Proving Quantum Mechanics Inequalities

[jameson2]

Homework Statement

I'm looking for help in proving a few quantum mechanics inequalities. I can't really get started on any of them, so just a few general tips would be helpful. For example:

Given a complete set of normalized discrete eigenstates |n> with eigenvalues q_n.
For any observable P the expectation value of P^2 in the state |n>, <n|P^2|n> satisfies |<n|P|m>|^2 \leq <n|P^2|n> where m is any state from the basis.

Homework Equations

The Attempt at a Solution

As I said, I'm not looking for an answer as I haven't even figured out how to attempt it, any hints would be great.

 

[fzero]

Try simplifying

\sum_m | \langle n | P | m\rangle |^2,

where the sum is over the complete basis.

 

[arkajad]

Alternatively: do you know the general inequality (it even has its name!)

|&lt;\psi |\phi&gt;|^2\leq\;\; &lt;\psi|\psi&gt;&lt;\phi|\phi&gt;?

If so, set |\psi&gt;=P|n&gt;,\, |\phi &gt;=|m&gt;.

 

[jameson2]

Ok I think I got it.
I also have this one which I'm not sure about:
For any obserable P and normalised state vector |\psi &gt;
&lt;\psi |P^2|\psi&gt; \geq &lt;\psi|P|\psi&gt;^2

Can you say that &lt;\psi |P^2|\psi&gt;=&lt;\psi |P|p_{\psi}\psi&gt;=p_{\psi}&lt;\psi |P|\psi&gt;=p_{\psi}^2&lt;\psi |\psi&gt; ?

 

[arkajad]

Can you say that ...

No.
But this is the same inequality as the first one, only written in the opposite direction and letters has been changed.

 

[jameson2]

But there are now state vectors instead of eigenstates, I assumed that makes a difference?

 

[arkajad]

In the first inequality you never use the fact that they are eigenvectors. It was irrelevant.