Questions about the projection operator

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The projection operator P = |a><a| satisfies the property P^2 = P, confirming that applying it twice yields the same result as applying it once. The eigenvalues of P are determined to be 0 and 1, derived from the eigenvalue equation P|psi> = p|psi>, where p represents the eigenvalue. Eigenvectors corresponding to the eigenvalue 0 are orthogonal to the subspace defined by |a>, while those corresponding to the eigenvalue 1 lie within that subspace. Understanding these properties is crucial for solving problems related to projection operators in quantum mechanics. Mastery of these concepts will aid in tackling similar quantum homework assignments effectively.
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A problem on my quantum homework assignment this week has to do with the projection operator P = |a><a|

I've been asked to show that P^2=P, and then give the eigenvalues of P and then to characterize its eigenvectors. The first part is easy enough:

P = |a><a|
so P^2 = |a><a||a><a| = |a><a|a><a| = |a>(1)<a| = |a><a| = P

It's the second part I'm having trouble with:

P|psi> = p|psi> where little p is the eigenvalue.

and coming the eigenvectors. Can anyone offer me a little help?
 
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Hint: Since P^2 v = P v, we have (P^2 - P) v =0.
 


The projection operator P is a fundamental concept in quantum mechanics, and it is important to understand it in order to solve problems like the one in your homework assignment. Let's break down the problem and go through each step.

First, you correctly showed that P^2 = P, which is a property of projection operators. This means that when you apply the projection operator twice, it gives the same result as applying it once. This is because the projection operator projects a vector onto a subspace, and when you project again, you are still within that subspace.

Next, you need to find the eigenvalues of P. To do this, we can use the definition of an eigenvalue: P|psi> = p|psi>. This means that when we apply the projection operator to a vector |psi>, we get back the same vector multiplied by some constant p. In other words, the vector |psi> is an eigenvector of P with eigenvalue p.

To find the eigenvalues, we can use the fact that P^2 = P and substitute it into the eigenvalue equation. This gives us P^2|psi> = P(p|psi>). Using the fact that P^2 = P, we can rewrite this as P|psi> = p^2|psi>. This shows that the eigenvalues of P are p^2, and since we know that P^2 = P, we can also say that the eigenvalues of P are 0 and 1.

Lastly, we need to characterize the eigenvectors of P. Again, we can use the eigenvalue equation P|psi> = p|psi> and substitute in the eigenvalues we found. This gives us two equations: P|psi> = 0|psi> and P|psi> = 1|psi>. These equations tell us that the eigenvectors of P are those that are projected to the zero vector (i.e. they are orthogonal to the subspace) and those that are projected to themselves (i.e. they are within the subspace).

I hope this helps you understand the projection operator and how to find its eigenvalues and eigenvectors. Remember to always start with the definition and use known properties to solve problems like this. Good luck with the rest of your quantum homework!
 
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