# Density Operators of Pure States

• I
• Dario56
In summary, quantum states are most often described by the wavefunction ##\Psi## and can also be described by density matrices. For a pure state, the density matrix is defined by the outer product of the wavefunction. The variables ##x_1^{'}x_2^{'}\dots x_n^{'}## and ##x_1x_2\dots x_n## represent different elements of the density matrix, with the latter being the diagonal elements. The wavefunction is related to the Hilbert-space vector ##|\psi \rangle## by an inner product, and the matrix elements of the statistical operator for a pure state can be calculated using the wavefunction in the coordinate representation.
Dario56
Quantum states are most often described by the wavefunction ,##\Psi##. Variable ,##\Psi(x_1x_2\dots x_n) \Psi^*(x_1x_2\dots x_n)## defines probability density function of the system.

Quantum states can also be described by the density matrices (operators). For a pure state, density matrix is defined by the outer product of the system's wavefunction, ##\Psi##: $$\gamma = |\Psi\rangle \langle\Psi| \tag{1}$$

In the textbook: Density Functional Theory; Parr, Yao; Chapter 2.2 Density Operators, it is written that any element of this matrix is obtained as follows: $$\gamma(x_1^{'}x_2^{'}\dots x_n^{'} , x_1x_2\dots x_n) = \Psi(x_1^{'}x_2^{'}\dots x_n^{'})\Psi^*(x_1x_2\dots x_n) \tag{2}$$

where ##x_1^{'}x_2^{'}\dots x_n^{'}## denote that if we set ##x_i^{'} = x_i## for every ##i## than we get diagonal element of the density matrix. Namely, ##\Psi(x_1x_2\dots x_n) \Psi^*(x_1x_2\dots x_n)##

I have two questions:

1. What do variables such as ##x_1^{'}x_2^{'}\dots x_n^{'}## actually represent and what it is their difference compared to ## x_1x_2\dots x_n ##? In the textbook, no good explanation is given and I couldn't find any source which explains it better.

2. Since density matrix is defined as the outer product of the wavefunction ##\Psi##, what do variables ##x_1^{'}x_2^{'}\dots x_n^{'}## and ##x_1x_2\dots x_n## have to do with the entry values of the matrix? If we want to calculate the outer product, we need to represent wavefunction ##|\Psi\rangle## in a certain basis set with coefficients forming a column vector which we use to calculate the outer product. Therefore, entries of the density matrix have to do with the basis set we use to represent the wavefunction ##|\Psi\rangle## and not with the values of ##x_1^{'}x_2^{'}\dots x_n^{'}## and ##x_1x_2\dots x_n##.

In fact, quantum states are described by "density operators" (I prefer to call them "statistical operators"). In the case of a pure state the statistical operator is a projection operator ##\hat{\rho}=|\psi \rangle \langle \psi|##, where ##|\psi \rangle## is a normalized vector in Hilbert space. Equivalently pure states are represented by unit rays in Hilbert space, i.e., ##|\psi ' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle## represents the same state as ##|\psi \rangle## for all ##\varphi \in \mathbb{R}##.

The wave function is related to ##|\psi \rangle## by
$$\psi(x_1,\ldots,x_n)=\langle x_1,\ldots,x_n|\psi \rangle,$$
where ##|x_1,\ldots, x_n \rangle## are the common (generalized) eigenvectors of the position operators ##\hat{x}_j## with eigenvalues ##x_j##.

It is important to distinguish between the abstract Hilbert-space vectors ##|\psi \rangle## and the wave functions ##\langle x_1,\ldots,x_n|\psi \rangle=\psi(x_1,\ldots,x_n)##. It's analogous to the importance to distinguish between a vector ##\vec{V}## in Euclidean space and its components with respect to a Cartesian basis ##V_j=\vec{e}_j \cdot \vec{V}##.

The matrix elements of the corresponding statistical operator of the pure state indeed are
$$\rho(x_1',\ldots,x_n'; x_1,\ldots x_n) = \langle x_1',\ldots,x_n'|\psi \rangle \langle \psi|x_1,\ldots,x_n) = \psi(x_1',\ldots,x_n') \psi^*(x_1,\ldots,x_n).$$

aaroman, bhobba, hutchphd and 1 other person
Dario56 said:
1. What do variables such as ##x_1^{'}x_2^{'}\dots x_n^{'}## actually represent and what it is their difference compared to ## x_1x_2\dots x_n ##? In the textbook, no good explanation is given and I couldn't find any source which explains it better.
They are just different. For example, if you have a single variable function ##f##, you can create a multivariable function$$g(x, y) = f(x)f(y)$$Some authors prefer ##x'## to ##y##.

I suspect no explanation is given as at this level the author assumes you have seen this a hundred times before!

bhobba and DrClaude
Dario56 said:
How can inner product give a function?
For every real number ##x \in \mathbb R##, we have a an eigenstate of the position operator in the Hilbert space, which we denote by ##\ket x##. This can be seen as a function$$\mathbb R \to H$$such that$$x \to \ket x$$Now take any state, ##\ket \psi## and we define a function:$$\psi: \mathbb R \to \mathbb C$$ such that$$\psi(x) = \langle x \ket \psi$$

Dario56, Peter Morgan, bhobba and 1 other person
vanhees71 said:
In fact, quantum states are described by "density operators" (I prefer to call them "statistical operators"). In the case of a pure state the statistical operator is a projection operator ##\hat{\rho}=|\psi \rangle \langle \psi|##, where ##|\psi \rangle## is a normalized vector in Hilbert space. Equivalently pure states are represented by unit rays in Hilbert space, i.e., ##|\psi ' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle## represents the same state as ##|\psi \rangle## for all ##\varphi \in \mathbb{R}##.

The wave function is related to ##|\psi \rangle## by
$$\psi(x_1,\ldots,x_n)=\langle x_1,\ldots,x_n|\psi \rangle,$$
where ##|x_1,\ldots, x_n \rangle## are the common (generalized) eigenvectors of the position operators ##\hat{x}_j## with eigenvalues ##x_j##.

It is important to distinguish between the abstract Hilbert-space vectors ##|\psi \rangle## and the wave functions ##\langle x_1,\ldots,x_n|\psi \rangle=\psi(x_1,\ldots,x_n)##. It's analogous to the importance to distinguish between a vector ##\vec{V}## in Euclidean space and its components with respect to a Cartesian basis ##V_j=\vec{e}_j \cdot \vec{V}##.

The matrix elements of the corresponding statistical operator of the pure state indeed are
$$\rho(x_1',\ldots,x_n'; x_1,\ldots x_n) = \langle x_1',\ldots,x_n'|\psi \rangle \langle \psi|x_1,\ldots,x_n) = \psi(x_1',\ldots,x_n') \psi^*(x_1,\ldots,x_n).$$
Yes, thank you. I think my problem with understanding has to do with the eigenvalue equation of the position operator: $$X|x\rangle = x|x\rangle$$

where ##|x\rangle## is a state of the particle with a definite position ##x##.

Determining ##|x\rangle## is important because projecting it on the ##|\psi\rangle## allows us to find the wavefunction in the coordinate representation, ##\psi(x)##: $$\langle x|\psi\rangle = \psi(x)$$

My question is:

What does state ##|x\rangle## actually represent? As far as I know, it should be a vector in the Hilbert space. However, result of the inner product is a scalar. This means that the wavefunction ##\psi(x)## can't be obtained in general terms from the inner product, we can only obtain its specific values at certain values of ##x## for which we know ##|x\rangle##. Reason for that is that functions aren't scalars. Did I say this correctly?

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Dario56 said:
Yes, thank you. I think my problem with understanding has to do with the eigenvalue equation of the position operator: $$X|x\rangle = x|x\rangle$$

where ##|x\rangle## is a state of the particle with a definite position ##x##.

Determining ##|x\rangle## is important because projecting it on the ##|\psi\rangle## allows us to find the wavefunction in the coordinate representation, ##\psi(x)##: $$\langle x|\psi\rangle = \psi(x)$$

My question is:

What does state ##|x\rangle## actually represent? As far as I know, it should be a vector in the Hilbert space. However, result of the inner product is a scalar. This means that the wavefunction ##\psi(x)## can't be obtained in general terms from the inner product, we can only obtain its specific values at certain values of ##x## for which we know ##|x\rangle##. Reason for that is that functions aren't scalars. Did I say this correctly?
You get that there is not a single ##\ket{x}##, but an infinite set of them? When you write
$$\psi(x) = \langle x|\psi\rangle$$
you are substituting for ##x## any possible value it can have.

This is the same a writing ##f(x) = x^2##.

DrClaude said:
You get that there is not a single ##\ket{x}##, but an infinite set of them? When you write
$$\psi(x) = \langle x|\psi\rangle$$
you are substituting for ##x## any possible value it can have.

This is the same a writing ##f(x) = x^2##.
I think got this correctly. To use the example you gave previously, you can't calculate $\psi(x)$ in general form like this: $$\psi(x) = \langle x|\psi\rangle = x^2$$ we can only pretend that this is a normalized state, but you can get for example: $$\psi(1) = \langle 1|\psi\rangle = 1$$
Reason is that the inner product is a scalar and we can only get specific values of ##\psi(x)## if we represent particular position ##x## with a vector ##|x\rangle## in the appropriate basis.

Also, to do the inner product, we need to expand both ##|x\rangle## and ##|\psi\rangle## in the same basis. As eigenstates of position operator are Dirac delta functions (distributions), I am not sure how good this functions are to expand ##|\psi \rangle##? These are very specific functions and I am not sure how good are these to represent anything else except states of perfectly defined position.

PeroK
Dario56 said:
I think got this correctly. To use the example you gave previously, you can't calculate $\psi(x)$ in general form like this: $$\psi(x) = \langle x|\psi\rangle = x^2$$ we can only pretend that this is a normalized state, but you can get for example: $$\psi(1) = \langle 1|\psi\rangle = 1$$
Reason is that the inner product is a scalar and we can only get specific values of ##\psi(x)## if we represent particular position ##x## with a vector ##|x\rangle## in the appropriate basis.
No.

When you write ##f(x)=x^2##, you define a function such that given a value of ##x##, the function returns ##x^2##. Now, when you write ##\psi(x) = \braket{x | \psi}##, it means that given a value of ##x## (i.e., a position), you take the corresponding bra ##\bra{x}## and return the inner product ##\braket{x | \psi}##. Since you can substitute any value for ##x##, ##\braket{x | \psi}## is a function of ##x##, which is the wave function.

Maybe this will other example will help. You want to figure what ##\psi## is at ##x=a##, then
$$\psi(a) = \braket{x=a | \psi}$$

Dario56 said:
Also, to do the inner product, we need to expand both ##|x\rangle## and ##|\psi\rangle## in the same basis. As eigenstates of position operator are Dirac delta functions (distributions), I am not sure how good this functions are to expand ##|\psi \rangle##? These are very specific functions and I am not sure how good are these to represent anything else except states of perfectly defined position.
The ##\{\ket{x}\}## form a basis for position. They are the eigenkets of the position operator ##\hat{X}##, as you noted above. But they are not valid states for a quantum system, e.g., ##\ket{\psi} = \ket{x}## is not a valid quantum state because it is not square integrable.

PeroK
Dario56 said:
Yes, thank you. I think my problem with understanding has to do with the eigenvalue equation of the position operator: $$X|x\rangle = x|x\rangle$$

where ##|x\rangle## is a state of the particle with a definite position ##x##.

Determining ##|x\rangle## is important because projecting it on the ##|\psi\rangle## allows us to find the wavefunction in the coordinate representation, ##\psi(x)##: $$\langle x|\psi\rangle = \psi(x)$$

My question is:

What does state ##|x\rangle## actually represent? As far as I know, it should be a vector in the Hilbert space. However, result of the inner product is a scalar. This means that the wavefunction ##\psi(x)## can't be obtained in general terms from the inner product, we can only obtain its specific values at certain values of ##x## for which we know ##|x\rangle##. Reason for that is that functions aren't scalars. Did I say this correctly?
It's not a vector in Hilbert space but a distribution. That's a bit subtle. The position operator has continuous "eigenvalues". Let's discuss a particle moving along a line. Then we have just one position operator ##\hat{x}##. The generalization to arbitrary dimensions is then straight forward. This position operator is only defined on a part of Hilbert space. Let's take the position representation, i.e., realize the Hilbert space as the space of square-integrable functions, ##\mathrm{L}^2(\mathbb{R})##, i.e., the Hilbert space consists of all functions ##\psi(x) \in \mathcal{C}## such that the integral
$$\int_{\mathbb{R}^3} \mathrm{d} x |\psi(x)|^2 \quad \text{exists.}$$
Then the scalar product of two such functions, defined by
$$\langle \psi_1|\psi_2 \rangle=\int_{\mathbb{R}} \mathrm{d}^3 x \psi_1^*(x) \psi_2(x),$$
also exists, and this defines the Hilbert space to describe the particle moving along a line described by the coordinate ##x##.

The position operator by definition is defined by
$$\hat{x} \psi(x)=x \psi(x),$$
i.e., it just multiplies the wave function by ##x##. Now take as an example ##\psi(x)=\sin x/x##, which for sure is square-integrable, but ##\hat{x} \psi(x) = x \psi(x)=\sin x## is not square integrable, i.e., this function is not in the domain of functions, where ##\hat{x}## is defined. You have to choose some subspace ##D_{\hat{x}}## of the Hilbert space, which is dense in the Hilbert space, where ##\hat{x}## is well defined. Dense means that you can describe any Hilbert-space member as a converging sequence of vectors in the dense subspace.

The "eigenvectors", in this case are "generalized functions" or distributions. The eigenfunction of ##\hat{x}## with eigenvalue ##x_0## in the position representation is a ##\delta## distribution,
$$u_{x_0}(x)=\delta(x-x_0).$$
It is also defined as a functional on the dense subspace ##D_{\hat{x}}##, for which you can choose the space of arbitrarily quickly falling functions, and on such functions the distribution is defined as
$$\int_{\mathbb{R}} \mathrm{d} x \delta(x-x_0) \psi(x)=\psi(x_0).$$

In the abstract formalism thus ##|x \rangle## represents a distribution on the dual space of a dense subspace of Hilbert space. Since this dense subspace is smaller than the Hilbert space its dual (the space of linear forms defined on the subspace) is larger. The dual of the Hilbert space itself is again the Hilbert space.

Formally you have
$$\langle x|x_0 \rangle=\delta(x-x_0) \quad \text{and} \quad \int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|=\hat{1}.$$
The wave function is then formally a "component" of the state vector ##|\psi \rangle##, i.e., ##\psi(x)=\langle x|\psi \rangle##. So you have a one-to-one mapping between the abstract Hilbert space and the Hilbert space of square-integrable functions ##|\psi \rangle \mapsto \psi(x)=\langle x|\psi \rangle##, and the other direction is given by "insertion of a unit operator"
$$\psi(x) \mapsto |\psi \rangle=\int_{\mathbb{R}} \mathrm{d} x |x \rangle \langle x|\psi \rangle=\int_{\mathbb{R}} \int \mathrm{d} x |x \rangle \psi(x).$$
If you are interested in a mathematical strict(er) formulation than usually presented in QM textbooks, have a look at Ballentine, Quantum Mechanics, where the socalled rigged-Hilbert-space formalism is introduced in an unformal way suited for physicists. Another good book is Galindo, Pascual, Quantum mechanics, 2 vols. A very good online-source is the dissertation by de la Madrid:

http://galaxy.cs.lamar.edu/~rafaelm/webdis.pdf

Peter Morgan and physicsworks
DrClaude said:
No.

When you write ##f(x)=x^2##, you define a function such that given a value of ##x##, the function returns ##x^2##. Now, when you write ##\psi(x) = \braket{x | \psi}##, it means that given a value of ##x## (i.e., a position), you take the corresponding bra ##\bra{x}## and return the inner product ##\braket{x | \psi}##. Since you can substitute any value for ##x##, ##\braket{x | \psi}## is a function of ##x##, which is the wave function.

Maybe this will other example will help. You want to figure what ##\psi## is at ##x=a##, then
$$\psi(a) = \braket{x=a | \psi}$$
That is what I mean. As far as I can see, you wrote the same thing as I did. Or am I missing something?
DrClaude said:
The ##\{\ket{x}\}## form a basis for position. They are the eigenkets of the position operator ##\hat{X}##, as you noted above. But they are not valid states for a quantum system, e.g., ##\ket{\psi} = \ket{x}## is not a valid quantum state because it is not square integrable.
Yes, that is clear. However, to do the inner product both ##|x\rangle## and ##|\psi\rangle## need to be written in the same basis set in the same way as when calculating dot product of two vectors in the Euclidean space they need to be written with the same basis vectors.

As basis set for position is made of Dirac functions ,which are very specific type of functions, I am not sure if those can be used as a basis set to describe ##|\psi\rangle## or in fact anything else than states of exact position.

When you say that eigenstates of position operator aren't valid quantum states, what does that imply exactly? Since these aren't square integrable, I guess that would mean they can't be normalized. However, if these are invalid how can we use them to calculate the wavefunction in the coordinate representation, ## \psi(x) ##?

The ##|x \rangle## are not a basis but a "generalized basis". The Hilbert space used in non-relativistic QM is a separable Hilbert space, i.e., all bases are countable. An example for a basis are the energy eigenstates of a harmonic oscillator, and indeed ##\mathrm{L}^2(\mathbb{R})## (wave mechanics) is isomorphic to the space ##\mathcal{l}^2## of "square-summable" sequences (matrix mechanics).

Dario56 said:
That is what I mean. As far as I can see, you wrote the same thing as I did. Or am I missing something?

The function you wrote is not square integrable, but otherwise I do agree with you.

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vanhees71 said:
The ##|x \rangle## are not a basis but a "generalized basis". The Hilbert space used in non-relativistic QM is a separable Hilbert space, i.e., all bases are countable. An example for a basis are the energy eigenstates of a harmonic oscillator, and indeed ##\mathrm{L}^2(\mathbb{R})## (wave mechanics) is isomorphic to the space ##\mathcal{l}^2## of "square-summable" sequences (matrix mechanics).
If I understood your answers accurately, there are uncountably infinite number of position eigenvectors which are given by the delta function if particle can be found at ##x_0## when measured. This is connected to the fact that position operator has continuous eigenvalues.

While I agree with this, I am not sure what does it have to do with the question of taking the inner product and basis. To define the inner product, we need to have ##|x\rangle## and ##|\psi \rangle## expressed in the same basis. As you mentioned, in this case we have generalized basis because of continuous eigenvalues of the position operator. However, since delta functions are rather limited, I am not sure how can we use these to describe ##|\psi \rangle## or anything else except states of perfectly defined position.

Also, since we have an uncountably infinite basis set (generalized basis), how can we calculate this inner product? I guess in that case, inner product is given by some integral and it has to do with your prior answer.

Let ##\psi(x)## be an square integrable function. Then there exist a vector ##\left|\psi\right\rangle## such that ##\left\langle x'\right.\left|\psi\right\rangle =\psi(x')##.

We can express ##\left|\psi\right\rangle## in the generalized position base by the following linear combination

$$\left|\psi\right\rangle =\int_{-\infty}^{\infty}\psi(x)\left|x\right\rangle dx$$

and the above result is easy to get $$\left\langle x'\right.\left|\psi\right\rangle =\int_{-\infty}^{\infty}\psi(x)\left\langle x'\right.\left|x\right\rangle dx=\int_{-\infty}^{\infty}\psi(x)\,\delta(x-x')\,dx=\psi(x')$$

vanhees71, Dario56 and PeroK
andresB said:
Let ##\psi(x)## be an square integrable function. Then there exist a vector ##\left|\psi\right\rangle## such that ##\left\langle x'\right.\left|\psi\right\rangle =\psi(x')##.

We can express ##\left|\psi\right\rangle## in the generalized position base by the following linear combination

$$\left|\psi\right\rangle =\int_{-\infty}^{\infty}\psi(x)\left|x\right\rangle dx$$

and the above result is easy to get $$\left\langle x'\right.\left|\psi\right\rangle =\int_{-\infty}^{\infty}\psi(x)\left\langle x'\right.\left|x\right\rangle dx=\int_{-\infty}^{\infty}\psi(x)\,\delta(x-x')\,dx=\psi(x')$$

Now, I can get back to my first questions.
vanhees71 said:
The wave function is related to ##|\psi \rangle## by
$$\psi(x_1,\ldots,x_n)=\langle x_1,\ldots,x_n|\psi \rangle,$$
where ##|x_1,\ldots, x_n \rangle## are the common (generalized) eigenvectors of the position operators ##\hat{x}_j## with eigenvalues ##x_j##.
As we discussed, wavefunction $\psi(x)$ is connected to the vector ##|\psi \rangle## by: $$\psi(x) = \langle x | \psi \rangle$$

I don't understand what does this equation mean: $$\psi(x_1,\ldots,x_n)=\langle x_1,\ldots,x_n|\psi \rangle$$

What operation is done here? We take the inner product of the ##n## different eigenvectors of the position operator with ##|\psi \rangle##? Since all first equations and questions come from the textbook: Density Functional Theory, Parr and Yang, I think ##n## represents number of electrons and corresponding multielectron wavefunction ##\psi(x_1,\dots,x_n)##.

Sorry if this sounds blunt, but you should improve your basic understanding of the math of QM before trying to read some advanced stuff.

Single particle wavefunctions can be obtained as I showed before. Let's now have n particles. The operator ##\hat{X}_{i}## can be used to measure the ##ith## particle position. It has an eigenvalue equation of the form

$$\hat{X}_{i}\left|x_{i}\right\rangle =x_{i}\left|x_{i}\right\rangle.$$

We now can construct a multiparticle Hilbert space using the tensor product. We start by defining the following ket

$$\left|x_{1},x_{2},...,x_{n}\right\rangle =\left|x_{1}\right\rangle \otimes\left|x_{2}\right\rangle \ldots\otimes\left|x_{n}\right\rangle.$$ The above ket represents a state where the 1st particle has the definite position ##x_{1},## the second particle has the definite position ##x_{2}##, and so on. This vector is, of course, of infinite norm, and the following is true

$$\left\langle x'_{1},x'_{2},...,x'_{n}\right.\left|x_{1},x_{2},...,x_{n}\right\rangle =\prod_{i=1}^{n}\delta(x_{i}-x'_{i}).$$

Moreover, ##\left|x_{1},x_{2},...,x_{n}\right\rangle## is a common eigenvector of all of the position operators (they form a commuting set),

$$\hat{X}_{1}\left|x_{1},x_{2},...,x_{n}\right\rangle =x_{1}\left|x_{1},x_{2},...,x_{n}\right\rangle,$$
$$\hat{X}_{2}\left|x_{1},x_{2},...,x_{n}\right\rangle =x_{2}\left|x_{1},x_{2},...,x_{n}\right\rangle,$$
$$\vdots$$
$$\hat{X}_{n}\left|x_{1},x_{2},...,x_{n}\right\rangle =x_{n}\left|x_{1},x_{2},...,x_{n}\right\rangle.$$

Given an square integrable function ##\psi(x_{1},x_{2},...,x_{n})## that goes from ##\mathbb{R}^{n}\rightarrow\mathbb{C},## then there exist a vector ##\left|\psi\right\rangle## belonging to the multiparticle Hilbert space such that $$\left\langle x'_{1},x'_{2},...,x'_{n}\right.\left|\psi\right\rangle =\psi(x'_{1},x'_{2},...,x'_{n}),$$

and we can write ##\left|\psi\right\rangle## as the following linear combination $$\left|\psi\right\rangle =\int_{\mathbb{R}^{n}}\psi(x_{1},x_{2},...,x_{n})\,\left|x_{1},x_{2},...,x_{n}\right\rangle \,d^{n}x.$$

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vanhees71 and PeroK
Dario56 said:
What does state ##|x\rangle## actually represent?

It is what is called a distribution or generalised function. It is often said that QM is formulated in a Hilbert space. That's fine for beginning students, but as usually practised by physicists, it is formulated in a Rigged Hilbert Space. The great polymath Von-Neumann formulated it that way, but most physicists/applied mathematicians used the generalised function approach pioneered by Dirac. It is conceptually simpler. At the beginner level, just think of the Dirac Delta Function, which is NOT part of a Hilbert Space, as a continuously differentiable function of a very small width but a very large height whose integral is one. It acts like the Dirac Delta function and is part of a Hilbert Space for all practical purposes. This can all be made rigorous but is advanced mathematically. The first step if you wish to go down that path is to study the theory of Distributions:
https://www.amazon.com/dp/0521558905/?tag=pfamazon01-20

It is very important in applied math and physics - it's something everyone should know but is often not taught in math/physics degrees, which is a pity.

Thanks
Bill

vanhees71
andresB said:
Sorry if this sounds blunt, but you should improve your basic understanding of the math of QM before trying to read some advanced stuff.

Single particle wavefunctions can be obtained as I showed before. Let's now have n particles. The operator ##\hat{X}_{i}## can be used to measure the ##ith## particle position. It has an eigenvalue equation of the form

$$\hat{X}_{i}\left|x_{i}\right\rangle =x_{i}\left|x_{i}\right\rangle.$$

We now can construct a multiparticle Hilbert space using the tensor product. We start by defining the following ket

$$\left|x_{1},x_{2},...,x_{n}\right\rangle =\left|x_{1}\right\rangle \otimes\left|x_{2}\right\rangle \ldots\otimes\left|x_{n}\right\rangle.$$ The above ket represents a state where the 1st particle has the definite position ##x_{1},## the second particle has the definite position ##x_{2}##, and so on. This vector is, of course, of infinite norm, and the following is true

$$\left\langle x'_{1},x'_{2},...,x'_{n}\right.\left|x_{1},x_{2},...,x_{n}\right\rangle =\prod_{i=1}^{n}\delta(x_{i}-x'_{i}).$$

Moreover, ##\left|x_{1},x_{2},...,x_{n}\right\rangle## is a common eigenvector of all of the position operators (they form a commuting set),

$$\hat{X}_{1}\left|x_{1},x_{2},...,x_{n}\right\rangle =x_{1}\left|x_{1},x_{2},...,x_{n}\right\rangle,$$
$$\hat{X}_{2}\left|x_{1},x_{2},...,x_{n}\right\rangle =x_{2}\left|x_{1},x_{2},...,x_{n}\right\rangle,$$
$$\vdots$$
$$\hat{X}_{n}\left|x_{1},x_{2},...,x_{n}\right\rangle =x_{n}\left|x_{1},x_{2},...,x_{n}\right\rangle.$$

Given an square integrable function ##\psi(x_{1},x_{2},...,x_{n})## that goes from ##\mathbb{R}^{n}\rightarrow\mathbb{C},## then there exist a vector ##\left|\psi\right\rangle## belonging to the multiparticle Hilbert space such that $$\left\langle x'_{1},x'_{2},...,x'_{n}\right.\left|\psi\right\rangle =\psi(x'_{1},x'_{2},...,x'_{n}),$$

and we can write ##\left|\psi\right\rangle## as the following linear combination $$\left|\psi\right\rangle =\int_{\mathbb{R}^{n}}\psi(x_{1},x_{2},...,x_{n})\,\left|x_{1},x_{2},...,x_{n}\right\rangle \,d^{n}x.$$
Well, to learn the math of the QM, one first needs to encounter it in the QM, if you learn QM by yourself as I do. It doesn't have sense other way around because how can one know what math exactly one needs if one didn't open QM textbook first. You first open QM textbook and than when you don't get the math, you open the math textbook.

If you learn on the university, that is a different matter, because you learn more systematically and with guidance. This is why learning curve and process might seem strange for people who learn by themselves especially in the eyes of the people who learn one the university.

In this case, I wasn't familiar with the notation of the tensor product used and you helped me to learn that.

Back to the question, I am not sure why is the norm of the ##|x_1,x_2,\dots,x_n \rangle## infinite? Vectors ##|x_1 \rangle, |x_2 \rangle, \dots ,|x_n \rangle## are a part of the generalized basis (which is uncountably infinite basis). This means that every component of these vectors is zero and only one has the value of one (this component corresponds to the position ##x_1, x_2, \dots, x_n## depending which basis vector we are describing). Taking their tensor product, of course, gives a vector which also must have a generalized basis. Since we only have zeros and ones in this vector and the number of basis vectors used in the tensor product is finite, why would such vector have the infinite norm?

Dario56 said:
Well, to learn the math of the QM, one first needs to encounter it in the QM, if you learn QM by yourself as I do. It doesn't have sense other way around because how can one know what math exactly one needs if one didn't open QM textbook first. You first open QM textbook and than when you don't get the math, you open the math textbook.

If you learn on the university, that is a different matter, because you learn more systematically and with guidance. This is why learning curve and process might seem strange for people who learn by themselves especially in the eyes of the people who learn one the university.

In this case, I wasn't familiar with the notation of the tensor product used and you helped me to learn that.
The solution is self-evident in that case, you have to start your study of QM from the basics. Density functional theory sounds like a hard topic, not the best source for a first encounter with QM.

Dario56 said:
Back to the question, I am not sure why is the norm of the ##|x_1,x_2,\dots,x_n \rangle## infinite? Vectors ##|x_1 \rangle, |x_2 \rangle, \dots ,|x_n \rangle## are a part of the generalized basis (which is uncountably infinite basis). This means that every component of these vectors is zero and only one has the value of one (this component corresponds to the position ##x_1, x_2, \dots, x_n## depending which basis vector we are describing). Taking their tensor product, of course, gives a vector which also must have a generalized basis. Since we only have zeros and ones in this vector and the number of basis vectors used in the tensor product is finite, why would such vector have the infinite norm?
Operators with a continuous spectrum must be normalized with a Dirac delta. I'm unsure what would be the best reference to give for this topic, but you can start with chapter 4 of Shankar. Messiah and Cohen tannoudji are also good.

vanhees71, bhobba and Dario56
andresB said:
The solution is self-evident in that case, you have to start your study of QM from the basics. Density functional theory sounds like a hard topic, not the best source for a first encounter with QM.

Operators with a continuous spectrum must be normalized with a Dirac delta. I'm unsure what would be the best reference to give for this topic, but you can start with chapter 4 of Shankar. Messiah and Cohen tannoudji are also good.
Thank you for giving the reference, I always appreciate that.

I did start with the basics of QM; postulates, failure of classical physics, linear algebra (eigenvectors, eigenvalues, basis etc.), expectation values, Hilbert space, Schrödinger equation... I have a solid foundation in the Hartree Fock, but DFT is more complex. It takes much more to learn it.

As we talked, when learning alone, you can miss some things when from the basics simply because of lack of guidance. But, I don't see that as a problem, you just go back or discuss a problem with others to see what you don't know. It is all part of the learning process.

bhobba and vanhees71
vanhees71 said:
In fact, quantum states are described by "density operators" (I prefer to call them "statistical operators"). In the case of a pure state the statistical operator is a projection operator ##\hat{\rho}=|\psi \rangle \langle \psi|##, where ##|\psi \rangle## is a normalized vector in Hilbert space. Equivalently pure states are represented by unit rays in Hilbert space, i.e., ##|\psi ' \rangle=\exp(\mathrm{i} \varphi) |\psi \rangle## represents the same state as ##|\psi \rangle## for all ##\varphi \in \mathbb{R}##.

The wave function is related to ##|\psi \rangle## by
$$\psi(x_1,\ldots,x_n)=\langle x_1,\ldots,x_n|\psi \rangle,$$
where ##|x_1,\ldots, x_n \rangle## are the common (generalized) eigenvectors of the position operators ##\hat{x}_j## with eigenvalues ##x_j##.

It is important to distinguish between the abstract Hilbert-space vectors ##|\psi \rangle## and the wave functions ##\langle x_1,\ldots,x_n|\psi \rangle=\psi(x_1,\ldots,x_n)##. It's analogous to the importance to distinguish between a vector ##\vec{V}## in Euclidean space and its components with respect to a Cartesian basis ##V_j=\vec{e}_j \cdot \vec{V}##.

The matrix elements of the corresponding statistical operator of the pure state indeed are
$$\rho(x_1',\ldots,x_n'; x_1,\ldots x_n) = \langle x_1',\ldots,x_n'|\psi \rangle \langle \psi|x_1,\ldots,x_n) = \psi(x_1',\ldots,x_n') \psi^*(x_1,\ldots,x_n).$$
After doing some reading and studying, here is where I got:

As we said, we express ##|\psi \rangle## in the generalized basis. Such basis set is uncountably infinite.

Density operator is defined as the outer product of the wavefunction ##|\psi\rangle## with itself: $$\gamma = |\psi \rangle \langle \psi |$$

If basis set is uncountably infinite than the number of entries of the corresponding density matrix should also be uncountably infinite given the fact how outer product of vectors is performed.

I see when you represented the density matrix, you claimed that it is ##n##x##n## matrix, where ##n## is the number of particles.

This is what I don't get.

Dario56 said:
This is what I don't get.

This is an advanced topic even for those with a math background like mine. In hindsight, I likely spent far more time on it than it was worth in understanding QM. Still, it is a legit area of study, and in itself, interesting mathematically - the cumulation of the work done by some of the 20th century's greatest mathematicians, such as Schwartz, Gelfand, and Grothendieck when he was working with Schwartz. But if you want the gory detail see the attached document.

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## 1. What is a density operator of a pure state?

A density operator of a pure state is a mathematical representation of a quantum state that describes the probability of finding a system in a particular state. It is also known as a density matrix and is represented by the Greek letter rho (ρ).

## 2. How is a density operator of a pure state different from a density operator of a mixed state?

A density operator of a pure state represents a system that is in a single, definite quantum state, while a density operator of a mixed state represents a system that is in a superposition of multiple quantum states. This means that a pure state has a single non-zero eigenvalue, while a mixed state has multiple non-zero eigenvalues.

## 3. How is a density operator of a pure state calculated?

A density operator of a pure state is calculated by taking the outer product of the state vector with itself. This means multiplying the state vector by its complex conjugate transpose. The result is a square matrix with the same dimension as the state vector.

## 4. What is the physical significance of the trace of a density operator of a pure state?

The trace of a density operator of a pure state is equal to 1, which represents the total probability of finding the system in any state. This means that the trace is a measure of the system's purity, with a higher trace indicating a purer state.

## 5. How is the density operator of a pure state used in quantum mechanics?

The density operator of a pure state is used in quantum mechanics to describe the state of a system and to make predictions about its behavior. It is also used in calculations of quantum entanglement and in the study of quantum information and quantum computing.

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