Thinking about the Schrodinger wave equation

  • #1
159
13

Main Question or Discussion Point

so H(psi) = E(psi) is the schrodinger equation such that psi is the eigen function of the hamiltonian operator, and since E is the eigen value of the hamiltonian, then this E is the measured E. This is what i understand so far, and i am building on this here. All psi must be eigen function of energy (by default). But the kinetic energy eigen function has to be a sinusoid (since it is only a sinusoid that will give the eigen value equation) and for momentum it is the complex exponential. Ok. But what if psi is not the eigen function for either the kinetic energy or momentum, does that mean that momentum is not measurable?

Sorry, i still dont know latex
 

Answers and Replies

  • #2
309
38
What you've written is the time-independent Schrodinger equation, but there is a more general time-dependent Schrodinger equation
http://en.wikipedia.org/wiki/Schrödinger_equation#Equation

You shouldn't assume anything about the solution being a sinusoid or whatever, the 3-D wave function for the hydrogen atom is not a sinusoid it involves associated Legendre polynomials for example, and it doesn't change from sinusoid to complex exponential just because you chose momentum instead of kinetic energy, you can express sinusoids in terms of complex exponentials and vice versa mathematically regardless, it's like a choice in the way you solve it really.
 
  • #3
159
13
What you've written is the time-independent Schrodinger equation, but there is a more general time-dependent Schrodinger equation
http://en.wikipedia.org/wiki/Schrödinger_equation#Equation

You shouldn't assume anything about the solution being a sinusoid or whatever, the 3-D wave function for the hydrogen atom is not a sinusoid it involves associated Legendre polynomials for example, and it doesn't change from sinusoid to complex exponential just because you chose momentum instead of kinetic energy, you can express sinusoids in terms of complex exponentials and vice versa mathematically regardless, it's like a choice in the way you solve it really.
I was talking of the time independent schrodineger equation, but in the realm of two dimensions, am i correct? And what is this thing about momentum can not be measured for example?
 
  • #4
309
38
Finding eigenfunctions of the kinetic energy operator, or of the momentum operator, are their own separate questions, and distinct questions from finding the eigenfunctions of the Hamiltonian operator, which amounts to finding solutions of the time-independent Schrodinger equation, which governs the dynamics of the wave function characterizing a pure state. If you can't solve the Schrodinger equation then you can't talk about the motion of the wave function of a system, ignoring a discussion of density matrices, so you have nothing to plug in to the kinetic energy or momentum operators. If you can solve it then you plug the solution into the K.E. or momentum operators in order to find the eigenvalues of those operators, i.e. the measurable values of those operators which is the point of solving those eigenfunction equations - to find the measurable eigenvalue, you already have the eigenfunction once you solved the Schrodinger equation. Hows that?
 
  • #5
DrClaude
Mentor
7,148
3,283
The solutions to the time-independent Schrödinger equation are eigenfunctions of energy,
$$
\hat{H} \psi_n = E_n \psi_n
$$
with energy ##E_n##. The set of all ##\psi_n## forms a complete basis, which means that the wave function (or state) of the system governed by the Hamiltonian ##\hat{H}## can be written as
$$
\Psi = \sum_n c_n \psi_n
$$
If ##[\hat{H}, \hat{P}^2] \neq 0##, that is, if the Hamitonian does not commute with the kinetic energy operator, then the ##\psi_n## are not momentum eigenfunctions. But the set of all kinetic energy eigefunctions ##\phi_k##, ##\hat{P}^2 \phi_k = \hbar^2 k^2 \phi_k##, also forms a complete basis, so that you can also write
$$
\Psi = \sum_k c_k \phi_k
$$
When making a measurement of momentum, after the measurement the system will be found in a single momentum eigenstate, ##\Psi = \phi_k##, and the probability of finding a particular state ##k## is ##| c_k |^2##.

Without considering a measurement, you can also calculate the expectation value of the system with wave function ##\Psi## and any operator ##\hat{O}##,
$$
\langle \hat{O} \rangle = \int \Psi^* \hat{O} \Psi d\tau
$$
where the integration runs over all space. By substituing ##\hat{O}## with ##\hat{P}##, you can figure out the expectation value of the momentum of state ##\Psi##, even though it is not a momentum eigenstate.
 
  • #6
159
13
The solutions to the time-independent Schrödinger equation are eigenfunctions of energy,
$$
\hat{H} \psi_n = E_n \psi_n
$$
with energy ##E_n##. The set of all ##\psi_n## forms a complete basis, which means that the wave function (or state) of the system governed by the Hamiltonian ##\hat{H}## can be written as
$$
\Psi = \sum_n c_n \psi_n
$$
If ##[\hat{H}, \hat{P}^2] \neq 0##, that is, if the Hamitonian does not commute with the kinetic energy operator, then the ##\psi_n## are not momentum eigenfunctions. But the set of all kinetic energy eigefunctions ##\phi_k##, ##\hat{P}^2 \phi_k = \hbar^2 k^2 \phi_k##, also forms a complete basis, so that you can also write
$$
\Psi = \sum_k c_k \phi_k
$$
When making a measurement of momentum, after the measurement the system will be found in a single momentum eigenstate, ##\Psi = \phi_k##, and the probability of finding a particular state ##k## is ##| c_k |^2##.

Without considering a measurement, you can also calculate the expectation value of the system with wave function ##\Psi## and any operator ##\hat{O}##,
$$
\langle \hat{O} \rangle = \int \Psi^* \hat{O} \Psi d\tau
$$
where the integration runs over all space. By substituing ##\hat{O}## with ##\hat{P}##, you can figure out the expectation value of the momentum of state ##\Psi##, even though it is not a momentum eigenstate.

Thank you, thank you, thank you, thank you :D weeks of misunderstanding some concepts are over!!! Please recommend a book to start quantum mechanics, i am still a beginner... Thank you again :D
 

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