What is the <x> for given wavefunction A*exp(-(\sqrt{}Cm/2h)x^{}2)?

In summary, the conversation discusses how to calculate <x> in the equation \Psi(x,t)=A*exp(-(\sqrt{}Cm/2h)x^{}2. The suggested method is to integrate the function over all space, but the individual is struggling with the integral. They are given a hint to use the anti-derivative of exp(ax^2)*x, and to make a substitution to simplify the integration. Another alternative method is suggested that involves a substitution and does not require integration.
  • #1
fredrick08
376
0

Homework Statement


calculate <x>, when [tex]\Psi[/tex](x,t)=A*exp(-([tex]\sqrt{}Cm[/tex]/2h)x[tex]^{}2[/tex]


Homework Equations


<x>=[tex]\int[/tex][tex]\Psi^{}*[/tex]x[tex]\Psi[/tex]dx over all space..
[tex]\int[/tex]exp(-[tex]\alpha[/tex]x[tex]^{}2[/tex])=[tex]\sqrt{}\pi/\alpha[/tex]


The Attempt at a Solution


ok know how to do this but how do i do the intergral... my maths isn't so good, and the book does it very very vague... i know <x>=0 but don't know how to prove it.
 
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  • #2
so far I've got down to

A2[tex]\int[/tex]exp(-([tex]\sqrt{}Cm[/tex]/2h)x2)x dx, fomr -a/2 to a/2
 
  • #3
Hi fredrick.
Just want to make sure, you aren't going to integrate the whole space, right?
Well, there are several ways of solving this problem.
So the easiest one is, well, do the math.
So it seems like you are stuck and don't know how to do the integration.

Just a very small hint, what is the anti derivative of exp(ax^2)*x?
not sure? Well, obviously, it is going to relate to exp(ax^2), right? There is no other way to get this term, right?
So it is probably something with exp(ax^2).
Know, try to differentiate your "guess" function. And remember the chain rule. And see if it is the same as exp(a*x^2)x
Once you get your guessed function right, I think you can find the anti derivative of your function.
And the rest is pretty much plug in

I'll tell the other way, which doesn't even involve doing all the dirty job after you finish your calculation :D
Good luck!
 
  • #4
Actually, if you have an x as well as an exp(x^2) as the integrand, there is no need to even worry about integrating exp(x^2).

Make a substitution u=ax^2, du=2ax and simply do it that way.
 

1. What is a QM expectation value of x?

A QM expectation value of x is a mathematical concept in quantum mechanics that represents the average value of the position of a particle in a given quantum state. It is calculated by taking the integral of the position operator multiplied by the wave function of the particle squared.

2. How is the QM expectation value of x different from the classical expectation value?

The QM expectation value of x takes into account the wave-like nature of particles in quantum mechanics, while the classical expectation value only considers the average position of a particle in a specific state. This means that the QM expectation value of x can give different results than the classical expectation value for the same system.

3. Why is the QM expectation value of x important?

The QM expectation value of x is important because it provides a way to predict the average position of a particle in a quantum state. It is also used in calculations for other properties, such as momentum and energy, and plays a crucial role in understanding the behavior of quantum systems.

4. How is the QM expectation value of x related to uncertainty in position?

The QM expectation value of x is related to uncertainty in position through the Heisenberg uncertainty principle. This principle states that the product of the uncertainties in position and momentum must always be greater than or equal to a certain value. This means that as the QM expectation value of x becomes more precise, the uncertainty in position increases.

5. Can the QM expectation value of x be negative?

Yes, the QM expectation value of x can be negative. This is because the wave function, which is squared in the calculation, can have negative values. However, the QM expectation value of x is a mathematical representation and does not necessarily correspond to the actual physical position of a particle, which is always positive.

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