0
$\begingroup$

I'm new to quantum mechanics, and I am beginning to study Dirac notation, but I do not understand the significance or meaning of the following equation:

$$\sum_n\left|e_n\right\rangle\left\langle e_n\right|=1$$

or for that matter, this equation as well:

$$\int\left|e_z\right\rangle\left\langle e_z\right| d z=1$$

But I do understand that:

$$\hat{P} \equiv \left|\alpha\right\rangle\left\langle \alpha\right|$$

is an operator that can find the projection of a vector, say $|\beta\rangle$, along the direction of $|\alpha\rangle$

Apologies if I similar question has been asked before, but I was not able to find one that directly answered my question.

Improve this question
$\endgroup$
10
  • 3
    $\begingroup$ What do you want to know exactly? Do you know what orthonormal bases are and that you can expand every vector in such a basis? If yes, just apply the LHS to a vector and see what comes out, then compare it with the identity operator applied to the same vector. If not, read about ONBs. $\endgroup$
    – Tobias Fünke
    Commented Mar 17 at 18:13
  • $\begingroup$ Yes I understand that, I don't understand the first equation I listed at all really, I read it as the sum of the inner products of every basis vector with itself in a vector space as being equal to one, but it makes no sense to me since an orthonormal vector has has a value of one when the inner product with itself is taken $\endgroup$
    – cookiecainsy
    Commented Mar 17 at 18:17
  • $\begingroup$ for example if a vector space had 3 basis vectors, all of which were orthonormal, wouldn't the first equation be equal to 3? $\endgroup$
    – cookiecainsy
    Commented Mar 17 at 18:19
  • $\begingroup$ I don't understand. Why don't you try what I've suggested in my comment? Note also that on the LHS there is no inner product and on the RHS the $1$ is meant to be the identity operator, not a number (which will become also clear if you do this simple exercise). $\endgroup$
    – Tobias Fünke
    Commented Mar 17 at 18:19
  • 1
    $\begingroup$ Why did you trace your 3x3 identity matrix? This is linear algebra, not QM… $\endgroup$
    – Cosmas Zachos
    Commented Mar 17 at 18:22

1 Answer 1

Reset to default
5
$\begingroup$

The equation $$\sum_n|e_n\rangle\langle e_n|=1$$ is an operator equation. That means $$\sum_n|e_n\rangle\langle e_n|\psi\rangle=|\psi\rangle$$ is true for every vector $|\psi\rangle$.

As you know, in the sum on the left side every term $|e_n\rangle\langle e_n|\psi\rangle$ is the projection of $|\psi\rangle$ along the direction of $|e_n\rangle$. So the equation says that the sum of all these projections completely recovers the vector $|\psi\rangle$ (hence the name "completeness relation").

Or in other words: There is no part of $|\psi\rangle$ missing, because the vectors $|e_n\rangle$ cover all possible directions of the Hilbert space.

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.