# Matrix rep of operator acting on bras

If A is a linear operator, and we have some ordered basis (but not necessarily orthonormal), then the element Aij of its matrix representation is just the ith component of A acting on the jth basis vector. We can also represent the action of A on a ket as the matrix product of A's matrix with the column matrix representing the ket.

We can also represent the action of A on a bra vector as matrix product of the row matrix of the bra with another matrix. If the basis was orthonormal, it would be the same matrix Aij as above. But if the basis isn't orthonormal, is it a different matrix?

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strangerep
We can also represent the action of A on a bra vector as matrix product of the row matrix of the bra with another matrix. If the basis was orthonormal, it would be the same matrix Aij as above.
No, I think you want ##A^\dagger## here. Depends on exactly what you're trying to do.
But if the basis isn't orthonormal, is it a different matrix?
Here's a bit more detail...

In the finite dimensional case -- where the kets are just column vectors -- the bras are row vectors. So one can pass from a ket to its corresponding bra by the ordinary Hermitian conjugate operation from linear algebra (i.e., conjugate transpose). E.g., if we write a ket ##v## (a column vector), then the corresponding bra is ##v^\dagger## (a row vector). The usual QM inner product in the Hilbert space of such column vectors is just ##w^\dagger v## (where ##w## is another column vector). So if ##A## is a matrix acting on ##v##, i.e., ##v' = Av##, then ##(v')^\dagger = v^\dagger A^\dagger## .

Also, $$w^\dagger A v ~=~ (A^\dagger w)^\dagger v$$ .

HTH.

No, I think you want ##A^\dagger## here. Depends on exactly what you're trying to do.
Maybe this'll be clearer: If $\langle \psi |$ is a bra vector and A a linear operator, the matrix representation of $\langle \psi |A$ in an orthonormal basis (basis vectors denoted by $|i\rangle$, with i acting as an index) will be the product of column vector with entries $\langle \psi |i\rangle$ with a square matrix with entries $\langle j|A|i\rangle$. This is the same matrix that comes up when we consider the matrix representation of A on a ket ie $A|\phi\rangle$.

Now, the same general thing happens when we consider the matrix representation of $\langle\psi |A$ and $A| \phi\rangle$ in some arbitrary, not necessarily orthonormal, basis -- just with different elements in the matrices. My question is whether or not the square matrix that appears in $\langle\psi |A$ is the same as the square matrix that appears in $A| \phi\rangle$.

The usual QM inner product in the Hilbert space of such column vectors is just ##w^\dagger v## (where ##w## is another column vector).
I don't think this is true. In general, there'll be a matrix between ##w^\dagger## and ##v##, which only becomes the identity matrix if the basis is orthonormal.

strangerep
Maybe this'll be clearer: If $\langle \psi |$ is a bra vector and A a linear operator, the matrix representation of $\langle \psi |A$ in an orthonormal basis (basis vectors denoted by $|i\rangle$, with i acting as an index) will be the product of column vector with entries $\langle \psi |i\rangle$ with a square matrix with entries $\langle j|A|i\rangle$. This is the same matrix that comes up when we consider the matrix representation of A on a ket ie $A|\phi\rangle$.
I think you're getting mixed up between abstract bra-ket notation, and its concrete representation using vectors and matrices in the finite dimensional case.

In general, there'll be a matrix between ##w^\dagger## and ##v##, which only becomes the identity matrix if the basis is orthonormal.
No. My notation is basis-independent.

I think you're getting mixed up between abstract bra-ket notation, and its concrete representation using vectors and matrices in the finite dimensional case.
What makes you say that?

No. My notation is basis-independent.
I don't think so. According to wikipedia, "[t]he general form of an inner product on $\mathbb{C}^n$ is given by:
$$\langle \mathbf{x}, \mathbf{y}\rangle := \mathbf{y}^*\mathbf{M}\mathbf{x}$$
where M is any Hermitian, positive definite matrix, and y* the conjugate transpose of y."

strangerep
According to wikipedia, "[t]he general form of an inner product on $\mathbb{C}^n$ is given by:
$$\langle \mathbf{x}, \mathbf{y}\rangle := \mathbf{y}^*\mathbf{M}\mathbf{x}$$
where M is any Hermitian, positive definite matrix, and y* the conjugate transpose of y."
There are many different inner products that can be defined on $\mathbb{C}^n$. This is distinct from the question of basis independence.

Sorry, I can't spare any more time to deconstruct and repair all these misunderstandings.
Maybe someone else can take over.

Look, if $|i\rangle$, i = 1 to N are basis vectors, then
$$\langle \phi|\psi\rangle = \sum_{i,j}\phi_i^* \psi_j \langle i|j \rangle$$.

If $\langle i|j \rangle =\delta_{ij}$, then

$$\langle \phi|\psi\rangle = \begin{pmatrix} \phi_1^* \ ... \ \phi_N^* \\ \end{pmatrix} \begin{pmatrix} \psi_1 \\ .\\ .\\ .\\ \psi_N \\ \end{pmatrix}$$.

But otherwise, there's going to be a Hermitian, positive definite matrix between the row and column vector.

The Hilbert space has a metric $g_{ij}$, such that $g_{ij}\psi^i \phi^j$ is the inner product. But it can just be handled by the actual inner product since linear functionals act on vectors, $\bar{\psi}^{i}\phi_{i}$. See this thread for more info Hilbert Space Metric

Hurkyl
Staff Emeritus
Gold Member
Writing [$] for the coordinate representation of$ relative to whatever basis you've chosen, then you always have

[wv] = [w][v]
[Av] = [A][v]
[wA] = [w][A]​

where A is an operator, v is a ket, and w is a bra.

The Hilbert space has a metric $g_{ij}$, such that $g_{ij}\psi^i \phi^j$ is the inner product. But it can just be handled by the actual inner product since linear functionals act on vectors, $\bar{\psi}^{i}\phi_{i}$. See this thread for more info Hilbert Space Metric
Unfortunately that thread was beyond my level, but I appreciate your help.

Writing [$] for the coordinate representation of$ relative to whatever basis you've chosen, then you always have

[wv] = [w][v]
[Av] = [A][v]
[wA] = [w][A]​

where A is an operator, v is a ket, and w is a bra.
I don't see how this can be true. As I said above,
$$\langle \phi|\psi\rangle = \sum_{i,j}\phi_i^* \psi_j \langle i|j \rangle$$
which is not equal to [w][v].

Hurkyl
Staff Emeritus
Gold Member
I don't see how this can be true. As I said above,
$$\langle \phi|\psi\rangle = \sum_{i,j}\phi_i^* \psi_j \langle i|j \rangle$$
which is not equal to [w][v].
Why do you think it's not equal?

What you're overlooking, presumably, is that $[v^\dagger] = [v]^*$ is not guaranteed. In fact, this is an identity if and only if your chosen basis is orthonormal.

Why do you think it's not equal?

What you're overlooking, presumably, is that $[v^\dagger] = [v]^*$ is not guaranteed. In fact, this is an identity if and only if your chosen basis is orthonormal.
But I think it can be proven that ##[v^\dagger]=[v]^*## in any basis.

Let ${|i\rangle}, \ i=1,...,N$ be our basis. $|v\rangle = \sum_i v_i|i\rangle$ and $(|v\rangle)^\dagger=\langle v|=\sum_i v_i^\dagger\langle i|$. Then
$$\langle v|w\rangle=\sum_i v_i^\dagger\langle i|w\rangle.$$
But
$$\langle v|w\rangle=\left(\langle w|v\rangle\right)^*=\left(\sum_i \langle w|i\rangle v_i\right)^*=\sum_i \langle w|i\rangle^* v_i^*=\sum_i v_i^*\langle i|w\rangle.$$
Therefore $v_i^\dagger =v_i^*$.

Hurkyl
Staff Emeritus
Gold Member
The coordinate representation of a bra is not relative to the basis $\langle i |$. I was in a hurry previously so I didn't have time to figure out what you were doing and that this was the point of confusion.

If $| i \rangle$ is a basis for the vector space of kets, then there is a dual basis $\langle \omega_i |$ for the vector space of bras. The dual basis is defined by the equation

$$\langle \omega_i | j \rangle = \delta_{ij}$$

The components of the row vector that is the coordinate representation of a bra are the coordinates relative to the dual basis $\langle \omega_i |$... not relative to the adjoint basis* $\langle i |$. The dual and adjoint bases are the same if and only if the original basis is orthonormal.

*: I do not know of a standard terminology for this notion

Of particular note is that the multiplication of a bra and a ket is an intrinsic property of vectors and dual vectors, and has absolutely nothing to do with a notion of inner product.

The inner product is a function of two kets. It is used to define a map that converts kets into bras. In the situation at hand, the place the inner product appears is not as the product $\langle i|$ with $| j \rangle$ -- instead, the place the inner product appears is as the definition of $\langle i|$.

Ok I understand now, thanks.

Final question: if the basis is not orthonormal, is ##[A^\dagger]=[A]^*## still true?

Hurkyl
Staff Emeritus