Is the Square of a Density Matrix Equal to the Density Matrix Itself?

[Mr confusion]

Homework Statement

to prove : square of density matrix= the density matrix itself (for a pure ensemble)

Homework Equations

density matrix=sum over P(i) ket(i) bra(i) where Pi = probability that random chosen system from ensemble shows state i.
summed over i , where P=1 for pure ensemble

The Attempt at a Solution

i thought it is giving the identity matrix, whose square is also I , hence the proof. but i am not sure about this.

 

[HallsofIvy]

No, the density matrix is NOT in general the identity matrix! There are many matrices with the property that A^2= A.

 

[Mr confusion]

then how will i prove it?

 

[Count Iblis]

then how will i prove it?

What are the P_i for a pure state?

 

[parton]

Ok, first we assume that the set \left \lbrace \left| i \right \rangle \right \rbrace forms a complete orthonormal base (otherwise if we have a general state \left| \psi \right \rangle we have to expand it into such a base). Further we just deal with the discrete case, the continuous one can be done similarly.

If \rho = \sum_{i} \left| i \right \rangle \left \langle i \right| is the density matrix, the square is simply given by:

\rho^{2} = \sum_{i,j} \left| i \right \rangle \langle i \vert j \rangle \left \langle j \right|

Now we use \left \langle i \vert j \right \rangle = \delta_{ij} and arrive at:

\rho^{2} = \sum_{i} \left| i \right \rangle \left \langle i \right| = \rho.

 

[Mr confusion]

What are the P_i for a pure state?

Pi for pure ensemble=1 since all are in same state, isn't it?

 

[Mr confusion]

Ok, first we assume that the set \left \lbrace \left| i \right \rangle \right \rbrace forms a complete orthonormal base (otherwise if we have a general state \left| \psi \right \rangle we have to expand it into such a base). Further we just deal with the discrete case, the continuous one can be done similarly.

If \rho = \sum_{i} \left| i \right \rangle \left \langle i \right| is the density matrix, the square is simply given by:

\rho^{2} = \sum_{i,j} \left| i \right \rangle \langle i \vert j \rangle \left \langle j \right|

Now we use \left \langle i \vert j \right \rangle = \delta_{ij} and arrive at:

\rho^{2} = \sum_{i} \left| i \right \rangle \left \langle i \right| = \rho.

i can just say -'COOL.' MANY thanks.
...sorry, just one more thing
how can i just keep them side by side, i mean\rho is a density matrix and to use it as a matrix i need to find its elements in some basis. how can i simply keep the abstract notation?

 

[Count Iblis]

Ok, first we assume that the set \left \lbrace \left| i \right \rangle \right \rbrace forms a complete orthonormal base (otherwise if we have a general state \left| \psi \right \rangle we have to expand it into such a base). Further we just deal with the discrete case, the continuous one can be done similarly.

If \rho = \sum_{i} \left| i \right \rangle \left \langle i \right| is the density matrix, the square is simply given by:

\rho^{2} = \sum_{i,j} \left| i \right \rangle \langle i \vert j \rangle \left \langle j \right|

Now we use \left \langle i \vert j \right \rangle = \delta_{ij} and arrive at:

\rho^{2} = \sum_{i} \left| i \right \rangle \left \langle i \right| = \rho.

This proof is wrong. What you've done is computing the square of the identity matrix which obviously is the identity matrix.

 

[Count Iblis]

Pi for pure ensemble=1 since all are in same state, isn't it?

No, for a pure state all the P_i are zero, except for one particular value of i (say i = r). So, the state is with certainty some particular state |r>.

 

[Mr confusion]

No, for a pure state all the P_i are zero, except for one particular value of i (say i = r). So, the state is with certainty some particular state |r>.

yes. i was wrong.
but does that mean there will be no summation?(for pure ensemble)
then how will i proceed?

 

[parton]

Sorry, the proof is of course wrong, such ugly things should not happen, very embarrassing...

In the pure case there is indeed no summation and we end up with one certain state \vert r \rangle.

If this state is normed, i.e. \langle r \vert r \rangle = 1 and your density matrix is \rho = \vert r \rangle \langle r \vert than you have the (trivial) relation:

\rho^{2} = \vert r \rangle \langle r \vert r \rangle \langle r \vert = \vert r \rangle \langle r \vert = \rho

 

[Mr confusion]

parton, to err is human.
i learned from your mistake. that will not alter my belief that u are a genius. thanks very much for helping me.
doing mistakes is a regular job for confused people like me.