# Inner products and Hilbert spaces

References: edX online course MIT 8.05.1x Week 4.

Sheldon Axler (2015), Linear Algebra Done Right, 3rd edition, Springer. Chapter 6.

An inner product defined on a vector space ${V}$ is a function that maps each ordered pair of vectors ${\left(u,v\right)}$ of ${V}$ to a number denoted by ${\left\langle u,v\right\rangle \in\mathbb{F}}$. The inner product satisfies the following axioms:

1. positivity: ${\left\langle v,v\right\rangle \ge0}$ for all ${v\in V}$.
2. definiteness: ${\left\langle v,v\right\rangle =0}$ if and only if ${v=0}$ (the zero vector).
3. additivity in the second slot: ${\left\langle u,v+w\right\rangle =\left\langle u,v\right\rangle +\left\langle u,w\right\rangle }$.
4. homogeneity in the second slot: ${\left\langle u,\lambda v\right\rangle =\lambda\left\langle u,v\right\rangle }$ where ${\lambda\in\mathbb{F}}$.
5. conjugate symmetry: ${\left\langle u,v\right\rangle =\left\langle v,u\right\rangle ^*}$.

[Axler requires additivity and homegeneity in the first slot rather than the second, but Zwiebach uses the conditions above, which are more usual for physics.]

The norm ${\left|v\right|}$ of a vector ${v}$ is defined as

$\displaystyle \left|v\right|^{2}\equiv\left\langle v,v\right\rangle \ \ \ \ \ (1)$

All the above applies to both real and complex vector spaces, although it should be noted that in the case of conjugate symmetry in a real space, ${\left\langle v,u\right\rangle ^*=\left\langle v,u\right\rangle }$ so in that case, the condition reduces to ${\left\langle u,v\right\rangle =\left\langle v,u\right\rangle }$.

Conditions 3 and 4 apply to the first slot as well, though with a slight difference for complex vector spaces. [These properties can actually be derived from the 5 axioms above, so aren’t listed as separate axioms]:

 $\displaystyle \left\langle u+v,w\right\rangle$ $\displaystyle =$ $\displaystyle \left\langle u,w\right\rangle +\left\langle v,w\right\rangle \ \ \ \ \ (2)$ $\displaystyle \left\langle \lambda u,v\right\rangle$ $\displaystyle =$ $\displaystyle \lambda^*\left\langle u,v\right\rangle \ \ \ \ \ (3)$

Two vectors are orthogonal if ${\left\langle u,v\right\rangle =\left\langle v,u\right\rangle =0}$. By this definition, the zero vector is orthogonal to all vectors, including itself.

The inner product is non-degenerate, meaning that any vector that is orthogonal to all vectors must be zero.

An orthogonal decomposition is defined for a vector ${u}$ as follows: Suppose ${u,v\in V}$ and ${v\ne0}$. We can write ${u}$ as

 $\displaystyle u$ $\displaystyle =$ $\displaystyle \frac{\left\langle u,v\right\rangle }{\left|v\right|^{2}}v+u-\frac{\left\langle u,v\right\rangle }{\left|v\right|^{2}}v\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle cv+w \ \ \ \ \ (5)$

where

 $\displaystyle c$ $\displaystyle =$ $\displaystyle \frac{\left\langle u,v\right\rangle }{\left|v\right|^{2}}\in\mathbb{F}\ \ \ \ \ (6)$ $\displaystyle w$ $\displaystyle =$ $\displaystyle u-\frac{\left\langle u,v\right\rangle }{\left|v\right|^{2}}v \ \ \ \ \ (7)$

From the definition, ${\left\langle w,v\right\rangle =0}$ so we’ve decomposed ${u}$ into a component ${cv}$ ‘parallel’ to ${v}$ and another component ${w}$ orthogonal to ${v}$.

There are a couple of important theorems for which we’ll run through the proofs:

Theorem 1 Pythagorean theorem. If ${u}$ and ${v}$ are orthogonal then

$\displaystyle \left|u+v\right|^{2}=\left|u\right|^{2}+\left|v\right|^{2} \ \ \ \ \ (8)$

Proof: From the definition of the norm

 $\displaystyle \left|u+v\right|^{2}$ $\displaystyle =$ $\displaystyle \left\langle u+v,u+v\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle u,u\right\rangle +\left\langle u,v\right\rangle +\left\langle v,u\right\rangle +\left\langle v,v\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle u,u\right\rangle +\left\langle v,v\right\rangle \ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|u\right|^{2}+\left|v\right|^{2} \ \ \ \ \ (12)$

where in the third line we used the orthogonality condition ${\left\langle u,v\right\rangle =\left\langle v,u\right\rangle =0}$. $\Box$

Theorem 2 Schwarz (or Cauchy-Schwarz) inequality. For all vectors ${u,v\in V}$

$\displaystyle \left|\left\langle u,v\right\rangle \right|\le\left|u\right|\left|v\right| \ \ \ \ \ (13)$

Proof: The inequality is obviously true (as an equality) if ${v=0}$, so we need to prove it for ${v\ne0}$. In that case we can form an orthogonal decomposition of ${u}$:

$\displaystyle u=\frac{\left\langle u,v\right\rangle }{\left|v\right|^{2}}v+w \ \ \ \ \ (14)$

Since ${v}$ and ${w}$ are orthogonal, we can apply the Pythagorean theorem:

 $\displaystyle \left|u\right|^{2}$ $\displaystyle =$ $\displaystyle \left|\frac{\left\langle u,v\right\rangle }{\left|v\right|^{2}}v\right|^{2}+\left|w\right|^{2}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|v\right|^{4}}\left|v\right|^{2}+\left|w\right|^{2}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|v\right|^{2}}+\left|w\right|^{2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle \ge$ $\displaystyle \frac{\left|\left\langle u,v\right\rangle \right|^{2}}{\left|v\right|^{2}} \ \ \ \ \ (18)$

Rearranging the last line and taking the positive square root (since norms are always non-negative) we have

$\displaystyle \left|\left\langle u,v\right\rangle \right|\le\left|u\right|\left|v\right| \ \ \ \ \ (19)$

$\Box$

Theorem 3 Triangle inequality. For all ${u,v\in V}$

$\displaystyle \left|u+v\right|\le\left|u\right|+\left|v\right| \ \ \ \ \ (20)$

Proof: As with the Schwarz inequality, we start with the square of the norm:

 $\displaystyle \left|u+v\right|^{2}$ $\displaystyle =$ $\displaystyle \left\langle u+v,u+v\right\rangle \ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle u,u\right\rangle +\left\langle v,v\right\rangle +\left\langle u,v\right\rangle +\left\langle v,u\right\rangle \ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left\langle u,u\right\rangle +\left\langle v,v\right\rangle +\left\langle u,v\right\rangle +\left\langle u,v\right\rangle ^*\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|u\right|^{2}+\left|v\right|^{2}+2\mathfrak{R}\left\langle u,v\right\rangle \ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle \le$ $\displaystyle \left|u\right|^{2}+\left|v\right|^{2}+2\left|\left\langle u,v\right\rangle \right| \ \ \ \ \ (25)$

The last line follows because ${\left\langle u,v\right\rangle }$ is a complex number, so

$\displaystyle \left|\left\langle u,v\right\rangle \right|=\sqrt{\left(\mathfrak{R}\left\langle u,v\right\rangle \right)^{2}+\left(\mathfrak{I}\left\langle u,v\right\rangle \right)^{2}}\ge\mathfrak{R}\left\langle u,v\right\rangle \ \ \ \ \ (26)$

We can now apply the Schwarz inequality to the last term in the last line to get

 $\displaystyle \left|u+v\right|^{2}$ $\displaystyle \le$ $\displaystyle \left|u\right|^{2}+\left|v\right|^{2}+2\left|u\right|\left|v\right|\ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\left|u\right|+\left|v\right|\right)^{2} \ \ \ \ \ (28)$

Taking the positive square root, we get

$\displaystyle \left|u+v\right|\le\left|u\right|+\left|v\right| \ \ \ \ \ (29)$

$\Box$

A finite-dimensional complex vector space with an inner product is a Hilbert space. An infinite-dimensional complex vector space with an inner product is also a Hilbert space if a completeness property holds. This property is a technical property which is always satisfied in quantum mechanics, so we can assume that any infinite-dimensional complex vector spaces we encounter in quantum theory are Hilbert spaces.

## 7 thoughts on “Inner products and Hilbert spaces”

1. Mark Weitzman

Actually I think the infinite dimensional spaces in QM frequently fail to be Hilbert Spaces (hence Dirac delta function etc.). But they are usually treated under an extension called rigged Hilbert Spaces. See Quantum Mechanics a Modern development by Ballentine. But essentially you are correct in that we can use the Hilbert space formalism in infinite dimensional QM spaces, if we are willing to overlook certain technicalities – like non-normalizable states etc.

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