Question: Suppose $C\_g$ is a smooth compact complex curve (of genus $g$), and let $J$ be its *Jacobian*. Recall that the Jacobian $J$ of a curve $C\_g$ is a complex torus that can by obtained by contractions of all rational curves on the $g$-th symmetric power of $C\_g$, e.i., $Sym^g(C\_g)$. Recall also that there is a *theta divisor* $\Theta$ in $J$, 
depending on a point $p\in C\_g$.
The divisor $\Theta$ is the image in $J$ of the set of points $(p,p\_1,...,p\_{g-1})$ with $p$ fixed.

**Question.** How to calculate the dimension of the set of divisors on $J$ linearly equivalent to $\Theta$? In other words, what is $dim( H^0(J,\cal O(\Theta)))$?

Answer: 
Since my comment was too cryptic, I will spell it out as an answer. 

A polarization on an abelian variety $A$ is an ample divisor $D$ (modulo linear equivalence). A polarization is principal if the self-intersection $D^g$ is equal to $g!$, where $g = \dim A$. It is well-known that the theta divisor is a principal polarization of the Jacobian. I am not sure what is a good reference, maybe Griffiths-Harris.

The Riemann-Roch theorem for abelian varieties (proved e.g. in Mumford) says that the Euler characteristic of the line bundle corresponding to a divisor $D$ is $D^g/g!$, so it is $1$ in the case of a principal polarization. Now, it is also shown in Mumford that only one $\dim H^i$ is non-zero, so if the divisor is ample it has to be $H^0$. Putting it all together $\dim H^0 =1$.

Edit: I was informed by email that my definition of polarization is too restrictive, so not quite right. The theta divisor defines a principal polarization anyway and this is proved in some of the other answers.
In general, let $L$ be a line bundle on a complex torus $X=V/ \Lambda$ of dimension $g$ and let $H$ be the hermitian form corresponding to the first Chern class $c\_1(L)$. The imaginary part $E:= \textrm{Im}(H)$ is an alternating form which is integer-valued on the lattice $\Lambda$. 

By elementary linear algebra there is a basis of $\Lambda$ with respect to which $E$ is given by the matrix $$\left(\begin{matrix}0 & D \cr - D & 0 \end{matrix}\right),$$ 
where $D=\textrm{diag}(d\_1, \ldots, d\_g)$ and the $d\_i$ are strictly positive integers satisfying $d\_i|d\_{i+1}$ for all $i=1, \ldots ,g-1$.

If $L$ is positive-definite then one shows that
$$h^0(X, L)=\textrm{Pf}(E)=\det(D).$$
The proof consists in explicitly writing a basis for $H^0(X, L)$ by using canonical theta functions, as in Sebastian's answer.

If $X=J(C)=H^0(\omega\_C)^\*/H\_1(C, \mathbb{Z})$ is the Jacobian of a smooth curve, then the theta divisor $\Theta$ is a *principal polarization*, i.e. $D$ is the identity matrix. This can be seen by taking a standard homology basis for $H\_1(C, \mathbb{Z})$. 

It follows $h^0(X, \Theta)=1$.

See [Birkenhake-Lange, Complex Abelian Varieties, Chapters 3 and 11] for further details.