The dual vector space in quantum mechanics

Question

I am a physicist in training, and was only taught about the dual vector space quite loosely within the context of quantum mechanics. I was told that the dual vector space to some ket space in which the kets are column vectors, consists of row vectors with elements from the same field.

Now I am reading the formal definition for a dual space as a "space of all linear functionals $f:V\rightarrow \mathbb{F}$".

Now I am happy with the idea that this itself forms a linear vector space. However I have not been able to convince myself- conceptually nor with a formal proof- that the dual space to a linear vector space consisting of some n dimensional column vectors is necessarily isomorphic to a linear vector space of row vectors.

If my understanding is correct, the loose notion of a dual space I was introduced to via QM is actually saying that "a linear functional $f:V\rightarrow \mathbb{F}$ acting on a linear vector space $V$ in which the elements are some n component column vectors with entries from a field $\mathbb{F}$, say {$v_i$} as the components of a vector, necessarily takes the form $\Sigma a_i v_i$ for any constants $a_i \in \mathbb{F}$. In this case there is clearly an isomorphism between the space of row vectors plus standard inner product (acting on the elemtns of the row vector space $V$) and the space of functionals on V.

However it is not clear to me that all linear functionals on the row vector space $V$ can be written in this form, and therefore that this does constitute the dual space to V.

Answer 1

Given a (finite dimensional, although generalizable to infinite dimensions in nice enough settings like QM) vector space $V$ with basis $(v_1, v_2,\ldots,v_n)$, a linear functional $f:V\to\Bbb F$ is uniquely determined by how it acts on the basis vectors $v_i$. That's because any vector $v\in V$ may be written as a linear combination $v = a_1v_1+\cdots a_nv_n$ with $a_i\in \Bbb F$, and $f$ is linear, so $$ f(v) = f(a_1v_1+\cdots a_nv_n) = a_1f(v_1) + \cdots +a_nf(v_n) $$ so $f(v)$ is completely determined by the $f(v_i)$, for any vector $v$. Thus any $f$ can be represented by a tuple (list of numbers) $(f_1,\ldots,f_n)\in \Bbb F^n$ where $f_i = f(v_i)$. Clearly, if two functions are different, then their corresponding tuples are also different.

On the other hand, any such tuple $(g_1, \ldots,g_n)$ may be used to define a function $g:V\to \Bbb F$. What's more, any such $g$ is linear.

So now we have a correspondence between the set of linear functionals $V\to \Bbb F$ and tuples in $\Bbb F^n$. These tuples are what becomes row vectors in the case where $V$ is a vector space of column vectors.

Answer 2

A standard axiom of Quantum Mechanics is that the underlying vector space is separable. That means that there is a countably infinite set of states $\{ s_1,s_2,\cdots\}$ such that all states can be approximated arbitrarily closely by a finite linear combination of such vectors. And you need a norm. (Actually you need an inner product because of the way that bras and kets are defined.) Such axioms are required for constructability, where answers must be obtained to some precision in a finite number of steps. The real numbers are constructible from the rational numbers in this sense, for example. And that can be built upon to reach these more complex objects. A non-constructible theory would not give a computable model. So Quantum vector spaces are all assumed to be separable, which leads to a construction of a countable basis from a countable dense subspace through the Gram-Schmidt process, for example. Arbitrary dimensions of spaces are not considered valid in a Quantum Model.

A completion of a separable inner product space to obtain a Hilbert space is considered valid, because every state vector in the system may be arbitrarily closely approximated by a finite linear combination of state vectors. So a Hilbert space with countable dimension is considered constructible, in the same way that the real numbers are considered constructible from the rational numbers. A Hilbert space $\mathcal{H}$ that is constructible has a countable orthonormal basis $\{ e_n \}_{n=1}^{\infty}$ such that every $x\in \mathcal{H}$ can be written in terms of this basis and the inner product $(\cdot,\cdot)$ as

$$ x = \sum_{n=1}^{\infty}(x,e_n) e_n. $$ You can see from this that $x$ is constructible. Not all functionals $F$ are considered on a Hilbert space, but only those that are constructible. That becomes an assumption of continuity for $F$. With continuity one obtains $$ F(x) = \sum_{n=1}^{\infty}(x,e_n)F(e_n) \\ = (x,\sum_{n=1}^{\infty}F(e_n)e_n). $$ It turns out that the assumptions imply that $\sum_{n}|F(e_n)|^2 < \infty$ and, therefore $F(x)=(x,y)$ for a unique $y\in\mathcal{H}$, which is the content of the Riesz representation theorem for continuous linear functionals on the Hilbert space $\mathcal{H}$. This is the representation you want for the infinite-dimensional separable case used in Quantum Mechanics by axiomatic assumption. If you think of $x$ as a column vector $$ x=\begin{pmatrix}(x,e_1)\\ (x,e_2)\\ \vdots \end{pmatrix} $$ then $F$ acts on this column vector as a row vector on the left. Or, if you like row vectors for $x$, then $F$ acts as a column vector on the right. These viewpoints are equivalent.

Answer 3

If I'm understanding you correctly, you want to see how all functionals on a Hilbert space can be represented by another vector via the inner product? This is known as the Riesz representation theorem. It states for for a Hilbert space $H$, the map $H \to H^*$ sending $v$ to $\langle v, \cdot \rangle$ (a linear function on $H$) is an isomorphism. In particular, given any functional $f$, there is a unique $v$ such that $f(\cdot) = \langle v, \cdot \rangle$ (and this satisfies some nice properties).

The proof of this is as follows. Given $0 \neq f \in H^*$, then $\ker f$ is a closed proper subspace of $H$ and so there exists $0 \neq x_0 \in (\ker f)^{\perp}$. Note that, since $f (x_0) \neq 0$, we can make a decomposition $$y = \underbrace{\left(y - \frac{f (y)}{f (x_0)} x_0\right)}_{ \in \; \ker f}+ \underbrace{\frac{f (y)}{f (x_0)} x_0}_{\in \;\text{span }\{x_0\}} \quad \forall y \in H,$$ which is an orthogonal decomposition (inner product between the two summands is zero). Since $$\langle y, x_0\rangle = \frac{f (y)}{f (x_0)} ||x_0||^2 \quad \forall y \in H,$$ it follows that $$x = \frac{\overline{f (x_0)}}{||x_0||^2} x_0$$ satisfies $\langle y, x\rangle = f (y)$ for all $y$. To see this, note \begin{align*} \langle y,x\rangle & = \left\langle y - \frac{f (y)}{f (x_0)} x_0 +\frac{f (y)}{f (x_0)} x_0, \frac{\overline{f(x_0)}}{||x_0||^2}x_0 \right\rangle \\ & = \left\langle \frac{f (y)}{f (x_0)} x_0 , \frac{\overline{f(x_0)}}{||x_0||^2}x_0 \right\rangle \\ & = \frac{f(y)}{||x_0||^2} \langle x_0,x_0 \rangle = f(y). \end{align*} And this is what we wanted to show.

It is worth noting that the above proof works for infinite-dimensional Hilbert spaces.