# Probability of finding particle in half of a box for a given state

• trelek2
In summary, the conversation is about solving the 1-dimensional time-independent Schrödinger equation for a particle confined to a box, with periodic boundary conditions. The eigenfunctions for this system are u_n(x) = Ae^(inπx/a) and the corresponding normalization constant A is found to be 1/√(2a). The probability of finding the particle in the region 0<x<a is calculated using the wave function ψ(x) = (1/√2)(u_1(x) + iu_2(x)), resulting in a value of 1/2 - 1/π. However, there is a concern about the boundary conditions being met, leading to the suggestion of using the
trelek2
Basically this is easy stuff I'm doing but i'n not sure whether i made a mistake or my answer is actually correct physically.

I'm solving the 1-dimensional TISE for a particle confined to a box -a<x<a.

I found that the eigenfunctions for this particle in a box with periodic boundary conditions are:$$u _{n}(x)=Ae ^{ \frac{in \pi x}{a} }$$ I'm pretty sure that this is correct.
I normalised A and got $$A= \frac{1}{ \sqrt{2a} }$$, which I'm also confident about.
And now I compute probability of finding particle in region 0<x<a when it is in state:
$$\psi (x)= \frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)]$$
I get that it is 1/2-1/pi which is about 0.2... Is that correct? Or should it be rather 1/2 simply.

trelek2 said:
Basically this is easy stuff I'm doing but i'n not sure whether i made a mistake or my answer is actually correct physically.

I'm solving the 1-dimensional TISE for a particle confined to a box -a<x<a.

I found that the eigenfunctions for this particle in a box with periodic boundary conditions are:$$u _{n}(x)=Ae ^{ \frac{in \pi x}{a} }$$ I'm pretty sure that this is correct.
I normalised A and got $$A= \frac{1}{ \sqrt{2a} }$$, which I'm also confident about.
And now I compute probability of finding particle in region 0<x<a when it is in state:
$$\psi (x)= \frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)]$$
I get that it is 1/2-1/pi which is about 0.2... Is that correct? Or should it be rather 1/2 simply.

Be careful that you are meeting the boundary conditions. I assume you are talking about an infinite square well. If this is the case, the wave functions that you use have to be 0 at x=a, and x=-a. I believe this gives the solutions: $$u _{n}(x)=\sqrt{\frac{2}{a}}sin ( \frac{n \pi}{2a} (x+a))$$

Now whatever $$\psi(x)$$ you use, the probability of finding the particle in region 0<x<a is $$\int_0^a |\psi(x)|^2 dx$$

I cannot confirm the accuracy of your calculations without reviewing your work. However, I can provide some general information about the probability of finding a particle in a given state.

In quantum mechanics, the probability of finding a particle in a certain region is given by the square of the wavefunction at that point. So for the given state \psi (x)= \frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)], the probability of finding the particle in the region 0<x<a would be:

P = |\psi(x)|^2 = (\frac{1}{ \sqrt{2} }[u _{1}(x)+iu _{2}(x)])^2 = \frac{1}{2}(|u _{1}(x)|^2 + |u _{2}(x)|^2)

Since the eigenfunctions for a particle in a box with periodic boundary conditions are orthogonal, meaning they have no overlap, the probability of finding the particle in the region 0<x<a would simply be 1/2, as you suspected.

It's important to note that the probability of finding a particle in a given region is dependent on the specific state of the particle. So while the overall probability of finding the particle in the region 0<x<a is 1/2, the probability distribution may vary for different states. Therefore, it's possible that your calculation of 1/2-1/pi is correct for a specific state, but not for all states.

In summary, the probability of finding a particle in a given region is determined by the square of the wavefunction at that point, and for a particle in a box with periodic boundary conditions, the probability of finding the particle in the region 0<x<a is 1/2 for all states. It's always a good idea to double check your calculations and make sure they align with known principles and concepts in physics.

## 1. What is the probability of finding a particle in half of a box for a given state?

The probability of finding a particle in half of a box for a given state depends on the specific state of the particle and the size of the box. It can be calculated using the wave function of the particle and the boundaries of the box.

## 2. Can the probability of finding a particle in half of a box be greater than 1?

No, the probability of finding a particle in any region cannot be greater than 1. This is because the total probability of finding the particle in any region must be equal to 1.

## 3. How does the size of the box affect the probability of finding a particle in half of the box?

The larger the box, the smaller the probability of finding a particle in half of the box for a given state. This is because the particle has a larger space to move around in and the probability of it being in a specific region decreases.

## 4. Does the shape of the box affect the probability of finding a particle in half of the box?

Yes, the shape of the box can affect the probability of finding a particle in half of the box for a given state. This is because the shape of the box can affect the boundary conditions and the wave function of the particle, thus altering the probability distribution.

## 5. How does the state of the particle affect the probability of finding it in half of the box?

The state of the particle, represented by its wave function, directly affects the probability of finding it in half of the box. A particle in a state with a higher probability density in a specific region will have a higher probability of being found in that region.

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