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I am confused about the difference between the terms "representation" and "basis" of a state or operator.

For example,

Let us have eigen-kets of Hamiltonian $H$ denoted by $|\phi_n\rangle $. These eigen-kets make a complete set of basis, $\sum_n |\phi_n\rangle\langle \phi_n|=1$. Using this identity, we can write any state of system $|\psi\rangle $ as $$ |\psi\rangle = \sum_n (\langle \phi_n|\psi\rangle )|\phi_n\rangle \tag{1} $$

Similarly, we have eigen-kets of position operator $r$ denoted by $|r\rangle$. They also form a complete set of basis, $\int dr |r\rangle\langle r|=1$. Next, we can write the above equation $(1)$ in terms of wavefunction by applying $\langle r|$ and using completeness identity $\int dr |r\rangle\langle r|=1$ $$ \langle \mathbf{r}|\psi\rangle = \sum_n \left(\int d\mathbf{r}'\langle \phi_n|\mathbf{r}'\rangle \langle \mathbf{r}'|\psi\rangle \right)\langle \mathbf{r}|\phi_n\rangle $$ $$ \psi(\mathbf{r}) = \sum_n \left(\int d\mathbf{r}' \phi_n^*(\mathbf{r}') \psi(\mathbf{r}') \right) \phi_n(\mathbf{r})\tag{2} $$

Questions:

  1. In equation $(1)$, by writing the state-ket $|\psi\rangle$ in terms of eigen-kets of $H$, do we say that we have written state-ket $|\psi\rangle$ in eigen-kets representation or eigen-kets basis?
  2. In equation $(2)$, we have used eigen-kets of both position and Hamiltonian operator. So, do we have state-function written in position representation/basis or Hamiltonian eigenfunctions representation/basis?
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4 Answers 4

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  1. In equation $(1)$, by writing the state-ket $|\psi\rangle$ in terms of eigen-kets of $H$, do we say that we have written state-ket $|\psi\rangle$ in eigen-kets representation or eigen-kets basis?

In this context the word "representation" is used synonymously with "basis". However, in my experience it is more common to use the word "basis" for finite or countable dimensional vectors (i.e. the "spin z basis" in the context of spin), whereas in the infinite, uncountable dimensional case both are common (e.g. "momentum basis" or "momentum representation").

  1. In equation $(2)$, we have used eigen-kets of both position and Hamiltonian operator. So, do we have state-function written in position representation/basis or Hamiltonian eigenfunctions representation/basis?

The fact that your wave function is in the end a function of $\vec{r}$ (namely $\psi(\vec{r})$) means it is in the position basis. In order to get one of the "components" of $\psi$, we need to plug in a position and that's it. The fact that in this case you have expanded the position-space wave function in terms of energy eigenstates is here just a confusing factor.

However, there are times when it is not only possible, but necessary, to write $\psi$ in terms of a simultaneous basis of at least two operators at once. Why? Sometimes, specifying just one eigenvalue (e.g. $x$ in your case) is sufficient to get the wave function of the system. But take for example three dimensions - then we need three coordinates, $x, y, z$ to specify which component of $\psi$ we want. $\psi$ is then simultaneously diagonalized in all three coordinates. Each coordinate is its own operator. Not only that, but if we want to model a particle with spin, we would need to specify the component of spin (for example, z-spin $S_z$ in the up direction) to get a specific complex number from $\psi$. This topic is called "simultaneous diagonalization" and it can only be done among operators that all commute:

$$0 = [x,y] = [x,z] = [y,z] = [x,S_z] = [y, S_z] = [z, S_z]$$

What this means is that there exist solutions $\psi$ to the eigenvalue problem for which the same $\psi$ is an eigenvector to all of these operators at once.

Simultaneous diagonalization is a fundamental topic in Quantum Mechanics which you will surely come across if not already.

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  • $\begingroup$ Thank you for the answer. About my second question, are you saying that we could write the wavefunction in position-representation and expand it in energy eigenfunctions because position and Hamiltonian operator commute with each other? $\endgroup$
    – Sana Ullah
    Commented Nov 18, 2022 at 15:37
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    $\begingroup$ That's a good question, I thought about adding it to the answer but it was getting long. No, actually you can't expand it in both! If you could, then there would be a full basis of position eigenstates which are each also energy eigenstates. But the only position eigenstates are dirac delta functions, which are not energy eigenstates. You can alternatively see this through the commutation relation: $[x,H]=1/2m [x,p^2] + [x, V(x)] = (1/2m) 2i\hbar p \neq 0$. The final step I calculated using the identity $[A,BC] = B[A,C] + [A, B] C$ and from $x$ commuting with any function $V(x)$. $\endgroup$ Commented Nov 18, 2022 at 16:00
  • $\begingroup$ Is representation not used in the abstract algebra sense here? Just curious. $\endgroup$ Commented Nov 18, 2022 at 16:43
  • $\begingroup$ @SillyGoose No, I don't think so. If the group operation is chosen to be $\psi(x_1) \oplus \psi(x_2) = \psi(x_1) + \psi(x_2)$ then there is no reason it should be closed under addition. If the group operation is chosen to be $\psi(x_1) \oplus \psi(x_2) = \psi(x_1 + x_2)$ we are closer, but nonetheless we have problems because of the domain, consider for example a spin 1/2 particle: $\psi(x_1, s=1/2) + \psi(x_2, s=1/2)$ would give us a state of spin 1 when the vector space only includes spin $\pm 1/2$. $\endgroup$ Commented Nov 18, 2022 at 18:01
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The term representation is in my experience often used when working with the expansion coefficients, and not with the full state composed of coefficients times basis, i.e working with $\psi(r)$ vs $\int dr |r\rangle \langle r|\psi\rangle$. As long as your equations contain explicit mention of the basis, I would say you work within a basis. If your equations only contain the coefficients with respect to that basis, I would say you work in the representation with respect to that basis.

(2) is not the same as (1), since $|\psi\rangle = \int dr |r\rangle \langle r|\psi\rangle = \int dr \ \psi(r)|r\rangle$. As such I would call $\psi(r)$ the position representation and you are working in a position representation in (2). You would work in the eigenstate representation if you worked with a set of discrete vectors with the expansion coefficients $c_n = \langle \phi_n |\psi\rangle $ as elements. In my experience people call that the eigenstate representation.

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  • $\begingroup$ very interesting. thank you so much for the answer. $\endgroup$
    – Sana Ullah
    Commented Nov 18, 2022 at 17:03
  • $\begingroup$ @SanaUllah This essentially tracks with the mathematical meaning of representation, in which we construct matrix representations of states and operators. The elements of the matrix representations of the states are essentially the expansion coefficients in some basis, etc. $\endgroup$
    – march
    Commented Nov 18, 2022 at 17:19
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  1. To my understanding, to be precise one would say you have written $| \Psi \rangle$ with respect to the eigenfunctions of the hamiltonian. In particular, this allows you to represent $|\Psi \rangle$ as a column vector (similar to how in Linear Algebra a matrix is said to represent a linear transformation). I think "represent" is used because kets, linear transformations, etc. exist as abstract mathematical objects WITHOUT mention of any basis. But, you can REPRESENT these objects with respect to some basis.
  2. To my understanding, you have now written $|\Psi\rangle$ with respect to the position basis. It may be helpful to think about how to transform from position space to momentum space, you use the Fourier transform. There is a reason that it is the Fourier transform!
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Basis is a collection of linearly independent vectors that span a vector space $V$; so the term cannot refer to the state $\left|\psi\right>$, which is a vector in $H$.

You can expand any vector in $v\in V$ w.r.t. some basis $(v_a)_{a\in I}$ for $V$, as in (1): the vector $\left|\psi\right>\in H$ is expanded w.r.t. $(\left|\phi_n\right>)_{n\in\mathbb{N}}$, which is a basis for $H$.

The collection of the coefficients $c_a=\left<v_a,v\right>$ that determine this expansion is a representation of $v$. If $I$ is finite, then $(c_a)_{a\in I}$ is a vector in $\mathbb{C}^{|I|}$, if it is infinite but countable, then $(c_a)_{a\in I}$ is a vector in $\ell^2$, and if $I$ is uncountable, then $(c_a)_{a\in I}$ is a vector in $L^2$.

So, in (1), the sequence $(\left<\phi_n|\psi\right>)_{n\in\mathbb{N}}$ is the "Hamiltonian representation" of $\left|\psi\right>\in H$, which is a vector in $\ell^2$. In (2) you have both a representation and an expansion:

  • the collection $(\psi(r))_{r\in\mathbb{R}}$, conventionally expressed as a map $r\mapsto\psi(r)$, is the "position representation" of $\left|\psi\right>\in H$, which is a vector in $L^2$.
  • similarly, for any $n\in\mathbb{N}$, the map $r\mapsto\phi_n(r)$ is the "position representation" of $\left|\phi_n\right>\in H$, which is also in $L^2$.
  • the collection of those maps, that is $(\phi_n)_{n\in\mathbb{N}}$, is a basis for $L^2$.
  • in the expansion (2), the sequence $\left(\int dr\phi_n^*(r)\psi(r)\right)_{n\in\mathbb{N}}$ is the "Hamiltonian representation" of $\psi\in L^2$, which is a sequence in $\ell^2$ and identical to $(\left<\phi_n|\psi\right>)_{n\in\mathbb{N}}$ (which was the "Hamiltonian representation" of $\left|\psi\right>\in H$).

You can do the same for any (hermitian) operator $A\colon H\to H$, starting with the collection $(A_u)_{u\in I^2}$, with $A_{(a,b)}=\left<\phi_a|A|\phi_b\right>$ for any $(a,b)\in I^2$. For finite $I$, the collection $(A_u)_{u\in I^2}$ is a matrix in $\mathbb{C}^{|I|\times |I|}$, and so on.

And some food for thought: in what kind of space belong the position and momentum eigenstates?

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