Psi[/itex] by solving Schrödinger's equation?
I know it's a lot of question marks up there, but students are known to get lost at this junction and I want to make sure I really understand what I'm doing. Thanks in advance for helping out!
I'm an undergrad, following my very first serious course in QM. We're following Griffith's book, and so far we're staying close to the text in terms of course structure.
Griffiths starts out his book by postulating that each and every state for any system [itex]\Psi[/itex] must be a solution to the Schrödinger equation. He then introduces two things. Firstly the statistical interpretation of this function. Secondly he introduces the notion of operators [itex]\hat{Q}[/itex] corresponding to observables [itex]Q[/itex] so that the integral over [itex]\Psi^* \hat{Q} \Psi[/itex] gives the expectancy value for the observable [itex]Q[/itex]. So far I understand everything.
After working through some example solutions of the Schrödinger equation, Griffiths moves into the formalism of QM: he recasts it into linear algebra. He shows us that the operators [itex]\hat{Q}[/itex] can be thought of as linear, hermitian operators that work upon functions in Hilbert space (and thus on the wave/state functions). He then introduces the notion of a determinate state: a state that is an eigenfunction of an operator. He proves that if a system exists in such a state, then the spread in the expectancy value is nil: any measurement will result in the same outcome. Additionally, since the operator is Hermitian the eigenvalues form a basis for the Hilbert space.
Now my question: is it true that not all of these eigenstates are admissible states to the problem at hand? All states must solve the Schrödinger equation. So while my wave function must be a linear combination of eigenstates, is it true that it is only a linear combination of eigenstates that solve the Schrödinger equation?
Does that mean that the "plan of attack" for quantum mechanical problems becomes the following? Instead of integrating the awful [itex]\Psi^* \hat{Q} \Psi[/itex] we find the eigenstates of [itex]\hat{Q}[/itex], express our wavefunction in this basis and use the coefficients of this projection to determine the expectancy value? Does that mean we still must find [itex]\Psi[/itex] by solving Schrödinger's equation?
I know it's a lot of question marks up there, but students are known to get lost at this junction and I want to make sure I really understand what I'm doing. Thanks in advance for helping out!