Is the Energy Expectation Value Always Real and Above a Minimum Potential?

[dyn]

Hi
A theorem states that if V(x , t) ≥ V₀ then <E> is real and <E> ≥V₀ for any normalizable state. The proof contains the following line

<E> = (ħ²/2m)∫∇ψ^*∇ψ d³x + ∫ Vψ^*ψ d³x ≥ ∫ V₀ψ^*ψ

Can anybody explain why that inequality is true ?
Thanks

 

[PeroK]

Hi
A theorem states that if V(x , t) ≥ V₀ then <E> is real and <E> ≥V₀ for any normalizable state. The proof contains the following line

<E> = (ħ²/2m)∫∇ψ^*∇ψ d³x + ∫ Vψ^*ψ d³x ≥ ∫ V₀ψ^*ψ

Can anybody explain why that inequality is true ?
Thanks

Why do you think?

Hint: when do we have $A= B + C \ge C$?

 

[dyn]

That would be if B ≥ 0 ; but why is the 1st integral ≥ 0 and why does the 2nd integral change from V to V₀ ?

 

[PeroK]

That would be if B ≥ 0 ; but why is the 1st integral ≥ 0 and why does the 2nd integral change from V to V₀ ?

What do you think? Show some effort!

 

[PeterDonis]

when do we have $A= B + C \ge C$?

The two $V$ integrals are not the same, so the form of the equation is not exactly $B + C \ge C$. It's more like $B + C \ge D$, where we are given that $C \ge D$.

 

[PeterDonis]

Can anybody explain why that inequality is true ?

Have you tried to evaluate the signs of each term? If each term is positive, and you are already given that $V \ge V_0$ for all $V$, what can you conclude?

 

[PeroK]

The two $V$ integrals are not the same, so the form of the equation is not exactly $B + C \ge C$. It's more like $B + C \ge D$, where we are given that $C \ge D$.

Or $B + C \ge C \ge C_0$

 

[PeterDonis]

That would be if B ≥ 0

Can you show that this is the case for what you gave in the OP?

 

[dyn]

Hi
A theorem states that if V(x , t) ≥ V₀ then <E> is real and <E> ≥V₀ for any normalizable state. The proof contains the following line

<E> = (ħ²/2m)∫∇ψ^*∇ψ d³x + ∫ Vψ^*ψ d³x ≥ ∫ V₀ψ^*ψ

Can anybody explain why that inequality is true ?
Thanks

The 1st integral is the integral of the modulus squared of ∇ψ which is always a positive quantity so that intgral will always be ≥ 0. It seems obvious that ∫Vψ^*ψ d³x ≥ ∫ V₀ψ^*ψ d³x but what is the actual reason , after all V₀ could be negative ?

 

[PeroK]

after all V₀ could be negative ?

Does that matter?

 

[Delta2]

It seems obvious that ∫Vψ*ψ d3x ≥ ∫ V0ψ*ψ d3x but what is the actual reason , after all V0 could be negative ?

Yes that is very obvious indeed for someone that has study integrals and inequalities and the reason is contained in any decent calculus textbook, but it might not be obvious to you. (don't get me wrong I know you probably have studied calculus but you missed that chapter on integrals and inequalities, these days the basics are not being taught well enough)

I ll give you a mini theorem (lemma) that is contained in any decent calculus textbook as I said and I think you should be able to infer the rest.
Theorem (on integrals and inequalities)
""If $f,g$ are two integratable functions in $[a,b]$and $f(x)\geq g(x)$ for all $x\in[a,b]$ then the following inequality holds $$\int_a^b f(x)dx\geq\int_a^b g(x) dx$$.""

 

[dyn]

The 1st integral is the integral of the modulus squared of ∇ψ which is always a positive quantity so that intgral will always be ≥ 0. It seems obvious that ∫Vψ^*ψ d³x ≥ ∫ V₀ψ^*ψ d³x but what is the actual reason , after all V₀ could be negative ?

Is the argument for the inequality as follows ?
∫ (V - V₀) (ψ^*ψ) d³x ≥ 0 because both brackets are always ≥ 0 so it is an everywhere positive integral then i just turn the integral into 2 terms and take the V₀ term over to the other side. Is that the correct reasoning ?

 

[dyn]

Does the my reasoning in #12 explain in simple terms the inequality lemma in #11 ?

 

[Delta2]

Does the my reasoning in #12 explain in simple terms the inequality lemma in #11 ?

Yes it does, that's the way the lemma in 11 is proved, that the function u=f-g is always positive, and the integral of a positive function is a positive number e.t.c.

 

[dyn]

Thank you everyone