Solve Physics Problem: Wave Function

[mlee]

hey who can help me with this physics problem?

A particle of mass m is in the state:
Ψ (x, t) = Aexp[-a(sqrt (mx^2) / h)-i (at / sqrt(m )) ]
where A and a are positive real constants.
a) Determine A.
b) What is the frequency ƒ associated with the wave function of this particle?
Explain your reasoning.
c) For what potential energy function U(x), does Ψ satisfy the Schrödinger
equation?
d) If we use the interpretation of [Ψ(x)]^2 dx as the probability that a particle of
mass m can be found in a region of width dx around the position x,
calculate the expected value (average value) of the position x.


Many thanx

 

[Palindrom]

A. Normalize it.
\int_{-\infty}^{+\infty} \psi^*(x) \psi(x) dx =1
B. Please don't make me do that. I have my own. It's probably just hard work.
C. Differentiate it and place it in Shroedinger's equation. You'll get the potential.
D. By definition:
<x>=\int \psi^*(x) x \psi(x)\,dx

 

[Nylex]

C. Derive it and place it in Shroedinger's equation. You'll get the potential.

By "derive", you mean "differentiate" or "take the derivative", right? Those things don't mean the same as "derive".

 

[Palindrom]

By "derive", you mean "differentiate" or "take the derivative", right? Those things don't mean the same as "derive".

That's what I meant. You'll have to forgive me, I'm not used to saying it in English.

 

[dextercioby]

A. Normalize it.
\int_{a}^{b} \psi dx =1
For a and b being - and + infinity.
B. Please don't make me do that. I have my own. It's probably just hard work.
C. Differentiate it and place it in Shroedinger's equation. You'll get the potential.
D. By definition:
<x>=\int \psi^*(x) x \psi(x)\,dx

1.Your edited your post and now it's "differentiate".
2.The first formula is incorrect;it has something missing:
\int_{-\infty}^{+\infty} \psi^*(x) \psi(x) dx =1
The rest is correct and i agree with you.

 

[Palindrom]

1.Your edited your post and now it's "differentiate".
2.The first formula is incorrect;it has something missing:
\int_{-\infty}^{+\infty} \psi^*(x) \psi(x) dx =1
The rest is correct and i agree with you.

Thanks for the correction, I edited.