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

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To calculate the expectation value <x²> for a particle in a box of width L, one must use the probability density function f(x). For a classical particle, f(x) is uniform, resulting in <x²> = ∫(0 to L) x²(1/L)dx. For a quantum particle, the probability density is derived from the wavefunction, leading to <x²> = ∫(0 to L) x²|ψ(x)|²dx. A participant in the discussion seeks confirmation on their calculation, which involves a specific formula for the nth state. Understanding both classical and quantum approaches is essential for accurately determining <x²>.
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)

&lt;x^2&gt;=\int_0^L{x^2f(x)dx}

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.

f(x)= \phi \phi^* \rightarrow &lt;x^2&gt;=\int_0^L{x^2\phi \phi^*dx}




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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