See main article: 1840 United States presidential election.
| Election Name: | 1840 United States presidential election in Louisiana |
| Country: | Louisiana |
| Type: | presidential |
| Ongoing: | no |
| Previous Election: | 1836 United States presidential election in Louisiana |
| Previous Year: | 1836 |
| Next Election: | 1844 United States presidential election in Louisiana |
| Next Year: | 1844 |
| Election Date: | October 30 – December 2, 1840 |
| Image1: | William Henry Harrison crop.jpg |
| Nominee1: | William Henry Harrison |
| Party1: | Whig Party (United States) |
| Home State1: | Ohio |
| Running Mate1: | John Tyler |
| Electoral Vote1: | 5 |
| Popular Vote1: | 11,296 |
| Percentage1: | 59.73% |
| Nominee2: | Martin Van Buren |
| Party2: | Democratic Party (United States) |
| Home State2: | New York |
| Running Mate2: | none |
| Electoral Vote2: | 0 |
| Popular Vote2: | 7,616 |
| Percentage2: | 40.27% |
| President | |
| Before Election: | Martin Van Buren |
| Before Party: | Democratic Party (United States) |
| Before Color: | FF3333 |
| After Election: | William Henry Harrison |
| After Party: | Whig Party (United States) |
| After Color: | FF3333 |
The 1840 United States presidential election in Louisiana took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose five representatives, or electors to the Electoral College, who voted for President and Vice President.
Louisiana voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Louisiana by a margin of 19.46%.
With 59.73% of the popular vote, Louisiana would prove to be Harrison's fourth strongest state after Kentucky, Vermont and Rhode Island.[1]