# B States as positive operators of unit trace

1. Oct 15, 2016

### cube137

I read that states are positive operators of unit trace - not elements of a vector space.

Is it referring to quantum states or all classical states?
I know operators are like minus, plus, square root and vectors are like rays in Hilbert space.. but why can't quantum states be vectors when in fact quantum state is rays in Hilbert space or vectors of unit 1?

Please give a gentler reply (take note of the B in the thread). Thank you.

2. Oct 15, 2016

### Mentz114

Without knowing what you actually read it's hard to give an answer.

States can be expressed as vectors in Hilbert space or density operators.
If there is a state, expressed as a Hilbert space vector $|\phi\rangle$ then the 'outer' product of the vector $|\phi\rangle \langle \phi |$ is a density operator. So there is an operator corresponding to a state vector.

But don't know if this is what you're asking.

3. Oct 16, 2016

### atyy

In the density matrix formalism, which can handle both pure and mixed states, the mixed or pure state is a positive operator with unity trace.

However, in the "quantum state is a ray in Hilbert space" idea, we only talk about pure states. Under the interpretation that a pure state is the state of a single system, we can derive the density matrix formalism.

So the two languages are the same.

4. Oct 16, 2016

### A. Neumaier

It refers to quantum states.

There are two different concepts of state, one a special case of the other, but phrased in a different language.

The general situation is that of a state described by a density operator $\rho$, a positive semidefinite Hermitian operator of trace 1 on a Hilbert space . (They are elements of another Hilbert space, the space of all Hermitian operators with the trace inner product.)

If such a density operator has rank 1 then it is called a pure state and has the form $\rho=\psi\psi^*$ with a norm 1 vector $\psi$ in the first Hilbert space. This vector (called a state vector or a wave function) is only determined by the state up to a phase, whereas the associated ray, the 1-dimensional vector space spanned by $\psi$, is determined uniquely (it is the range of $\rho$). Thus pure states are equivalently described by a ray in a Hilbert space.

The description by rays is less general since most states (e.g., for an infinite-dimensional Hilbert space all states at finite temperature, and hence all preparable position or momentum dependent states) are not pure. Unpolarized states of a photon are also not pure.

However, pure states are often used as simplified idealizations of general states. For example, in much (not all) of chemistry, only the electronic ground state of molecules is significantly populated, and in the absence of degeneracy, this is a pure state. Quantum information theory also works mostly with pure states, and leaves the non-idealized situation to specialists in quantum optics.

This is not true. Operations like plus, minus, sqrt, are neither operators nor rays, and not even alike rays in any sense.

Vectors are also not rays but a ray is the collection of all vectors parallel to a given one.

One can add state vectors and gets another state vector, but adding two distinct rays produces a 2-diemnsional subspace and not a ray.

Last edited: Oct 16, 2016
5. Oct 16, 2016

### vanhees71

To translate to the physicists' Dirac notation, a pure state is written as
$$\hat{\rho}=|\psi \rangle \langle \psi|.$$
I guess that's what A. Neumaier means in his mathematician's notation by $\psi \psi^*$ which is more a shorthand notation (in my opinion even a bit unclear) for the matrix elements wrt. an orthonormal basis $|k \rangle$:
$$\rho_{jk} = \langle j|\hat{\rho} k \rangle=\langle j|\psi \rangle \langle \psi|k \rangle=\psi_j \psi_k^*.$$

6. Oct 16, 2016

### A. Neumaier

what is unclear? The ordinary matrix product of a nonzero $n\times 1$ matrix $\psi$ (i.e., a vector of length $n$) and its conjugate transpose $\psi^*$ is a Hermitian $n\times n$ matrix $\rho$ of rank 1 with entries $\rho_{jk}=\psi_j\bar\psi_k$, and this is the precise content of what I wrote.

Writing additional bars and pointed brackets does not add any information but only obscures the underlying simplicity of matrix calculus.

7. Oct 16, 2016

### vanhees71

Well, the Dirac notation has its merits. Contrary to what you write, I think it simplifies calculations a lot. I had a big advantage in the functional analysis recitations compared to the poor pure mathematicians who had to deal with the typical unstructured notations the mathematicians prefer. Some don't like the Dirac notation for some reason, but it's a very intuitive way to write QT in terms of the representation free abstract-vector representation. This is the one point I disagree with Weinberg concerning notation, but he seems to have an aversion against Dirac and thus also doesn't like his notation ;-)).

Your posting is a good example, where unstructured mathematicians' notation leads to misunderstandings: You didn't even mention that you work with components and matrix notation! That's why I (partially) misunderstood your posting. Mathematics is often much more complicated than physics only because they don't distinguish their symbols clearly. It's, e.g., great to have a vector written with different symbols than scalars.

8. Oct 16, 2016

### A. Neumaier

I used coordinates only because you used it, since I wanted to show that the same information is encoded in far fewer letters. My notation is coordinate-free, borrowed from linear algebra (in its coordinate-free form with linear mappings and abstract vectors). $\psi$ is a state vector in an arbitrary Hilbert space, $\psi^*$ the linear functional defined by the inner product $\psi^*\phi$, and $\rho$ is the operator mapping $\phi$ to $\rho\phi=\psi\psi^*\phi$. It couldn't be more concise and more component-free.

Dirac notation is useful primarily when more than one label is needed to label states. It is also ambiguous, for how do you know whether |u> is a position eigenstate at x=u, a momentum eigenstate at p=u, or a general state vector u in Dirac notation? The context and the choice of letters usually tell what is meant, but with context you can also resolve ambiguities in mathematical notation.

What is easier to read depends on what you practiced more.

9. Oct 17, 2016

### vanhees71

Yes, but it's confusing for physicists to use this notation. That's all I'm saying.