Inner product - positive or positive semidefinite?

[dyn]

Hi

In QM the inner product satisfies < a | a > ≥ 0 with equality if and only if a = 0.

Is this positive definite or positive semidefinite because i have seen it described as both

Thanks

 

[anuttarasammyak]

Inner products $<a|b>$ is a complex number in general.
Inner product with its conjugate $<a|a>$ is positive and usually to be normalized. I am not sure
whether we include $|a>=0$ as a state vector which allows semidefinite nature.

 

[Demystifier]

https://math.stackexchange.com/questions/1733726/positive-semi-definite-vs-positive-definite

 

[vanhees71]

In the standard definition of a Hilbert space the scalar product is positive definite, i.e., if $\langle a|a \rangle=0$, the vector $|a \rangle=|0 \rangle$ is the null vector.

That's why, e.g., in its realization as the Hilbert space of square integrable functions (as used in "wave mechanics" a la Schrödinger) a function represents the null vector if it is zero except on a zero-measure null [edit: corrected in view of #5; and once more in view of #7] set.

 

[martinbn]

In the standard definition of a Hilbert space the scalar product is positive definite, i.e., if $\langle a|a \rangle=0$, the vector $|a \rangle=|0 \rangle$ is the null vector.

That's why, e.g., in its realization as the Hilbert space of square integrable functions (as used in "wave mechanics" a la Schrödinger) a function represents the null vector if it is zero except on a zero set.

You mean except on a zero measure set.

 

[vanhees71]

Thanks, I wasn't sure about the correct english term for "Nullmenge" in the sense of Lebesgue-integration theory.

 

[martinbn]

Thanks, I wasn't sure about the correct english term for "Nullmenge" in the sense of Lebesgue-integration theory.

Ah, i see! It is null set.

 

[vanhees71]

So much for the reliability of online dictionaries :-(. BTW does anybody know a good one for mathematics/physics vocabulary? One translated it with "empty set", which doesn't make sense in this context or with "zero set" which may be wrong either.

 

[dyn]

https://math.stackexchange.com/questions/1733726/positive-semi-definite-vs-positive-definite

The definition given here shows that in terms of QM the inner product can be defined as positive definite or positive semi-definite which might account for the fact that Shankar calls it positive semidefinite while Sakurai calls it positive definite

 

[Demystifier]

BTW does anybody know a good one for mathematics/physics vocabulary?

https://www.physicsforums.com/threads/collection-of-science-jokes-p2.847743/post-5988541

 

[vanhees71]

The definition given here shows that in terms of QM the inner product can be defined as positive definite or positive semi-definite which might account for the fact that Shankar calls it positive semidefinite while Sakurai calls it positive definite

You mix the positive definiteness of the scalar product of a Hilbert space, and by definition a Hilbert space has a positive definite scalar product and the positive definiteness of a self-adjoint operator.

 

[martinbn]

You mix the positive definiteness of the scalar product of a Hilbert space, and by definition a Hilbert space has a positive definite scalar product and the positive definiteness of a self-adjoint operator.

I looked at Shankar. He does call it semidefinite.

 

[vanhees71]

What does Shankar call semidefinite? The scalar product? I'd be surprised!

 

[martinbn]

What does Shankar call semidefinite? The scalar product? I'd be surprised!

Yes.

 

[vanhees71]

That's very strange. Then QT doesn't work the usual way!

 

[martinbn]

That's very strange. Then QT doesn't work the usual way!

No, his inner product is the same, he just calls it semidefinite.

 

[vanhees71]

Ok, an inner product is positive semidefinite if for all vectors $|v \rangle$ you have $\langle v|v \rangle \geq 0$. Take a finite-dimensional space. Then you can choose a basis such that
$$\langle v|v \rangle=\sum_{j=1}^d g_{jk} v_j^* v_k,$$
and you must have $(g_{jk})=\mathrm{diag}(1,1,\ldots,1,0,0,0\ldots)$. In that case the inner product does not induce a metric.

That's why, for a scalar product you define it to be positive definite, i.e., it's positive semidefinite and in addition you have $\langle v|v \rangle = 0 \Leftrightarrow |v \rangle=|0 \rangle$ (where $|0 \rangle$ is the null vector).

 

[martinbn]

Ok, an inner product is positive semidefinite if for all vectors $|v \rangle$ you have $\langle v|v \rangle \geq 0$. Take a finite-dimensional space. Then you can choose a basis such that
$$\langle v|v \rangle=\sum_{j=1}^d g_{jk} v_j^* v_k,$$
and you must have $(g_{jk})=\mathrm{diag}(1,1,\ldots,1,0,0,0\ldots)$. In that case the inner product does not induce a metric.

That's why, for a scalar product you define it to be positive definite, i.e., it's positive semidefinite and in addition you have $\langle v|v \rangle = 0 \Leftrightarrow |v \rangle=|0 \rangle$ (where $|0 \rangle$ is the null vector).

Yes, but you are not listening to what I am saying. His inner product is positive definite, he includes the condition $\langle v|v\rangle = 0$ iff $v=0$. He just calls it positive semi-definite.

 

[vanhees71]

So it's positive definite. That seems to be a pretty confusing textbook then.

 

[dyn]

The Mathematical methods book by Riley & Hobson also states that positive semi-definite is < a | a > ≥ 0 with < a | a > = 0 implying a = 0

 

[vanhees71]

Ok, what then for those authors is "positive definite"? In the usual textbooks you have $\langle a|a \rangle \geq 0$ as the condition for a "positive semidefinite" sesquilinear form. If in addition $\langle a| a \rangle =0 \Leftrightarrow |a \rangle=|0 \rangle$ is fulfilled the sequilinear form is called "positive definite".

As usual, standard conventions followed by the vast majority of textbooks and scientific papers, make more sense than private nomenclature of some singular authors ;-).

 

[dyn]

It seems that some physics textbooks can be less that strictly accurate with their mathematical terminology !