Why does expectation values are always nonnegative?

[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?

 

[bhobba]

Why does the expectation values of some operators, such as 'number' operator $a^{\dag}a$ and atomic population operator $\sigma^{\dag}\sigma$, are always nonnegative? Can we prove this from a mathematical point?

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

 

[zhdx]

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

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

 

[bhobba]

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

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

Its the basis of similar operators in QFT.

Thanks
Bill

 

[zhdx]

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

Its the basis of similar operators in QFT.

Thanks
Bill

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)?

 

[mfb]

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.

 

[zhdx]

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.

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

 

[kith]

Wikipedia has a short proof that all eigenvalues being positive is equivalent to positive definiteness:
https://en.wikipedia.org/wiki/Positive-definite_matrix#Characterizations

 

[zhdx]

Wikipedia has a short proof that all eigenvalues being positive is equivalent to positive definiteness:
https://en.wikipedia.org/wiki/Positive-definite_matrix#Characterizations

Thanks! Can this proof be generalized to the positive semidefinite case? That is, if all eigenvalues are non-negative, then the matrix is positive semidefinite.

 

[kith]

Thanks! Can this proof be generalized to the positive semidefinite case? That is, if all eigenvalues are non-negative, then the matrix is positive semidefinite.

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.

 

[Fredrik]

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".

 

[micromass]

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.

All semidefinite matrices are Hermitian.

 

[micromass]

The proof for operators of the form $A^\dagger A$ is very easy. For all state vectors $x$, we have
$$\langle x,A^\dagger Ax\rangle =\langle A^{\dagger\dagger}x,Ax\rangle =\langle Ax,Ax\rangle\geq 0.$$

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.

 

[zhdx]

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 with saying that the spectrum of $B$ (thus if I understand QM well: the set of all outcomes) is nonnegative.

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

 

[micromass]

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

For matrices, see Roman, page 250: https://www.amazon.com/dp/0387728287/?tag=pfamazon01-20
For bounded operators: see Reed & Simon, chapter VI.4: https://www.amazon.com/dp/0125850506/?tag=pfamazon01-20
For unbounded operators, see 5.6.21 of Kadison & Ringrose: https://www.amazon.com/dp/0821808192/?tag=pfamazon01-20

 

[zhdx]

For matrices, see Roman, page 250: https://www.amazon.com/dp/0387728287/?tag=pfamazon01-20
For bounded operators: see Reed & Simon, chapter VI.4: https://www.amazon.com/dp/0125850506/?tag=pfamazon01-20
For unbounded operators, see 5.6.21 of Kadison & Ringrose: https://www.amazon.com/dp/0821808192/?tag=pfamazon01-20

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

 

[micromass]

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

Yes.

 

[vanhees71]

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

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!

 

[zhdx]

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!

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.

 

[bhobba]

The position and momentum are not positive definite. So the expectation of them can be positive, zero or negative. Similar observable include 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.

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

Thanks
Bill

 

[rjshscs11]

then what about the expectation value of energy in second excited state?(it is negative)

 

[Vanadium 50]

then what about the expectation value of energy in second excited state?(it is negative)

No, it isn't. It's 5/2 hbar omega.

 

[rjshscs11]

But energy of H-atom is given by this relation = - 13.6/n^2

 

[mfb]

Relative to an arbitrary choice of zero. You can choose the ground state to be zero, for example, then no negative energies occur.

The original question was about operators that don't have this ambiguity.