Quantum Computing - projection operators

[Peter_Newman]

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]

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.

 

[PeterDonis]

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)?

 

[Peter_Newman]

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?).

 

[PeterDonis]

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?

 

[PeterDonis]

$P_2P_1$ 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 $P_1 P_2 = 0$.

 

[PeterDonis]

Since these are orthogonal

The problem statement does not require that $P_1$ and $P_2$ are orthogonal.

 

[Peter_Newman]

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 $P_1 P_2 = P_2 P_1$

 

[PeterDonis]

can not quite see how far this property can help me

What does $P^2 = P$ look like if $P = P_1 P_2$?

 

[Peter_Newman]

If $P=P_1P_2$ so $P^2=(P_1P_2)(P_1P_2)=P_1^2P_2^2$

 

[PeterDonis]

If $P=P_1P_2$ so $P^2=(P_1P_2)(P_1P_2)$

Right.

$(P_1P_2)(P_1P_2)=P_1^2P_2^2$

How are you obtaining this? The ordering of the factors on the LHS is different than on the RHS, since $P_1^2 = (P_1 P_1)$ and $P_2^2 = (P_2 P_2)$.

 

[Peter_Newman]

$P^2=P_1P_2P_1P_2=P_1^2P_2^2=P_1P_2=P$

 

[PeterDonis]

$P^2=P_1P_2P_1P_2=P_1^2P_2^2=P_1P_2=P$

Ok, so does this answer your question in the OP?

 

[Peter_Newman]

Ok, so does this answer your question in the OP?

Yes, almost. I just do not quite know how to express that linguistically ...

 

[PeterDonis]

I just do not quite know how to express that linguistically

Why do you need to? The equations are clear and straightforward.

 

[Peter_Newman]

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!

 

[PeterDonis]

thanks for the support!

You're welcome!