Question on Dirac notation with operator [closed]

Question

What does $\langle\psi|A|\phi\rangle$ mean if $A$ is some operator like how does $A$ acts on these two vectors $\phi$ and $\psi$ and what is it equal to and also does $A$ act on both vectors or just one vector?

Answer 1

Both $|\phi\rangle$ and $\langle\psi|$ are vectors in a vector space V and its dual, which we usually denote as $V^*$. $A$ is an operator that acts on vectros, belonging to that vector space and $A^{\dagger}$ is the adjoint (i.e. transpose and with the elements starred/complex conjugate) of that operator, acting on the dual space. Having said that, $A$ can act on only one of the vectors $|\psi\rangle$ and $|\phi\rangle$ and one either acts with $A$ on the ket $|\phi\rangle$ (element of the vector space), or one can act with $A^{\dagger}$ on the ket $\langle\psi|$ (element of the dual vector space).

If the vector space (and its dual) are finite dimensional, $A$ will have eigenvectors and eigenvalues. If the eigenvalues are all distinct, then the eigenvectors span the vector space. I will not go into the technical details of why this is true, unless you comment saying so. Despite that, the practical use of that is that it allows us to write every random vector as a linear combination of vectors that are eigenvectors of the operator $A$. Let those eigenvectors be denoted by $|\alpha_i\rangle$. Then, $$A|\alpha_i\rangle=\alpha_i|\alpha_i\rangle$$ where $\alpha_i$ is the corresponding eigenvalue. So, this is how the perator acts on arbitrary vectors. First you write them as a linear combination of the operator's eigenvectors and then you let the operator act on the eigenvectors returning $\alpha_i$.

So, to apply all these things to your particular example, we do what we discussed earlier (i.e. write the two vectors as linear combinations of the eigenvectors) $$|\psi\rangle=\sum_i\beta_i|\alpha_i\rangle$$ $$|\phi\rangle=\sum_j\gamma_j|\alpha_j\rangle$$ Then, we let $A$ act on $\phi$ like this $$A|\phi\rangle=A\sum_j\gamma_j|\alpha_j\rangle=\sum_j\gamma_jA|\alpha_j\rangle =\sum_j\gamma_j\alpha_j|\alpha_j\rangle$$ and then dot it (i.e. perform the inner product) to $\langle\psi|$ like $$\langle\psi|A|\phi\rangle= \Big(\sum_i\beta_i^*\langle\alpha_i|\Big) \Big(\sum_j\gamma_j\alpha_i|\alpha_j\rangle\Big)= \sum_{i,j}\beta_i^*\gamma_j\alpha_j\langle\alpha_i|\alpha_j\rangle$$ where $\langle\alpha_i|\alpha_j\rangle=\delta_{i,j}$ provided that the basis is orthonormal, which we assume they are... You can use the Kronecker $\delta$ property and kill one of the sums, identifying in this way the indices $i$ with $j$ or vice versa and this would be the final form of your expression.

If there are any questions, please comment

Answer 2

I will assume that you already know what a ket and a bra are.

In the braket notation the usual Kronecker delta is defined as,

$$\langle i|j\rangle = \delta^{ij}$$

The sum of the outer products of a orthonormal unit basis gives the completeness relation given by,

$$|1\rangle\langle 1| + |2\rangle\langle 2| + |3\rangle\langle 3| = \mathbb{I}$$

where $\mathbb{I}$ means $3$x$3$ unit matrix, whose elements are the $\delta^{ij}$. Any set of orthonormal unit vector that satisfy the completeness relation form a basis, i.e, they are linearly independent and span the space.

Therefore, any vector in this space can be written as,

$$|A\rangle = a|1\rangle + b|2\rangle + c|3\rangle$$ and similarly,

$$\langle B| = \langle 1|e + \langle 2|f + \langle 3|g$$

Now considering the outer product $M \equiv|A\rangle\langle B|$. Its element standing in the ith row and jth column, $M^{ij}$, is the same as,

$$M^{ij} = \langle i|M|j\rangle$$

So, for instance, $M^{12} = \langle 1|A \rangle\langle B|2\rangle = af$

Other examples in quantum mechanics, are the position and momentum operators. In the configuration space,

$$\langle x|X|x^{\prime}\rangle = x\delta(x - x^{\prime})$$

$$\langle x|P|x^{\prime}\rangle = \delta(x - x^{\prime})\dfrac{d}{dx}$$