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

• I
• dyn
In summary, the 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
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

dyn said:
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##?

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

dyn said:
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!

PeroK said:
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##.

Delta2
dyn said:
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?

PeterDonis said:
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##

dyn said:
That would be if B ≥ 0
Can you show that this is the case for what you gave in the OP?

dyn said:
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 ?

dyn said:
after all V0 could be negative ?
Does that matter?

dyn said:
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 $$\int_a^b f(x)dx\geq\int_a^b g(x) dx$$.""

vanhees71, malawi_glenn and dyn
dyn said:
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 ?

Delta2 and PeroK
Does the my reasoning in #12 explain in simple terms the inequality lemma in #11 ?

dyn said:
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.

PeroK and dyn
Thank you everyone

Delta2

## What is the expectation value of energy?

The expectation value of energy is a concept in quantum mechanics that represents the average value of the energy of a system in a given state. It is calculated by taking the sum of the products of the energy values and their corresponding probabilities.

## Why is the expectation value of energy important?

The expectation value of energy is important because it allows us to make predictions about the behavior of a system. It can help us determine the most likely energy state of a system and how it will evolve over time.

## How is the expectation value of energy related to the uncertainty principle?

The expectation value of energy and the uncertainty principle are related through the Heisenberg uncertainty principle, which states that the product of the uncertainties in the position and momentum of a particle cannot be smaller than a certain value. This means that the more precisely we know the energy of a system, the less we know about its position and momentum, and vice versa.

## Can the expectation value of energy be measured?

No, the expectation value of energy is a theoretical concept and cannot be directly measured. However, it can be calculated using mathematical equations and can be used to make predictions about the behavior of a system.

## How does the expectation value of energy change with time?

The expectation value of energy can change with time if the system is in a state that is not an energy eigenstate. In this case, the energy of the system will oscillate between different values, with the expectation value representing the average energy over time.

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