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For simplest operator $\textit{M}$, I could write it as $|k\rangle\langle m|$. \begin{equation} \mathit{M} = |k\rangle\langle m| \end{equation} When operating on a state $|n\rangle$, I could write as: \begin{equation} \mathit{M}|n\rangle=|k\rangle\langle m|n\rangle=\langle m|n\rangle|k\rangle \end{equation} From that, I could interpret the operator M as transforming the state $|n\rangle$ to $|k\rangle$ with $\langle m|n\rangle$ as coefficient. However, how does $|n\rangle$ relate to $|k\rangle$ by the inner product $\langle m|n\rangle$ in Hilbert space? For example, the identity operator I is just:

\begin{equation} \mathit{I}|n\rangle = \sum_{j=1}^{N} |j\rangle \langle j|n\rangle = \sum_{j=1}^{N} \langle j|n\rangle |j\rangle \end{equation}

It is just projection of $|n\rangle$ on the basis $|j\rangle$. For $\langle m|n\rangle|k\rangle$, I have no idea how they are related in Hilbert space.

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Your simple operator is a kind of projection onto the one-dimensional space spanned by $|k\rangle$.
The inner product $\langle m,n\rangle$ is the cosine of the angle between m and n scaled by their lengths. If they are all unit vectors, it is the cosine of the angle between them. And if k is a unit vector, this cosine gives the length of $M |n\rangle$. That is the relation you have asked for. It is not exactly the usual projection, since you have introduced this angle with $|m\rangle$.

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  • $\begingroup$ Your answer solved my confusion. Btw is it a typo that the length of vector k should be <m,n>? $\endgroup$
    – Simon219
    Dec 7 '20 at 6:34
  • $\begingroup$ How exactly is the operator introduced by OP, $M=\vert k\rangle \langle m \vert$, a projection operator? $M^2\neq M$ unless $m=k$. :/ $\endgroup$
    – Dvij D.C.
    Dec 7 '20 at 7:04
  • $\begingroup$ I did not mean to say the operator was a projection operator, just that M of a vector yielded a kind of projection onto a one-dimensional subspace. @Simon219 there is an art to being vague to a very precise degree: "gives" is actually a slightly vague word. You can easily derive the precise formula yourself. All I wanted to do was help clear up your confusion by providing "the big picture". And you already clearly know how to write a projection operator's formula. $\endgroup$ Dec 7 '20 at 7:10
  • $\begingroup$ Your "projection" of a vector onto the subspace spanned by $\vert k \rangle$ would be non-zero even if the vector is orthogonal to $\vert k\rangle$. There is no earthly way in which this can be called a projection. $\endgroup$
    – Dvij D.C.
    Dec 7 '20 at 7:15
  • $\begingroup$ Your point is well taken, but in a curved universe, where I am living, it is hard to see straight, and one might think of this a kind of projection. From certain skew angles. $\endgroup$ Dec 7 '20 at 7:18
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  • $\langle j \vert n \rangle \vert j \rangle$ 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 $\mathbb{P}_j=\vert j \rangle\langle j \vert$ and you can see that $\mathbb{P}_j\vert n\rangle$ is indeed $\langle j \vert n \rangle \vert j \rangle$. You can verify that $\mathbb{P}_j=\vert j \rangle\langle j \vert$ is a projection operator by observing that $\mathbb{P}^2_j=\vert j \rangle\langle j \vert j \rangle\langle j \vert = \vert j \rangle\langle j \vert = \mathbb{P}_j$. That it is a projection operator along $\vert j \rangle$ is verified by observing that $\mathbb{P}_j\vert j \rangle = \vert j \rangle\langle j \vert j\rangle = \vert j \rangle$.

  • 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 $\vert j\rangle$ and $\vert i\rangle $ are orthogonal then $\mathbb{P}_j\vert i\rangle$ vanishes. You can also see that $\sum_j\mathbb{P}_j=\mathbb{I}$ 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_j\mathbb{P}_j\vert n\rangle=\sum_j\langle m\vert j \rangle\langle j \vert n\rangle= \sum_j \delta_{mj}\delta_{jn}=\delta_{mn}\implies \sum_j\mathbb{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.

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Little Base:

Suppose $\{|\phi_i\rangle\}$ form a basis in LVS $\mathcal{V}$. Then any vector $|\psi\rangle$ can be written as a linear combination of basis. $$|\psi\rangle=\sum_i\langle \phi_i|\psi_i\rangle|\phi_i\rangle=\sum_ic_i|\phi_i\rangle$$

The combination $\{|\phi_i\rangle\langle\phi_j|\} $ form a basis for operators in LVS $\mathcal{V}$. Thus any operator can be written as $$\Omega=\sum_i\sum_j|\phi_i\rangle\Omega_{ij}\langle\phi_j|$$ where $\Omega_{ij}=\langle\phi_i|\Omega|\phi_j\rangle$ are matrix element of the operator.

Now let's see How this operator acts on vector in this space. $$\Omega|\psi\rangle=\sum_{i,j,k}|\phi_i\rangle\Omega_{ij}c_k\langle\phi_j|\phi_k\rangle$$ Using the orthonormality condition for the basis set $$\langle\phi_i|\phi_j\rangle=\delta_{ij}$$ $$\Omega|\psi\rangle=\sum_{i,j,k}|\phi_i\rangle\Omega_{ij}c_k\delta_{jk}=\sum_{i,j}|\phi_i\rangle\Omega_{ij}c_j$$

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 $LVS$.


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

Essence of Algebra By 3Blue1Brown

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