Matrix rep of operator acting on bras

[dEdt]

If A is a linear operator, and we have some ordered basis (but not necessarily orthonormal), then the element A_ij 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 A_ij as above. But if the basis isn't orthonormal, is it a different matrix?

 

[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 A_ij 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.

 

[dEdt]

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.

 

[dEdt]

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.

 

[dEdt]

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.

 

[jfy4]

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]

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.

 

[dEdt]

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]

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.

 

[dEdt]

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]

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 $\langle i|$.

 

[dEdt]

Ok I understand now, thanks.

 

[dEdt]

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

 

[Hurkyl]

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

I highly doubt it.