QM operator and double slit experiment doubt basics

santo35
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well i am a starter in QM, i have 2 big doubts ! let me first tell what i understood ,
there is phsi which defines a state of a system, phsi times x is a position operator and phsi 's derivative of x multiplied by i h is its momentum operator ...
well then i operator these in phsi and what do i get in return, i mean what is the actual meaning of the result of operating phsi with position and momentum operator ?

This is in regard with the modern physics QM lecture by Susskind lecture 4, when he derives the double slit experiment, he uses phsi as exponential function for the electron after it passes through the slit , how well is he justified in assuming that ?
 
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Do you mean psi (##\psi##)? That is just a symbol, frequently used for a wave function.
A usual symbol for the position operator is X. "psi times x" does not make sense in that context.
well then i operator these in phsi and what do i get in return, i mean what is the actual meaning of the result of operating phsi with position and momentum operator ?
I don't understand that question.
how well is he justified in assuming that ?
I don't think you describe the textbook content accurately here, but the experimental results agree very well with the predictions based on the calculations done there.
 
yea i ment psi, and the position operator X... well what the position operator does to a vector funstion is that take the function and multiplies by its variable rite? like X|f(x)> -> x*f(x) ... that's how he explains position operator in previous lectures.
well my question is , i operate X on psi , and what do i get in return? well or basically what is the use of position operator ?
its eigen values gives us the probablity of finding the position of the particle at 'x' ?
PS thanks for your reply !
 
and what do i get in return?
In position space, the wave function multiplied with the position.
what is the use of position operator ?
You can use it to calculate the expected ("average") position, for example.
##\langle X \rangle = \langle \psi | X | \psi \rangle##
its eigen values gives us the probablity of finding the position of the particle at 'x' ?
There are no "physical" eigenfunctions for X. They would correspond to particles at a single, exact location.
 
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