# Trace of an operator

I have read that if I have two vectors $|0\rangle$ and $|1\rangle$, then$$\mathrm{tr} ( |0\rangle \langle1|)=\langle1|0\rangle.$$

I also know that for an operator $A$ $$\mathrm{tr} ( A) = \sum_{i} \langle i|A|i\rangle$$ where $|i\rangle$ is an arbitrary basis. Well, if I was to choose my operator $A =|0\rangle \langle1|$ and use the $|0\rangle$, $|1\rangle$ basis to evaluate the trace, I will get

$$\mathrm{tr} (|0\rangle \langle1|) = \langle 0|0\rangle \langle1|0\rangle + \langle1|0\rangle \langle1|1\rangle$$

and assuming the basis vectors are normalised I get

$$\mathrm{tr} (|0\rangle \langle1|) = 2\langle1|0\rangle$$ which is disagreement with what I have read. Where am I going wrong?

• Isnt $\langle1|0\rangle$ = $0$? Oct 26, 2016 at 18:40
• The formula you are using for the trace only works if the $|i \rangle$ are orthonormal. In that case $\langle 1 | 0 \rangle = 0$ and your equation reduces to $0 = 0$, which is correct. Oct 26, 2016 at 18:44
• You should be careful with results in Dirac notation, because orthonormality is very often implicitly assumed. Oct 26, 2016 at 18:44

There is nothing wrong once you realize that $\langle 0|1\rangle=0$ so that the claimed result and your explicit calculation do actually match. More generally, the trace of an operator $| i\rangle \langle j|$ is $$Tr| i\rangle \langle j| =\sum_k \langle k|i\rangle \langle j | k\rangle=\sum_k\delta_{ki} \delta_{jk}=\delta_{ij}=\langle i|j\rangle.$$ where I have been working with an orthonormal basis of vectors $|i\rangle$ and and their dual $\langle i |$.
• Do $|0\rangle$ and $|1\rangle$ have to be orthogonal? I have not made that assumption Oct 26, 2016 at 19:09
• you need to assume so if you want to get the claimed result that the trace is $\langle 1|0\rangle$. Of course you could assume otherwise but then the result would not be the one you wanted to obtain. With no orthogonality you just get your expression at the bottom of your post, not the one at the top. Oct 26, 2016 at 19:12