To find the expectation value of momentum

Also, thefactor of 1/2 is not correct.In summary, the conversation discusses a one-dimensional wavefunction for a particle at time t=0, described by equation 1 with real positive constants k and a. The equations for normalization are also mentioned, as well as the attempt to find <p subscript(x)> using integration. The conversation concludes with a question about the necessity of normalization before attempting to answer the question.
  • #1
johnny1
1
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Homework Statement



At time t=0 a particle is described by a one dimensional wavefunction
(capital)ψ(x,0)= (2a/)^(1/4) e^(-ikx)e^(-ax^2)
(three lines)=(2a/)^(1/4) e^(-ikx-ax^2)--------equation 1
k and a are real positive constants

Homework Equations



I think this is the one
<p subscript(x)> = ∫(from-inf to inf)[(capitalψ)*(x,t)(-i(hbar)(∂/∂x))capitalψ(x,t)]dx
for normalization ∫from-inf to inf e^-x^2 dx =√π
and ∫from-inf to inf e^-ibx dx=√π *e^(-b^2/4) where b is a real constant

The Attempt at a Solution



----
I verified that equation 1 is normalized
by completing the square and letting t=u/a I then integrate upon which I arrive at √(π/a) *e^(k^2/4a)

I know that *e(k^2/4a) is a phase factor which is irrelevant to the equation
so
integration of e^(-ikx-ax^2)=√(π/a)

since ∫-inf to inf IΨI^2dx=(Ψ*)(Ψ)=Ae(-ax^2)*Ae(-ax^2)=(A^2)e(-2ax^2)
and if e(-2ax^2)=√(π/2a)
then A^2=√(π/2a)
hence A=(2a/π)^(1/4)

Was it necessary to normalize
before I begin in attempting to answer the question

attempting to find<p subscript(x)>
<psubscriptx> =∫(Ψ*)(-ihbar(∂/∂x))(Ψ)=(e(-ax^2))(-ihbar((1/2)e^(-4ax^2))(e(-ax^2))
and here is where I am really worried
I don't know if for <psubscriptx> I am plugging in correctly
any ideas or advice?

----

Homework Statement


Homework Equations


The Attempt at a Solution

 
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  • #2
johnny1 said:
I know that *e(k^2/4a) is a phase factor which is irrelevant to the equation

That is a real quantity, so it's not just a phase factor.
Was it necessary to normalize
before I begin in attempting to answer the question

It looks like it.

attempting to find<p subscript(x)>
<psubscriptx> =∫(Ψ*)(-ihbar(∂/∂x))(Ψ)=(e(-ax^2))(-ihbar((1/2)e^(-4ax^2))(e(-ax^2))

You didn't compute the derivative in the last expression.
 

What does it mean to find the expectation value of momentum?

The expectation value of momentum is a mathematical concept used in quantum mechanics to calculate the average value of a physical quantity, specifically the momentum of a particle. It is calculated by taking the product of the probability of a particle having a certain momentum and the momentum itself, and then summing up all possible values. It is often denoted by p with a bar on top, or ⟨p⟩.

Why is it important to find the expectation value of momentum?

The expectation value of momentum is important because it provides a way to predict the most likely outcome for the momentum of a particle in a given system. It also helps to understand the overall behavior and properties of a quantum system, which is crucial in fields such as quantum mechanics, quantum chemistry, and particle physics.

How do you calculate the expectation value of momentum?

The expectation value of momentum can be calculated using the following formula:
⟨p⟩ = ∑ pn * ⟨pn
where pn is the probability of the particle having a certain momentum ⟨pn. This formula can also be expressed in integral form for continuous systems:
⟨p⟩ = ∫ Ψ*(p) * p * Ψ(p) dp
where Ψ(p) is the wavefunction of the particle.

What units is the expectation value of momentum measured in?

The expectation value of momentum is measured in the same units as momentum, which is typically kg * m/s in the SI system. However, in quantum mechanics, it is common to use the reduced Planck's constant to represent momentum instead of kg * m/s, so the units of the expectation value of momentum can also be seen as /m.

Can the expectation value of momentum be negative?

Yes, the expectation value of momentum can be negative. This is because it represents the average value of momentum for a particle in a given system, and the momentum of a particle can be negative depending on its direction and the reference frame. However, in most cases, the expectation value of momentum is positive or zero, as particles tend to have higher probabilities of having positive momenta.

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