# Why does expectation values are always nonnegative?

1. Apr 6, 2015

### zhdx

Why does the expectation values of some operators, such as 'number' operator $a^\dagger a$ and atomic population operator $\sigma^\dagger\sigma$, are always nonnegative? Can we prove this from a mathematical point? For example, are these operators positive semidefinite?

Last edited: Apr 6, 2015
2. Apr 6, 2015

### bhobba

Its because the outcome of the observation is a number so obviously is always positive.

If you are talking about the number operator of the harmonic oscillator yes you can prove that as any text will explain eg:
http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator

Thanks
Bill

3. Apr 6, 2015

### zhdx

Thank you for your reply. It does so that the obervation result is nonnegative. I just want to find a mathematical proof.

4. Apr 6, 2015

### bhobba

Did you see the proof in the case of the harmonic oscillator?

Its the basis of similar operators in QFT.

Thanks
Bill

5. Apr 6, 2015

### zhdx

Thanks. I've read the proof for this specific case. What I mean is a more general case. For example, the operator $(a^\dagger)^2 a^2 \sigma^\dagger\sigma$ for a coupled cavity-atom system. Is the expectation for this operator always nonnegative for any state (include the mixed state)?

Last edited: Apr 6, 2015
6. Apr 6, 2015

### Staff: Mentor

A measurement cannot give a negative result, so the expectation value has to be positive - otherwise you made a mistake with the calculations or the operator definition.

7. Apr 6, 2015

### zhdx

Thanks for your reply. The measurement results of many operators can be negative. For example, the position and the momentum of a particle.

Last edited: Apr 6, 2015
8. Apr 6, 2015

9. Apr 6, 2015

### zhdx

10. Apr 6, 2015

### kith

I don't know if the proof can be generalized because I haven't looked into it in detail but the wikipedia article says that the statement is true at least for Hermitean matrices. See this section.

Last edited: Apr 6, 2015
11. Apr 6, 2015

### Fredrik

Staff Emeritus
The proof for operators of the form $A^\dagger A$ is very easy. For all state vectors $x$, we have
$$\langle x,A^\dagger Ax\rangle =\langle A^{\dagger\dagger}x,Ax\rangle =\langle Ax,Ax\rangle\geq 0.$$
The requirement that $\langle x,x\rangle\geq 0$ for all $x$ is part of the definition of "inner product", and therefore part of the definition of "Hilbert space".

12. Apr 6, 2015

### micromass

Staff Emeritus
All semidefinite matrices are Hermitian.

13. Apr 6, 2015

### micromass

Staff Emeritus
And there is an interesting converse too. If for all $x$, we have $\langle x,Bx\rangle \geq 0$, then there is an operator $A$ such that $B = A^\dagger A$. And all of this is equivalent tfor Hermitian operators) with saying that the spectrum of $B$ (thus if I understand QM well: the set of all outcomes) is nonnegative.

Last edited: Apr 6, 2015
14. Apr 6, 2015

### zhdx

Thanks you! Can you help me with the proof of the theorem you mentioned?

15. Apr 6, 2015

### micromass

Staff Emeritus
Last edited by a moderator: May 7, 2017
16. Apr 6, 2015

### zhdx

Thank you very much! Can we say that the operator of the form $A^\dagger A$ is a positive semidefinite operator?

Last edited by a moderator: May 7, 2017
17. Apr 6, 2015

### micromass

Staff Emeritus
Yes.

18. Apr 7, 2015

### vanhees71

I don't understand this question. Why should the expectation value of a position or momentum (vector component) be positive definite? It's of course not and there's no reason why it should!

19. Apr 7, 2015

### zhdx

The position and momentum are not positive definite. So the expectation of them can be positive, zero or negative. Similar observable includes the energy, which is dependent on the zero point we choose. However, the operator of the form $A^\dagger A$, such as 'number' operator $a^\dagger a$, is positive semidefinite. So the expectation values are always nonnegative. This is also a physically reasonable result.

Last edited: Apr 7, 2015
20. Apr 7, 2015

### bhobba

Then I fail to understand your question - you have answered it yourself.

Thanks
Bill