# Meaning of operator in ket bra notation in Hilbert space

For simplest operator $$\textit{M}$$, I could write it as $$|k\rangle\langle m|$$. $$$$\mathit{M} = |k\rangle\langle m|$$$$ When operating on a state $$|n\rangle$$, I could write as: $$$$\mathit{M}|n\rangle=|k\rangle\langle m|n\rangle=\langle m|n\rangle|k\rangle$$$$ From that, I could interpret the operator M as transforming the state $$|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:

$$$$\mathit{I}|n\rangle = \sum_{j=1}^{N} |j\rangle \langle j|n\rangle = \sum_{j=1}^{N} \langle j|n\rangle |j\rangle$$$$

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.

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 $$|m\rangle$$.

• Your answer solved my confusion. Btw is it a typo that the length of vector k should be <m,n>? Dec 7 '20 at 6:34
• How exactly is the operator introduced by OP, $M=\vert k\rangle \langle m \vert$, a projection operator? $M^2\neq M$ unless $m=k$. :/ Dec 7 '20 at 7:04
• 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. Dec 7 '20 at 7:10
• 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. Dec 7 '20 at 7:15
• 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. Dec 7 '20 at 7:18
• $$\langle j \vert n \rangle \vert j \rangle$$ 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.

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