Using the normalise wavefunction, calculate momentum squared

In summary, the wavefunction for a particle in a stationary state of a one dimensional infinite potential well is given by \Psi_{1}(x,t)=A\cos(\frac{\pi x}{2a})e^{-i\frac{Et}{\hbar}} for -a\leq x\leq a, and the expectation value of momentum squared is calculated by squaring the momentum operator and integrating it between the two wavefunctions. The normalization condition requires solving for A to make the wavefunction normalized. An alternative method is to operate one momentum operator on the wavefunction and the other on its conjugate.
  • #1
AStaunton
105
1
[problem is:

the wavefunction for a particle in a stationary state of a one dimensional infinite potential well is given by:

[tex]\Psi_{1}(x,t)=A\cos(\frac{\pi x}{2a})e^{-i\frac{Et}{\hbar}}[/tex] for [tex]-a\leq x\leq a[/tex]

= 0 otherwise.

using the normalised wavefunction calculate the expectation value of momentum squared:

[tex]\langle p^{2}\rangle[/tex]


My attemted solution:

i know that to find momentum p, the operator is:

[tex]\bar{p}=-\hbar\bar{\nabla}^{2}[/tex]

so we say:

[tex]\langle p\rangle=\int\Psi^{\star}\bar{p}\Psi dx[/tex]

is it simply a matter of squaring p at this stage to get p^2?

Also, it says to use the normalised wave function, have I done this already in my integration step or is there something else needs be done?

Andrew
 
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  • #2
Just square the momentum operator first, then integrate between the two Psi's (I'm completely new to this forum, so I can't write tex code yet. bear with me)
 
  • #3
in terms of squaring the momentum operator, who is this done?

ie...the first part simple goes to (i^2hbar^2) but what does it mean to square the laplacian?
 
  • #4
In position representation (one dimension), the momentum operator

[tex]\hat{P}\rightarrow-i\hbar\dfrac{d}{dx}[/tex].

You want to square this. Squaring a derivative operator just turns it into a second derivative. The process is then taking

[tex]\langle\hat{P}^2\rangle=\langle\Psi_1|\hat{P}^2|\Psi_1\rangle[/tex].

As for the normalization condition, what they want you to do is first normalize the wave function, i.e. solve for [tex]A[/tex] such that

[tex]\langle\Psi_1|\Psi_1\rangle=1[/tex].

Sometimes an even easier way of taking <P^2> is to let one P operate on [tex]\langle\psi|[/tex] and the other on [tex]|\psi\rangle[/tex] so that you only have to work with first derivatives. In this case it won't make much of a difference, but it is a nice trick to have up your sleeve.
 
Last edited:
  • #5
,

To calculate the momentum squared, we first need to find the expectation value of momentum. This can be done using the normalised wavefunction by taking the integral of the wavefunction multiplied by the momentum operator, as you have correctly stated.

However, to find the expectation value of momentum squared, we need to square the momentum operator before taking the integral. This is because the expectation value of a squared quantity is not equal to the square of the expectation value.

So, the correct expression for calculating the expectation value of momentum squared would be:

\langle p^{2}\rangle = \int \Psi^{\star} \bar{p}^{2} \Psi dx

where \bar{p}^{2} = (-\hbar\bar{\nabla}^{2})^{2} = \hbar^{2}\bar{\nabla}^{4}.

As for using the normalised wavefunction, you have already used it in your integration step. The normalisation condition ensures that the wavefunction is properly scaled and thus, any calculations using it will give physically meaningful results.

I hope this helps clarify your approach to solving this problem. Keep up the good work!
 

1. What is the purpose of normalizing a wavefunction?

The purpose of normalizing a wavefunction is to ensure that the total probability of finding a particle in any region of space is equal to 1. This allows for meaningful calculations of physical quantities, such as momentum squared, to be performed.

2. How do you normalize a wavefunction?

To normalize a wavefunction, you must first calculate the integral of the absolute square of the wavefunction over all space. Then, you divide the wavefunction by the square root of this integral. This will result in a normalized wavefunction with a total probability of 1.

3. What is the significance of calculating momentum squared?

Momentum squared is an important physical quantity that can provide information about the motion and behavior of a particle. In quantum mechanics, it is often used to calculate the uncertainty in a particle's position, as well as its kinetic energy.

4. Can momentum squared ever be negative?

No, momentum squared cannot be negative. This is because momentum is a vector quantity, and when squared, it will always result in a positive value. However, the direction of the momentum can be negative, which is represented by a negative sign in front of the momentum squared term.

5. Are there any limitations to using the normalized wavefunction to calculate momentum squared?

There are a few limitations to using the normalized wavefunction to calculate momentum squared. Firstly, it assumes that the wavefunction is well-behaved and can be normalized. Secondly, it only applies to systems with a discrete energy spectrum. Lastly, it is only valid for non-relativistic particles.

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