I Quantum Computing - projection operators

Assume ##P_1## and ##P_2## are two projection operators. I want to show that if their commutator ##[P_1,P_2]=0##, then their product ##P_1P_2## is also a projection operator.

My first idea was:

$$P_1=|u_1\rangle\langle u_1|, P_2=|u_2\rangle\langle u_2|$$
$$P_1P_2= |u_1\rangle\langle u_1|u_2\rangle\langle u_2|\neq 0$$
the second expression is not zero if ##\langle u_1|u_2\rangle## are not orthononal.

But I do not really get on with this task, which is why I hope for some advice.
 

hilbert2

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An expression like ##P_1 = |u_1\rangle\langle u_1 |## is for a projection operator that projects any vector on a 1-dimensional line in the state vector space. A more general projection is

##P = \sum\limits_{k} |u_k \rangle\langle u_k |##

where the vectors ##|u_k \rangle## form an orthonormal set.
 
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the second expression is not zero
You're not trying to show that the second expression is zero. You're trying to show that ##P_1 P_2## is a projection operator if ##P_1## and ##P_2## commute.

Do you know what defines a projection operator? That is, if ##P## is a projection operator, can you write down a statement that must be true of it (without writing it out in terms of bras and kets)?
 
My knowledge of projection operators is similar to what @hilbert2 summarized in his post. I somehow lack the relation to the commutation operator. It would be very nice if you could help me a bit.

So the commutator operator comes to mind. At least I have snapped that at some point:

$$[A,B]=AB-BA$$

So suppose I have the two projection operators ##P_1## and ##P_2##. Since these are orthogonal, the product of these two is 0, which I wanted to say in my first post. But now the connection to what is missing in the task of my input posts (##P_2P_1## would also have to be 0 and therefore can not be a projection operator?).
 
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My knowledge of projection operators is similar to what @hilbert2 summarized in his post.
Did you know that a projection operator ##P## has the property that ##P^2 = P##? In other words, applying it twice is the same thing as applying it once. (This should make sense if you think about what a projection operator means.)

Can you see how this fact might be helpful?
 
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##P_2P_1## would also have to be 0
No, that's not true. ##[P_1, P_2] = P_1 P_2 - P_2 P_1##. So ##[P_1, P_2] = 0## just implies ##P_1 P_2 = P_2 P_1##. It does not imply that ##P_1 P_2 = 0##.
 
No, that's not true. ##[P_1, P_2] = P_1 P_2 - P_2 P_1##. So ##[P_1, P_2] = 0## just implies ##P_1 P_2 = P_2 P_1##. It does not imply that ##P_1 P_2 = 0##.
Ok, that would have been my alternative hypothesis.

Did you know that a projection operator ##P## has the property that ##P^2 = P##? In other words, applying it twice is the same thing as applying it once. (This should make sense if you think about what a projection operator means.)

Can you see how this fact might be helpful?
Yes, I have heard of this property before. Unfortunately, I can not quite see how far this property can help me with ##P_1 P_2 = P_2 P_1##
 
If ##P=P_1P_2## so ##P^2=(P_1P_2)(P_1P_2)=P_1^2P_2^2##
 
##P^2=P_1P_2P_1P_2=P_1^2P_2^2=P_1P_2=P##
 
Yes, I agree with that a bit, if I see ##P^2=P_1P_2P_1P_2=P_1^2P_2^2=P_1P_2=P##
Here is the above mentioned product already in it ...

@PeterDonis by the way, thanks for the support!
 

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