Why Must the Expectation Value of H Be at Least the Lowest Energy Eigenvalue?

[ehrenfest]

Shankar 163

Homework Statement

Show that for any normalized |psi>, <psi|H|psi> is greater than or equal to E_0, where E_0 is the lowest energy eigenvalue. (Hint: Expand |psi> in the eigenbasis of H.)

Homework Equations

The Attempt at a Solution

I think the question assumes the V = 0, so H = P^2/2m. The eigenvalues for the equation P^2/2m|p> = E|p> are then p = +/- (2mE)^(1/2) and the eigenkets are of the form | p = + (2mE)^(1/2)> and | p = (2mE)^(1/2)> (or in energy terms the eigenvalues are of the form |E,->, |E,+> and the eigenkets are the same as the momentum ones. I'm not really sure how you can expand anything with these though...

 

[lalbatros]

No, this statement is valid for any Hamiltonian.
To spare your pencil, assume the system has only two basis states and develop <psi|H|psi>, you will see how obvious this result is.
(take the basis states as H-eigenvectors for simplicity, then imagine th consequence if you don't know these eigenvectors but would like to find the fundamental state, very useful statement indeed!)

 

[chanvincent]

This is how we expand $$\Psi$$ in terms of the eigenbasis of H :

Use the fact that the eigenvalues of H which is $$\{\Psi_0,\Psi_1,\Psi_2 ...\}}$$ forms the basis of a Hilbert space. Notice Hibert space is completed.
That means any vector $$\Psi$$ could be written as a linearly combinition of $$\Psi_0,\Psi_1,\Psi_2 ...$$
i.e. $$\Psi = c_0\Psi_0+c_1\Psi_1+c_2\Psi_2+...$$.

The problem should be trivial from this point ( use the orthogonality of the eigenvector, although the eigenvectors in degenerate state are not orthogonal, they do not affect the result. )

 

[lalbatros]

Of course the end result is

<H> = p1 H1 + p2 H2 + ...

where pi is the probability of being in state i,
and 0<pi<1
and p1 + p2 + ... = 1

 

[ehrenfest]

OK. So we use the time-independent version of the Schrodinger equation H * psi = E * psi.

So $$|\psi> = p0 * |H0> +p1 * |H1 > + ...= a0 * |E0> + a1 * | E1> + ...$$

p (and my choice of a) are arbitrary letters, right? You did not use p for momentum did you lalbatros?

Now, $$<\psi| H | /psi> = (E0*a0 + E1*a1 + ...)</psi|/psi> = E0*a0 + E1*a1 + ...$$

So the minimum of that expression would occur when a0 = 1 and ai = 0 for all i > 1.

Does that explanation work?

 

[Dick]

Your notation is confusing. I'll confuse the waters a little more by posting ugly, non-TeX. Which I prefer to garbled TeX. |psi>=sum(ci*|Ei>). ci are the amplitudes, |Ei> is the eigenstate corresponding to eigenvalue Ei. So <psi|H|psi>=sum(ci*conjugate(ci)*<Ei|H|Ei>)=sum(ci*conjugate(ci)*Ei*<Ei|Ei>).
Since we'll take |Ei> normalized <Ei|Ei>=1.
So <psi|H|psi>=sum(pi*Ei) where pi=ci*conjugate(ci). Since we'll also take |psi> normalized, sum(pi)=1.

So if sum(pi)=1, pi>=0, do you believe sum(pi*Ei)>=E0 where E0 is the smallest of the Ei's? Can you prove it?

 

[ehrenfest]

I believe it.

sum(pi*Ei) >= sum(pi*E0) = sum(pi)*E0 = E0

 

[ehrenfest]

How do you go from |psi>=sum(ci*|Ei>) to <psi|H|psi> = sum(ci*conjugate(ci)*<Ei|H|Ei>)?

Shouldn't it be something like <psi|H|psi> = sum(conjugate(ci)*<Ei|) H sum(ci*|Ei>) just by replacing |psi> with sum(ci*|Ei>) and < psi| with sum(conjugate(ci)*<Ei|)?

 

[Dick]

We've taken the |Ei> to be orthonormal, so <Ei|Ej>=delta(i,j) And H(ci*|Ei>)=ci*Ei*|Ei>. And your proof is perfect.

 

[ehrenfest]

I see. Thanks.

 

[Dick]

You really do? You're not kidding, right?