How to preserve the original order of elements in an unnested array?
WITH ORDINALITY in Postgres 9.4 or later
The new feature simplifies this class of problems. The above query can now simply be:
SELECT *
FROM   regexp_split_to_table('I think Postgres is nifty', ' ') WITH ORDINALITY x(word, rn);
Or, applied to a table:
SELECT *
FROM   tbl t, regexp_split_to_table(t.my_column, ' ') WITH ORDINALITY x(word, rn);
Details:

PostgreSQL unnest() with element number

About the implicit LATERAL join:

What is the difference between LATERAL and a subquery in PostgreSQL?

Postgres 9.3 or older - and more general explanation
For a single string
You can apply the window function row_number() to remember the order of elements. However, with the usual row_number() OVER (ORDER BY col) you get numbers according to the sort order, not the original position in the string.
You could simply omit ORDER BY to get the position &quot;as is&quot;:
SELECT *, row_number() OVER () AS rn
FROM   regexp_split_to_table('I think Postgres is nifty', ' ') AS x(word);
Performance of regexp_split_to_table() degrades with long strings.  unnest(string_to_array(...)) scales better:
SELECT *, row_number() OVER () AS rn
FROM   unnest(string_to_array('I think Postgres is nifty', ' ')) AS x(word);
However, while this normally works and I have never seen it break in simple queries, Postgres asserts nothing as to the order of rows without an explicit ORDER BY.
To guarantee ordinal numbers of elements in the original string, use generate_subscript() (improved with comment by @deszo):
SELECT arr[rn] AS word, rn
FROM   (
   SELECT *, generate_subscripts(arr, 1) AS rn
   FROM   string_to_array('I think Postgres is nifty', ' ') AS x(arr)
   ) y;
For a table of strings
Add PARTITION BY id to the OVER clause ...
Demo table:
CREATE TEMP TABLE strings(string text);
INSERT INTO strings VALUES
  ('I think Postgres is nifty')
 ,('And it keeps getting better');

I use ctid as ad-hoc substitute for a primary key. If you have one (or any unique column) use that instead.
SELECT *, row_number() OVER (PARTITION BY ctid) AS rn
FROM  (
   SELECT ctid, unnest(string_to_array(string, ' ')) AS word
   FROM   strings
   ) x;

This works without any distinct ID:
SELECT arr[rn] AS word, rn
FROM  (
   SELECT *, generate_subscripts(arr, 1) AS rn
   FROM  (
      SELECT string_to_array(string, ' ') AS arr
      FROM   strings
      ) x
   ) y;

SQL Fiddle.
Answer to question
SELECT z.arr, z.rn, z.word, d.meaning   -- , partofspeech -- ?
FROM  (
   SELECT *, arr[rn] AS word
   FROM  (
      SELECT *, generate_subscripts(arr, 1) AS rn
      FROM  (
         SELECT string_to_array(string, ' ') AS arr
         FROM   strings
         ) x
      ) y
   ) z
JOIN   dictionary d ON d.wordname = z.word
ORDER  BY z.arr, z.rn;