Meaning of operator in ket bra notation in Hilbert space

• $langle j vert n rangle vert j rangle$ is indeed the projection of $vert n rangle$ onto the vector $vert j rangle$ (assuming that $vert j rangle$ is normalized). The projection operator along $vert j rangle$ is $mathbb{P}_j=vert j ranglelangle j vert$ and you can see that $mathbb{P}_jvert nrangle$ is indeed $langle j vert n rangle vert j rangle$. You can verify that $mathbb{P}_j=vert j ranglelangle j vert$ is a projection operator by observing that $mathbb{P}^2_j=vert j ranglelangle j vert j ranglelangle j vert = vert j ranglelangle j vert = mathbb{P}_j$. That it is a projection operator along $vert j rangle$ is verified by observing that $mathbb{P}_jvert j rangle = vert j ranglelangle j vert jrangle = vert j rangle$.

• There are many other intuitive verifications you can perform and see that this is a sensible definition of projection. For example, you can see that if $vert jrangle$ and $vert irangle $ are orthogonal then $mathbb{P}_jvert irangle$ vanishes. You can also see that $sum_jmathbb{P}_j=mathbb{I}$ where ${j}$ form a complete orthonormal basis. Proof: Consider that $m,n$ label the same orthonormal basis that is labeled by ${j}$. Then, we can write begin{align*}langle m vert sum_jmathbb{P}_jvert nrangle=sum_jlangle mvert j ranglelangle j vert nrangle= sum_j delta_{mj}delta_{jn}=delta_{mn}implies sum_jmathbb{P}_j=mathbb{I}end{align*}

• So, to answer your question: As I said, $langle j vert n rangle vert j rangle$ is indeed the projection of $vert n rangle$ onto the vector $vert j rangle$ (assuming that $vert j rangle$ is normalized). More geometrically, $langle j vert n rangle vert j rangle$ is the vector in direction of $vert j rangle$ whose magnitude is equal to the inner product of $vert j rangle$ with $vert n rangle$.

Finally, I should mention that the operator $M$ you wrote is not a Hermitian operator and thus, it wouldn't be the kind of operator that shows up often in quantum mechanics, or at the least, it wouldn't correspond to an observable.

Little Base:

Suppose ${|phi_irangle}$ form a basis in LVS $mathcal{V}$. Then any vector $|psirangle$ can be written as a linear combination of basis. $$|psirangle=sum_ilangle phi_i|psi_irangle|phi_irangle=sum_ic_i|phi_irangle$$

The combination ${|phi_iranglelanglephi_j|} $ form a basis for operators in LVS $mathcal{V}$. Thus any operator can be written as $$Omega=sum_isum_j|phi_irangleOmega_{ij}langlephi_j|$$ where $Omega_{ij}=langlephi_i|Omega|phi_jrangle$ are matrix element of the operator.

Now let's see How this operator acts on vector in this space. $$Omega|psirangle=sum_{i,j,k}|phi_irangleOmega_{ij}c_klanglephi_j|phi_krangle$$ Using the orthonormality condition for the basis set $$langlephi_i|phi_jrangle=delta_{ij}$$ $$Omega|psirangle=sum_{i,j,k}|phi_irangleOmega_{ij}c_kdelta_{jk}=sum_{i,j}|phi_irangleOmega_{ij}c_j$$

Now we know from our linear algebra course that matrix multiplication is defined as $$C=AB$$ $$c_{ij}=sum_ka_{ik}b_{kj}$$

That looks like what we have found (just use the fact that $B$ has only one column). There is nothing new going on it's just simple matrix multiplication. It's just a generalization to all the $LVS$.

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A Very good visualization for matrix multiplication and transformation can be found here.

Essence of Algebra By 3Blue1Brown