Trace of a commutator is zero, since trace(AB)=trace(BA). Therefore, if we apply same thing to the commutator of position and momentum operator which equals -i*h; we don't obtain zero..!! Where is the ambiguity?.
Dear Hrushikesh, This is an interesting question, If we calculate [ x , p ] in terms of well known creation and annihilation operators in normal unit, i.e. [ x , p ] = i [ a , a+ ] and represent a+ and a operators in matrix form (as in any quantum mechanics book) a simple calculation shows that (a a+ - a+a) is unit matrix 1 with infinite dimension, then [ x , p ] = i1 and its trace will tend to infinite, but interestingly if you calculate the trace of a a+ and a+a separately you will find them equal to a sum of natural number series i.e. (1+2+3+....) , then Tr( a a+) = Tr( a+a) !! this may only occur where we use the infinite dimensional space representations such as Hilbert space, where trace is a sum of infinite series that in spite of the validity of Tr( a a+) = Tr( a+a) we have [ x , p ] = i1 . :)
The dimension of X and P operators is ∞. and so on their Trace. When you calculate :
Trace (XP) - Trace (PX) = -i hbar Trace(I)
∞ - ∞ = ∞
which is not a wrong mathematical expression. The R.H.S ∞ is because the identity matrix, I, 's dimension are ∞ too.
In other word let Trace (XP) = Trace (PX) = A you may conclude from A-A=/ 0 that A should be infinity.
Dear Hrushikesh, Oleksandr has already answered your question,and the Sahar Arabzadeh make a example for you.If you have two operators A and B are bounded. The relation trace(AB)=trace(BA) is right.If the two operators A and B are not bounded,we did not can defint trace of the operator.
Dear Yonghong, it sounds that here is some misunderstanding about the trace of matrix representation of an operator and "bounded" and "unbounded" operators in this thread. The bounded and unbounded operators indicates a precise properties of operators in a Banach space. However if you equate the bounded operators with finite dimensional matrix representation, then for a well known counter example of your statement, recall the definition of "Density matrix" operator in a Hilbert space, i.e. " a Hermitian matrix (possibly infinite dimensional) with trace 1".Therefore not only trace calculation of infinite dimensional matrices is possible but also there are bounded operators with infinite dimensional matrix representation.
Dear Manouchehr Amiri,as you say ,but Trace[X,P]=ihbar*Trace(I)≠Trace (XP) - Trace (PX) = 0,can you give me a advance?
@ Yonghong,If you follow my first post and work out the matrix form of a+a and aa+ you will arrive to this result that in spite of the Tr( a a+) = Tr( a+a) we have the equation [ a , a+] = 1 .
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