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## Homework Statement

This problem is about the momentum squared operator. First, I state how I saw the derivation for the momentum operator. Then I state how I attempt to (and fail to) derive the momentum squared operator using the same methods.

## Homework Equations

<p> = ∫ ψ*(ħ/i ∂/∂x)ψ dx

The construction for this definition is straightforward. We start with the fact that

<p> = m d/dt <x>

= m ∂/∂t ∫ ψ* x ψ dx

= m ∫ x ∂/∂t ψ*ψ dx

= m ∫ x ∂/∂x [iħ/2m ψ*(∂ψ/∂x)-(∂ψ*/∂x)ψ] dx (from the probability flux)

Here, we use the substitution x(∂/∂x)f(x) = ∂/∂x[xf(x)]-f(x) and say that since square integrable functions (that is, normalizable functions) must have derivatives that vanish, the first part is zero. We're left with "-f(x)", which is:

= m(iħ/2m) ∫ -[ψ*(∂ψ/∂x)-(∂ψ*/∂x)ψ] dx

= ħi/2 ∫ [(∂ψ*/∂x)ψ-ψ*(∂ψ/∂x)] dx

= ħi/2 ∫ (∂/∂x)ψ*ψ -2ψ*(∂ψ/∂x) dx

Again, the derivative vanishes.

= ħ/i ∫ ψ*(∂ψ/∂x) dx

Hence, we define the momentum operator as:

p = ħ/i (∂/∂x)

## The Attempt at a Solution

To obtain the momentum squared, I go through the following steps (the first few steps are about the same):

<p> = m

^{2}d/dt <x

^{2}>

= m

^{2}∂/∂t ∫ ψ* x

^{2}ψ dx

= m

^{2}∫ x

^{2}∂/∂t ψ*ψ dx

= m

^{2}∫ x

^{2}∂/∂x [iħ/2m ψ*(∂ψ/∂x)-(∂ψ*/∂x)ψ] dx (from the probability flux)

From here, I use:

x

^{2}(∂/∂x)f(x) = ∂/∂x[x

^{2}f(x)]-2xf(x)

**(1)**

The first part vanishes. We're left with 2xf(x)

= m

^{2}(iħ/2m) ∫ 2x [ψ*(∂ψ/∂x)-(∂ψ*/∂x)ψ] dx

= mħi ∫ x [ψ*(∂ψ/∂x)-(∂ψ*/∂x)ψ] dx

= mħi ∫ x [(∂/∂x)[ψ*ψ]-2(∂ψ*/∂x)ψ] dx

using another x(∂/∂x)f(x) = ∂/∂x[xf(x)]-f(x) type substitution:

= mħi ∫ -[ψ*ψ]-2x(∂ψ*/∂x)ψ] dx

= -mħi (1+∫ 2x(∂ψ*/∂x)ψ] dx)

It's obviously wrong since it's imaginary. Going back to step

**(1)**, which I bolded, and making the substitution

x(∂/∂x)g(x) = ∂/∂x[xg(x)]-g(x), where g(x) = (∂/∂x)g(x)=f(x), I'm left with 2g(x)=2∫f(x)dx

= 2mħi ∫∫ [(∂/∂x)[ψ*ψ]-2(∂ψ*/∂x)ψ] dx

^{2}

I don't understand why this happens. I get an imaginary answer every time. :(

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