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Last time, even after showing my sincere attempt, nobody replied.So, I am posting this problem 2nd time.Do not worry.The problem is probably done.Just I want to be sure of it.

PROBLEM 1. 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> ?

Similar to the previous problem,I am thinking on this problem for a long time.In fact,there is a problem like this but that demands to show

PROBLEM 2. Neither <xp>=∫ψ* x(ħ/i)(∂/∂x) ψ dx nor <px>=∫ψ*(ħ/i)(∂/∂x) xψ dx is correct as they lead to imaginary values.But

[∫ψ* x(ħ/i)(∂/∂x) ψ dx + ∫ψ*(ħ/i)(∂/∂x) xψ dx] leads to an acceptable real eigenvalue...

The standard procedure to do this problem is to take the conjugate of <xp> or <px> and to check if <xp>=<xp>* and <px>=<px>*

But as for PROBLEM 2,I got after calculation <xp>=<xp>*+iħ and <px>=<px>*-iħ

So, it is clear that they are not correct relations.But when added together, the iħ s cancel and [∫ψ* x(ħ/i)(∂/∂x) ψ dx + ∫ψ*(ħ/i)(∂/∂x) xψ dx] leads to an acceptable real eigenvalue

But for PROBLEM 1 which appears in university booklet,it still remained a problem.

The first part is identical for both problems.

But when I tried to do with the 1st problem,first note that the question has written

the expectation value of xp,but the operator sandwitched between ψ* and ψ is xpx.Also the integral came as real!!!

∫ψ* x (ħ/i)(∂/∂x) xψ dx =∫ψ* xpx ψ dx= (ψ, xpx ψ) = (xpx ψ,ψ) [as xpx is Hermitian]

But, (xpx ψ,ψ)=(ψ, xpx ψ)* by definition of scalar products...

It follows that (ψ, xpx ψ)=(ψ, xpx ψ)*

Thus the given integral is real...

This is where I am having doubt if the question printed on that page is correct at all.

Again,can anyone please say if the question is wrong or I am wrong anywhere? If the question itself is misprinted, then there is no meaning of wasting more time for this problem.

**1. Homework Statement**PROBLEM 1. 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> ?

Similar to the previous problem,I am thinking on this problem for a long time.In fact,there is a problem like this but that demands to show

PROBLEM 2. Neither <xp>=∫ψ* x(ħ/i)(∂/∂x) ψ dx nor <px>=∫ψ*(ħ/i)(∂/∂x) xψ dx is correct as they lead to imaginary values.But

[∫ψ* x(ħ/i)(∂/∂x) ψ dx + ∫ψ*(ħ/i)(∂/∂x) xψ dx] leads to an acceptable real eigenvalue...

**2. Homework Equations****3. The Attempt at a Solution**The standard procedure to do this problem is to take the conjugate of <xp> or <px> and to check if <xp>=<xp>* and <px>=<px>*

But as for PROBLEM 2,I got after calculation <xp>=<xp>*+iħ and <px>=<px>*-iħ

So, it is clear that they are not correct relations.But when added together, the iħ s cancel and [∫ψ* x(ħ/i)(∂/∂x) ψ dx + ∫ψ*(ħ/i)(∂/∂x) xψ dx] leads to an acceptable real eigenvalue

But for PROBLEM 1 which appears in university booklet,it still remained a problem.

The first part is identical for both problems.

But when I tried to do with the 1st problem,first note that the question has written

the expectation value of xp,but the operator sandwitched between ψ* and ψ is xpx.Also the integral came as real!!!

∫ψ* x (ħ/i)(∂/∂x) xψ dx =∫ψ* xpx ψ dx= (ψ, xpx ψ) = (xpx ψ,ψ) [as xpx is Hermitian]

But, (xpx ψ,ψ)=(ψ, xpx ψ)* by definition of scalar products...

It follows that (ψ, xpx ψ)=(ψ, xpx ψ)*

Thus the given integral is real...

This is where I am having doubt if the question printed on that page is correct at all.

Again,can anyone please say if the question is wrong or I am wrong anywhere? If the question itself is misprinted, then there is no meaning of wasting more time for this problem.

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