# B Mixed states basic question

#### jk22

Summary
Can mixed states be seen as many particles and formulas
Considering a mixture $\sum p_i|\Psi_i\rangle\langle\Psi_i|$

This does not describe an ensemble of quantum systems since the particle number is defined by $\Psi_i$.

The question is in the continuous wave-mechanical formalism where I don't understand what object the density matrix is : I know $\langle\Psi_i|x\rangle=\Psi_i(x)=\int \Psi_i(x')\delta(x-x')dx'$. It seems that here I could exchange the order but what happen to the braces ? Is it $|\Psi_i\rangle\langle\Psi_i|=\Psi_i(x)\int\Psi_i(x')\delta(x'-x'')[\circ]dx'$ ?

Where $\circ$ means it's the place for a function in $x''$ ?

Why is then the sum different than a single term ?

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#### PeterDonis

Mentor
This does not describe an ensemble of quantum systems
It can. Whether or not it does depends on who is using the mixed state and for what purpose.

the particle number is defined by $\Psi_i$.
This makes no sense. Particle number is an operator, not a state.

#### jk22

This makes no sense. Particle number is an operator, not a state.
I think it's the eigenvalues of that operator, but is it in quantum field theory ?
What I meant is old wavemechanics where the structure of the wavefunction depends on this number : $\Psi(\vec{x}_1,...\vec{x}_n)$ with n the number of particles

But basically I understood : it because the sum of $\sum f_i(x)f_i(x')\neq g(x)g(x')$

Last edited:

#### vanhees71

Gold Member
I think you should read a good book about quantum mechanics first. I'd recommend Sakurai, Modern Quantum mechanics.

In the "first-quantization formalism" you deal by definition with situations where the particle number is fixed. That's by the way, why this formalism doesn't work well for interacting relativistic quantum systems, and that's why today we only use relativistic quantum field theory to deal with relativistic quantum systems.

So now let's quickly get the first-quantization formalism for $N$ spinless dinstinguishable particles right. Here the Hilbert space is a product space $\mathcal{H}=\mathcal{H}_1 \otimes \mathcal{H}_2 \otimes \cdots \otimes \mathcal{H}_N$, where $N$ is the number of particles, which is fixed by definition once and for all. No particle can be destroyed and no new particles can be created in any way (by definition, and thus dealing with non-relativistic particles only, i.e., particles where the interaction energies are way less than the rest energies $m_i c^2$ of all particles involved).

Now one particular basis of this $N$-particle Hilbert space are the common (generalized) position eigenvectors of the positions of the particles,
$$|\vec{x}_1,\ldots,\vec{x}_N \rangle = |\vec{x}_1 \rangle \otimes |\vec{x}_2 \rangle \otimes \cdots \otimes |\vec{x}_N \rangle.$$
For a given normalized vector $|\Psi \rangle$ the wave function is defined as
$$\Psi(\vec{x}_1,\ldots,\vec{x}_N)=\langle \vec{x}_1,\ldots,\vec{x}_N|\Psi \rangle.$$
A statistical operator, as any operator is, as any operator, is described in the position representation by
$$\rho(\vec{x}_1,\ldots \vec{x}_N;\vec{x}_1',\ldots,\vec{x}_N')=\langle \vec{x}_1,\ldots,\vec{x}_N|\hat{\rho}|\vec{x}_1',\ldots,\vec{x}_N' \rangle.$$
If your state is given by
$$\hat{\rho}=\sum_i p_i |\Psi_i \rangle \langle \Psi_i |,$$
from the general formula and the above definition of the wave functions you get
$$\rho(\vec{x}_1,\ldots,\vec{x}_N;\vec{x}_1',\ldots,\vec{x}_N')=\sum_i p_i \Psi_i(\vec{x}_1,\ldots,\vec{x}_N) \Psi_i^*(\vec{x}_1',\ldots,\vec{x}_N').$$

#### PeterDonis

Mentor
I think it's the eigenvalues of that operator
A state is not an eigenvalue of an operator. It could be an eigenvector of an operator.

is it in quantum field theory ?
Do you mean, is particle number an operator in QFT? Yes.

What I meant is old wavemechanics where the structure of the wavefunction depends on this number : $\Psi(\vec{x}_1,...\vec{x}_n)$ with n the number of particles
In ordinary (non-relativistic) QM, the number of particles is fixed, yes, and there is no such thing as a particle number operator. But the state $\Psi(\vec{x}_1,...\vec{x}_n)$ can still describe an ensemble of systems instead of a single system; each system in the ensemble is a system of $n$ particles.

"Mixed states basic question"

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