- #1

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Sorry, i still dont know latex

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- Thread starter Ahmad Kishki
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- #1

- 159

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Sorry, i still dont know latex

- #2

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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

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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

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- #5

DrClaude

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$$

\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

$$

\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

- #7

DrClaude

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