# Operators in finite dimension Hilbert space

• I
• Sebas4
In summary, the matrix representation of an operator in a finite dimensional Hilbert space is dependent on the basis chosen, but the operator itself is independent of any specific basis. This means that the function of the operator remains the same regardless of the basis, but the numerical values of the matrix elements will vary depending on the chosen basis. This concept is similar to how a vector in a vector space can be represented in different bases, but the vector itself remains the same object.f

#### Sebas4

TL;DR Summary
What does the following sentence mean: the matrix representation of operators are basis dependent but the operator itself is bases independent.
I have a question about operators in finite dimension Hilbert space.

I will describe the context before asking the question.
Assume we have two quantum states $| \Psi_{1} \rangle$ and $| \Psi_{2} \rangle$.
Both of the quantum states are elements of the Hilbert space, thus $| \Psi_{1} \rangle$, $| \Psi_{2} \rangle$ $\in$ $H$.
Both quantum states are described in different basis. We can write:
$$|\Psi_{1}\rangle = \sum_{i = 1}^{n}a_{i}| \psi_{i}\rangle$$ and $$|\Psi_{2}\rangle = \sum_{i = 1}^{n}b_{i}| \psi_{i}\rangle$$.
The coëfficients, $a_{i}$ or $b_{i}$ are basis dependent.

Next, the dictation tells us that the matrix representation of operators depend on bases but that the operator itself is bases independent.

My question what does this mean, the matrix representation of operators are bases dependent but the operator itself is bases independent.

You can write the Hamiltonian ##\hat{H}## or some other operator ##\hat{O}## in finite-dimensional space as a matrix with elements ##\langle\psi_i \left.\right|\hat{H}\left|\right.\psi_j \rangle##. These have two indices, ##i## and ##j##, and depend on how you choose the orthonormal set of basis vectors ##\left\{\left|\right.\psi_k \rangle\right\}##. The operator is the same no matter what basis set you use or whether you even refer to any particular basis at all.

It's not much different from the situation how, say, the real number with decimal representation 7.5 is the same mathematical object even if you convert to binary, hexadecimal or some other base. It's just written in different way.

• vanhees71 and topsquark
My question what does this mean, the matrix representation of operators are bases dependent but the operator itself is bases independent.
This is similar to the case for vectors. A vector in a vector space is a well-defined object and can be represented in any basis. The coefficients associated with this representation are different for each basis.

• vanhees71 and topsquark
So, the elements of $\widehat{H}$ are given by $\langle \psi_{a} | \widehat{H} | \psi_{b} \rangle$ with $| \psi_{a} \rangle$ and $| \psi_{b} \rangle$ are two basis vectors of the set (containing basis vectors) $\left\{ | \psi_{k} \rangle \right\}$. $\langle \psi_{a} | \widehat{H} | \psi_{b} \rangle$ is the element of the operator $\widehat{H}$ in the basis set $\left\{ | \psi_{k} \rangle \right\}$. But what is the meaning of itself? The operator itself is basis independent. Is the meaning that the operator $\widehat{H}$ has the same function but the matrix elements of the operator depends on the basis?

So, the elements of $\widehat{H}$ are given by $\langle \psi_{a} | \widehat{H} | \psi_{b} \rangle$ with $| \psi_{a} \rangle$ and $| \psi_{b} \rangle$ are two basis vectors of the set (containing basis vectors) $\left\{ | \psi_{k} \rangle \right\}$. $\langle \psi_{a} | \widehat{H} | \psi_{b} \rangle$ is the element of the operator $\widehat{H}$ in the basis set $\left\{ | \psi_{k} \rangle \right\}$. But what is the meaning of itself? The operator itself is basis independent. Is the meaning that the operator $\widehat{H}$ has the same function but the matrix elements of the operator depends on the basis?
Have you studied linear algebra?

Here's a simple example. In the usual "z-basis", the spin operators take the form:
$$S_z= \frac{\hbar}{2} \begin{bmatrix} 1& 0\\ 0&-1 \end{bmatrix}$$
$$S_x= \frac{\hbar}{2} \begin{bmatrix} 0& 1\\ 1&0 \end{bmatrix}$$
$$S_y= \frac{\hbar}{2} \begin{bmatrix} 0& i\\ -i&0 \end{bmatrix}$$
And, in the this case the basis vectors are the eigenstate of ##S_z## corresponding to spin-up and spin-down in the z-direction. But, if we change to the x-basis, where the basis vectors are the eigenstates of ##S_x##, then in this basis we have:
$$S_x= \frac{\hbar}{2} \begin{bmatrix} 1& 0\\ 0&-1 \end{bmatrix}$$

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• topsquark
Summary: What does the following sentence mean: the matrix representation of operators are basis dependent but the operator itself is bases independent.

I have a question about operators in finite dimension Hilbert space.

I will describe the context before asking the question.
Assume we have two quantum states $| \Psi_{1} \rangle$ and $| \Psi_{2} \rangle$.
Both of the quantum states are elements of the Hilbert space, thus $| \Psi_{1} \rangle$, $| \Psi_{2} \rangle$ $\in$ $H$.
Both quantum states are described in different basis. We can write:
$$|\Psi_{1}\rangle = \sum_{i = 1}^{n}a_{i}| \psi_{i}\rangle$$ and $$|\Psi_{2}\rangle = \sum_{i = 1}^{n}b_{i}| \psi_{i}\rangle$$.
The coëfficients, $a_{i}$ or $b_{i}$ are basis dependent.

Next, the dictation tells us that the matrix representation of operators depend on bases but that the operator itself is bases independent.

My question what does this mean, the matrix representation of operators are bases dependent but the operator itself is bases independent.
Have looked at a definition of an operator? Presumably yes, what is the definition in your textbook?

• topsquark
So, the elements of $\widehat{H}$ are given by $\langle \psi_{a} | \widehat{H} | \psi_{b} \rangle$ with $| \psi_{a} \rangle$ and $| \psi_{b} \rangle$ are two basis vectors of the set (containing basis vectors) $\left\{ | \psi_{k} \rangle \right\}$. $\langle \psi_{a} | \widehat{H} | \psi_{b} \rangle$ is the element of the operator $\widehat{H}$ in the basis set $\left\{ | \psi_{k} \rangle \right\}$. But what is the meaning of itself? The operator itself is basis independent. Is the meaning that the operator $\widehat{H}$ has the same function but the matrix elements of the operator depends on the basis?
It's much the same as how a point in space is the same point no matter how you define your ##xyz## coordinate system to label it with coordinates.