Proving Real Value of Hard Quantum Question Homework

[neelakash]

Homework Statement

Prove that

Neither

<xp>=∫ψ* x(ħ/i)(∂/∂x) ψ dx

nor <xp>=∫ψ* x(ħ/i)(∂/∂x)xψ dx

is acceptable because both lead to imaginary value.Show that

<xp>=∫ψ* x(ħ/i)(∂/∂x) ψ dx + ∫ψ* x(ħ/i)(∂/∂x)xψ dx leads to real value.Does

<xp>=<x><p> ?

Homework Equations

The Attempt at a Solution

Taking * of proposed <xp>=∫ψ* x(ħ/i)(∂/∂x) ψ dx and carrying out by parts integral,I get

<xp>*= -iħ +<xp> ≠ <xp> .Hence, given expression for <xp> is not real where it should be so.

[What I am woried about,even <xp> comes out to be imaginary, <xp>-<px>=iħ is still OK.Because <xp>*=<px>]

Also,my attempts to show the second proposed expression leads to imaginary value failed:

<xp>=∫ψ* x(ħ/i)(∂/∂x)xψ dx

=> <xp>*=∫ψ x(-ħ/i) (∂/∂x)xψ* dx

=><xp>*=(iħ) ∫ψx (∂/∂x)xψ* dx =(iħ) [-∫xψ*(ψ+ x(∂ψ/∂x)) dx + 0]

where I assumed [x²ψ*ψ] gives zero in both +∞ and -∞ limit.

[I am not sure at this point too.As the term involves a factor x²]

Can anyone suggest if I am going through the correct way?If I am doing wrong in integration, please show it.

 

[neelakash]

Given proposed expressions for $$\langle\ x \ p_x \rangle=\int^\infty_{\ - \infty}\psi\ast\ x \frac{\hbar}{i}$$$$\frac{\partial\psi}{\partial\ x }$$$$\ dx$$

Another given expression for $$\langle\ x \ p_x \rangle=\int^\infty_{\ - \infty}\psi\ast\ x \frac{\hbar}{i}$$$$\frac{\partial(\ x \psi)}{\partial\ x }$$$$\ dx$$

We are to show that neither is correct but the sum of the integrations is the correct expression for <xp_x>

I started with taking * of $$\langle\ x \ p_x \rangle$$ of the first expression:

$$\langle\ x \ p_x \rangle\ast=\frac{-\hbar}{i}\int_{\ - \infty}^\infty\psi\ x \frac{\partial\psi\ast}{\partial\ x }$$$$\ dx=$$$$\ i \hbar\ [\ - \int_{\ - \infty}^{\infty}\frac{\partial(\psi\ x )}{\partial\ x }\psi\ast\ dx\ + \(\psi\ast\ x \psi)_{\ - \infty}^{\infty}]$$

Last term=0

$$\langle\ x \ p_x \rangle\ast=\ i \hbar\[\ - \int_{\ - \infty}^\infty\psi\ast\ (\psi + \ x \frac{\partial\psi}{\partial\ x})\ dx \]=\ - \ i \hbar\ + \langle\ x \ p_x \rangle$$

But this is not $$\langle\ x \ p_x \rangle$$.So, we conclude given expression of $$\langle\ x \ p_x \rangle$$ is not correct because the expectation value must be a real quantity Q such that Q*=Q

Note an interesting thing! Even if the expression is not correct,it correctly leads to <xp>-<px>=iħ as <xp>*=<px>

But I am in trouble with the second integration.It appears that the conjugate of the second integral should be equal to iħ so that the sum of the two integrals be <xp>

Can anyone please suggest anything?