How Do You Calculate the Expectation Value <x²> for a Particle in a Box?

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SUMMARY

The expectation value for a particle in a box can be calculated using the probability density function f(x). For a classical particle, f(x) is defined as 1/L, leading to the formula =∫₀ᴸ{x²(1/L)dx}. In the case of a quantum particle, the probability density function is derived from the wavefunction, represented as f(x)=φφ*. The integral for the quantum case becomes =∫₀ᴸ{x²φφ*dx}, which is essential for determining the expectation value in quantum mechanics.

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CollectiveRocker
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If the expectation value <x> of a particle trapped in a box L wide is L/2, which means its average position in the middle of the box. Find the expectation value <x squared>. How do I go about doing this? I am really confused.
 
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CollectiveRocker said:
If the expectation value <x> of a particle trapped in a box L wide is L/2, which means its average position in the middle of the box. Find the expectation value <x squared>. How do I go about doing this? I am really confused.

The expectation value of x^2 can be calculated if you know the probability density function of x. If it is f(x)

[tex]<x^2>=\int_0^L{x^2f(x)dx}[/tex]

For a classical particle f(x) = 1/L as it has equal probability of being anywhere in the box.

If it is a quantum particle, find out how the wavefunction of that particle, trapped in a box extending from x=0 to x=L, looks like. The probability density function is the square of the absolute value of the wavefunction.

[tex]f(x)= \phi \phi^* \rightarrow <x^2>=\int_0^L{x^2\phi \phi^*dx}[/tex]




ehild
 
I have the same problem as one part of a larger question, and was hoping to check my answer. Did you get a2(1/3 + 3/(2*pi2*n2) for the nth state?
 

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