Wave Function of a Proton in a Box with a Slide

[Einstein's Cat]

Let's say that there's a box of height, h, length, l, and width, w, and within the box there's a proton. Thus it's certain that the proton would be inside the box.

A slide is then put halfway along the length of the box. Thus the proability that the proton is in one side of the box is 0.5.

What would the wave function of these two systems be and how would they change with time? Furthermore say the slide is put xl along the length of the box where x is greater than 0 but smaller than 1 what would the wave function of that system be?

If possible could references to useful paper please be given.

 

[Irene Kaminkowa]

Let's consider the ground state.
The wave function in the given 3D rigid box
$$\Psi_{1,1,1} (x,y,z) = \sqrt{\frac{8}{V}} sin\left ( \frac{\pi x}{l} \right )sin\left ( \frac{\pi y}{w} \right )sin\left ( \frac{\pi z}{h} \right )$$
Probability to find the particle at (x,y,z)
$$P_{1,1,1}(x,y,z)=|\Psi (x,y,z)|^2=\frac{8}{V}sin^2\left ( \frac{\pi x}{l} \right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )$$
And then, the box is divided.
$$P_{1,1,1}(0\leq x\leq l,y,z)=0.5P_{1,1,1}(0\leq x\leq l/2,y,z) + 0.5P_{1,1,1}(l/2\leq x\leq l,y,z)$$
The probability in the left half-box
$$P_{1,1,1}(0\leq x\leq l/2,y,z)=\frac{16}{V}sin^2\left ( \frac{2\pi x}{l} \right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )$$
where V = lwh
What is the probability in the right half-box?

 

[Einstein's Cat]

Let's consider the ground state.
The wave function in the given 3D rigid box
$$\Psi_{1,1,1} (x,y,z) = \sqrt{\frac{8}{V}} sin\left ( \frac{\pi x}{l} \right )sin\left ( \frac{\pi y}{w} \right )sin\left ( \frac{\pi z}{h} \right )$$
Probability to find the particle at (x,y,z)
$$P_{1,1,1}(x,y,z)=|\Psi (x,y,z)|^2=\frac{8}{V}sin^2\left ( \frac{\pi x}{l} \right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )$$
And then, the box is divided.
$$P_{1,1,1}(0\leq x\leq l,y,z)=0.5P_{1,1,1}(0\leq x\leq l/2,y,z) + 0.5P_{1,1,1}(l/2\leq x\leq l,y,z)$$
The probability in the left half-box
$$P_{1,1,1}(0\leq x\leq l/2,y,z)=\frac{16}{V}sin^2\left ( \frac{2\pi x}{l} \right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )$$
where V = lwh
What is the probability in the right half-box?

It is the value you gave last subtracted from 1 I believe

 

[Irene Kaminkowa]

Not quite
$$P_{1,1,1}(l/2\leq x\leq l,y,z)=\frac{16}{V}sin^2\left ( \frac{2\pi (x-l/2)}{l} \right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )=\frac{16}{V}sin^2\left ( \frac{2\pi x}{l} -\pi\right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )=\frac{16}{V}sin^2\left ( \frac{2\pi x}{l}\right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )$$

 

[Einstein's Cat]

Not quite
$$P_{1,1,1}(l/2\leq x\leq l,y,z)=\frac{16}{V}sin^2\left ( \frac{2\pi (x-l/2)}{l} \right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )=\frac{16}{V}sin^2\left ( \frac{2\pi x}{l} -\pi\right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )=\frac{16}{V}sin^2\left ( \frac{2\pi x}{l}\right )sin^2\left ( \frac{\pi y}{w} \right )sin^2\left ( \frac{\pi z}{h} \right )$$

Thank you for your help; however what would the wave function of the box be where the slide is at xl along the box?

 

[Irene Kaminkowa]

To solve this problem, we are to find the probabilities P₁ for 0<=x<=L and P₂ for L<=x<=1, where 0<=L<=1 is the slider's position.
The result probability is the sum of P₁ and P₂ with appropriate weights.

 

[Paul Colby]

A slide is then put halfway along the length of the box. Thus the proability that the proton is in one side of the box is 0.5.

The box is represented is a potential constraining the proton. As the slide is inserted the potential changes with time. This will cause energy to be imparted to the particle since the eigenvalues become time dependent.

 

[Nugatory]

The box is represented is a potential constraining the proton. As the slide is inserted the potential changes with time. This will cause energy to be imparted to the particle since the eigenvalues become time dependent.

Yep - although in the context of a B-level thread I'd expect that OP would be best off starting with the time-independent solution: what is the wave function of a particle in a box (idealized infinite potential well) with a barrier (idealized delta function potential) in the middle. One more simplification to make it a one-dimensional problem and we'd have a standard first-year exercise.