# What is Quantum state

1. Jan 26, 2013

### shounakbhatta

Hello,

I have been reading many things on Quantum state but unable to understand what actually it is. Can anybody explain me?

Thanks

2. Jan 26, 2013

### Simon Bridge

Depends a lot on context.
A "state", in general, is the collection of numbers needed to completely describe the physics you are interested in. Thus, the state of an ideal gas is described by 4 numbers: P,V,T, and N. Mathematically it is those four numbers.

The QM state is everything you need to record to know about the statistics of whatever it is you are looking at. For an electron in an atom, that would be 4 numbers again.

The concept of a state is an abstraction, though, so what it means physically depends on the thing that has the state, and what you are doing with it.

3. Jan 27, 2013

### shounakbhatta

Actually, I was trying to understand a wave function; it states that it's a a probability amplitude in quantum mechanics describing the quantum state of a particle and how it behaves.

So hence quantum state. So if u can plz.explain...........

4. Jan 27, 2013

### andrien

A quantum state carries all the relevant information about the system.Square modulus of a wave function gives probability density.A quantum state is just abstract thing used by Dirac.Any quantum system is composed of different quantum pure states,when you take measurement you get a quantum state which will be state of system after measurement.The wave function belong to Schrodinger and Quantum state to mostly Dirac.There is just different notation.

5. Jan 27, 2013

### Simon Bridge

The wave-function encodes the statistics of whatever is being considered.
As andrien says, it's square-modulus is a probability density function. From the wave-function you can make predictions about the outcome of measurements made on the system, how the system evolves with time, and how it interacts with other systems with their wavefunctions. It's wholly abstract.

6. Jan 27, 2013

### tom.stoer

A state $|\psi\rangle$ in Dirac Notation is a vector in an abstract Hilbert space. Wave functions are projections using specific states (e.g. position or momentum eigenstates) living in the dual Hilbert space.

$$\psi(x) = \langle x | \psi \rangle$$

7. Jan 27, 2013

### shounakbhatta

Hello Simon,

Thanks for the detailed answer. If you can please explain me 'how it interacts with other systems '. What it actually means?

8. Jan 27, 2013

### Simon Bridge

It means exactly what it says. A particle in some state may interact with another particle in some state ... that's pretty much what particle physics is.

The questions are very general so the answers are also general.
Did you have a particular example in mind?

9. Jan 27, 2013

### ZealScience

Wave function is a little bit different from quantum state. Although I prefer the more general definition by Dirac's Bra-ket vector, wave function is easier to interpret.

Basically, as you have said, wave functions are probability amplitudes that carries complete information about a particle (well, mostly postional; to build other representations we need more complicated QM). They are postulated to square to probability (this is a fundamental postulation of QM, which by definition cannot be verified).

In order to describe the particle the wave functions must follows certain operator equations. E.g the Schrodinger's equation. Whereas the quantisation of states follows from boundary conditions. (In some cases, it is quite analogous to solving EOM of classical standing waves, which would make it more understandable if you know them).

While the state eigenvectors collapse in Hilbert space, the corresponding situation in wave functions is that they collapse into δ-functions which make it orthogonal to all functions except a particular point, making the eigenvalue specific to an observable.

To sum up, wave functions describe probability amplitude by postulation, and they must follow the operator equations specific to certain description. And the quantisation follows directly from the constraints to the system.

10. Jan 27, 2013

### vanhees71

The question about the notion of "quantum state" is a pretty tricky issue. One always has to put a disclaimer on any explanation concerning this subject that this is the view of the author, and I have sometimes the impression that there are as many interpretations of quantum mechanics as there are quantum physicists. Most practitioners are "shutup-and-calculate people", which is not the worst of the interpretations around. It just says that quantum theory is a handful of simple rules to calculate probabilities. Don't ask about any further interpretation.

I myself belong to the followers of the ensemble interpretation or minimal statistical interpretation. This is the interpretation that the state is an abstract prescription of an (often idealized) equivalence class of reprodcucible preparations of a quantum system which has only a meaning for ensembles of such well-defined prepared systems that are otherwise totally independent from each other.

A complete determination of the state of a particle admits the representation of such an ensemble in terms of a ray in an appropriate Hilbert space, which can be represented by any normalized state vektor $|\psi \rangle$ belonging to this ray. The normalization to one fixes this state vector still only up to a phase factor, but these phase factors have no physical significance whatsoever.

The next step is to interpret the statistical properties included in this state vector. To this end one has to think about the representation of observables. These are represented by essentially self-adjoint operators, defined on a dense subspace of Hilbert space. The possible values an observable can take is given by the spectrum of this self-adjoint operator. In general there exists not a proper eigenvector to a spectral value but only "generalized" eigenvectors, belonging to the dual of the definition range of the operator. We don't go into this quite formal thing here. It's most conveniently formalized in terms of the socalled "rigged Hilbert space" (see, e.g., the textbook by Galindo and Pascual on this).

Anyway, Born's postulate then states that the probability to find a value $a$ of an observable $A$, represented by the self-adjoint operator $\hat{A}$ is given by
$$P_{\psi}(a)=\int \mathrm{d} \beta |\langle{a,\beta}|\psi \rangle|^2,$$
where $|a,\beta \rangle$ is a complete set of orthonormalized (generalized) eigenvectors of $\hat{A}$ for the spectral value (generalized eigenvalue) $a$, and $\beta$ is one or some finite set of parameters, labeling the different states to the same generalized eigenvalue.

One can also define complete measurements, i.e., one considers several independent observables $A_1,\ldots,A_n$ of compatible observables which have only one-dimensional common (generalized) eigenvectors $|a_1,\ldots,a_n$. Then the probability to measure a possible set of values $(a_1,\ldots,a_n)$ is simply given by
$$P_{\psi}(a_1,\ldots,a_n)=|\langle a_1,\ldots,a_n|\psi \rangle|^2.$$
According to the minimal statistical interpretation this is the only meaning such a (pure) state has. Of course, the previous definition in the case of degenerate eigenstates is included in this definition, because one only has to integrate/sum over all possible values of the Observables $A_2,\ldots,A_n$ to get the probability distribution for $A_1$:
$$P_{\psi}(a_1)=\int \mathrm{d} a_2 \cdots \mathrm{d} a_{n} P_{\psi}(a_1,a_2,\ldots,a_n).$$
Within quantum theory it is the most comprehensive knowledge we can have about a quantum system, i.e., when we know that the quantum system is described by such a pure state. We principally cannot know more than the statistical content encoded in the state according to the above explained Born's rule.

Whether or not quantum theory is a complete theory and whether there is a more complete deterministic theory with hidden variables is not yet clear. If this is, however, the case, this theory must be very weird (perhaps even weirder than quantum theory itself), because it would have to be a non-local theory, as seen by the empirically very well established violation of Bell's inequality, but that's another subject, also often discussed in this forum.

11. Jan 27, 2013

### Maui

Some theorists will disagree(as they have done in the past), but the latest reasearch has already discovered that a quantum state has a measureable form of physical existence(and is soon to be put to use).

Have a look at this:

"A single-atom electron spin qubit in silicon"

http://www.nature.com/nature/journal/v489/n7417/full/nature11449.html

Then this:

"Landmark in quantum computing"

Theoretical and experimental physics compete for new discoveries using different methodology and various discoveries at times surprize one or both camps.

As for how states interact, they must have some form of existence, otherwise it'd be pure magic that we have tables and chairs composed of numbers and statistics.

Last edited by a moderator: May 6, 2017
12. Jan 28, 2013

### vanhees71

Can you point at the place in the above mentioned nature article, where there is a contradiction to the probabilistic interpretation according to the Born rule of the quantum state? I don't think that recent research contradicts quantum theory. To the contrary the better the quantum optiticians and quantum engineers can perform their measurements on the foundational issues of quantum theory the better it is confirmed. See, e.g., the great work awarded with the Nobel prize 2012.

What do you mean by the phrase "the states must have some form of existence"? As far as I know, and as I experience in my environment, I don't see vectors in Hilbert space or Statistical operators but tables and chairs, and the abstract mathematical objects of quantum theory are our means to describe those and other entities, no more nor less!

Last edited by a moderator: May 6, 2017
13. Jan 28, 2013

### Maui

Short answer - a qubit is a TWO state quantum system that is now put in practice(as opposed to theory and the practice requires that the system in a two state have a measureable existence).

From the article - "We use electron spin resonance to drive Rabi oscillations, and a Hahn echo pulse sequence reveals a spin coherence time exceeding 200 µs. This time should be even longer in isotopically enriched 28Si samples."

From their video of the experiment:

"Now imagine you can make a bit that can be in the zero and the one state at the same time. That's called a quantum bit or qubit. And the simplest example in nature would just a single electron. It has a magnetic dipole called spin, which is like the needle of a compass but because it's a quantum needle, it can be pointing up and down at the same time.

I didn't say it contradicts quantum theory, only certain interpretations of it(and there are many).

Some of the electronics inside your monitor utilizes quantum effects and principles so these effects can be said to be experienced to a large extent. Chairs and tables belong to a different realm and require an interpretation which is a separate issue but this new angle of looking at nature gives one the ability to consider a new perspective to the usual solipsistic "brain in a vat, reality is a 75-year long dream, etc." consipracy interpretations.

Last edited: Jan 28, 2013
14. Jan 28, 2013

### vanhees71

I don't see, where a two-level quantum system can contradict the Born interpretation. I've just tried to discuss a very nice experiment, concerning the uncertainty relation, done with neutron-spin measurements.

Unfortunately so far, nobody has answered to this. I'd really like to discuss this in view of different interpretations, because I think this is a pretty nice very fundamental example, which really has been measured in a real experiment!

Spin-1/2 experiments have always been the most simple examples for the application of Born's rule in good textbooks of quantum theory (e.g., Sakurai, Modern Quantum Mechanics, Addison-Wesley; J. J. Schwinger, Quantum Mechanics, Symbolism for Atomic Measurements, Springer). Nowadays, indeed one can do these experiments, and quantum theory in the minimal interpretation gives the right prediction of their outcome.

I'd have to analyze the cited paper in nature about the qu-bit in detail. I'd like to do that, if I'd only understood what you imply by your statement. Of course such a two-level system exists and nowadays one can measure many of those (spin-1/2 systems in Stern-Gerlach-like settings; polarization of single- and multi-photon states, including entanglement etc. etc.). Never has quantum theory denied the existence of such systems, and I don't know any interpretation which does so. Quantum theory is a theory about real objects in nature, what else should a physical theory be about?

The statement you quote from the video is, of course plain wrong. Let's take this example of the electron spin. An electron has $s=1/2$ and thus the spin-state space is two-dimensional. The standard basis is given by the eigenbasis of $\hat{\sigma}_z$ with the eigenvalues $\pm 1/2$.

Then there are only two sensible statements about the preparation of an electron's spin:

(1) the electron is prepared such that it is in a certain eigenstate of $\hat{\sigma}_z$, which is feasible since 1923 by using a Stern-Gerlach apparatus. Then this electron has a definite $\sigma_z$ component, taking the corresponding definitive value, i.e., either $+1/2$ or $-1/2$, never both values "at the same time". The corresponding eigenstates are, as it must be for eigenstates of a self-adjoint operator with different eigenvalues, orthogonal to each other, and thus the preparation in one of the spin-z eigenstates excludes it to take the other value.

(2) The electron is not prepared in a certain eigenstate of $\hat{\sigma}_z$. Then, in general, it's spin state is described by a statistical operator, i.e., a positive semidefinite hermitian matrix $\hat{R}$ with $\mathrm{Tr} \hat{R}=1.$ It can also be in a pure state, but that's included in the general case, because then you just set $\hat{R}=|\psi \rangle \langle \psi|.$ Then the electron has no definite $\sigma_z$ component. Measuring $\sigma_z$ has the probability to find either $\sigma_z=+1/2$ or $\sigma_z=-1/2$ with the probabilities given by Born's rule:
$$P_{R}(\sigma_z)=\langle \sigma_z|\hat{R}|\sigma_z \rangle.$$

In neither of the two cases, it makes the slightest sense to claim, the particle's spin-z-component has "both values a the same time". It's simply contradicting quantum theory, no matter which interpretation you adapt.

Last edited: Jan 28, 2013
15. Jan 28, 2013

### shounakbhatta

Thank you everyone for the replies. It has been really enlightening to go through the answers.

One question: the probability amplitude that we speak, is it the same amplitude that is the max.absolute value reached by the height of the wave?

Thanks.

16. Jan 28, 2013

### f95toli

Qubits have been around for a very long time now (I am not sure how long....25-30 years?)
and 2-state systems in general have been around since the birth of QM, a good example would be the Stern-Gerlach experiment (there is a reason for why most of the experiments used to characherize qubits bears the names of people like Rabi and Hahn).

There is nothing in this that has any )relevance at all when discussing interpretations.

17. Jan 28, 2013

### Maui

How so? Superpositions of probability amplitudes is not the same as superposition of actual, measureable states. Or did you not read the thread so far?

There are actual experiements that demonstrate quantum behavior at the macro scale and one was awared the Break-through of the year 2010 prize:

http://en.wikipedia.org/wiki/Quantum_machine

A very informative to the general audience video presentation of the experiment by the author:

If the above doesn't say anything about interpretations, then nothing has ever shed light at all and as far as i am concerned, the debb interpretation has already been falsified by demonstrating quantum superposition of states(i.e. that the pilot wave is a myth).

Last edited by a moderator: Sep 25, 2014
18. Jan 28, 2013

### f95toli

But all of these experiments can be interpreted within the framework of orthodox QM, or any interpretation you like.

And macroscopic quantum effects have also been around for ages (part of my PhD was on macroscopic quantum tunnelling in superconducting systems), so this again is nothing new.

19. Jan 28, 2013

### Maui

In historical plan the main motivation for postulating that the wavefunction isn't real has been the logical contradiction of having mutually exclusive properties at the same time. We are now seeing that that logical contradiction is actually part of how nature works at the bottom and every similar experiment in the last couple of years confirms it time after time. I'd say that the claim of the author of the standard interpretaion(and which is largely accepted as true) that there is no quantum world is, to put it mildly, weird given what is known today.

20. Jan 28, 2013

### bhobba

One of the fundamental axioms of quantum mechanics is that given any observable R, while we cannot predict what the outcome of a measurement will be we can determine its average or expectation value E(R). It is E(R) = Tr(PR) where P is a positive definite operator of unit trace. This P, by definition, is called the state of the system and given the state and an observable we can always determine the expected value of measurements using that observable.

States of the form |u><u| are called pure while the rest are called mixed. It can be shown that any mixed state is the sum of pure states - but not necessarily uniquely.

Another thing that can be shown is from the assumption the state of a system changes only infinitesimally in an infinitesimal amount of time after an observation (this is called the continuity assumption) the system, after the observation, will be in a pure state |u><u| where |u> is the eigenvector of the observable R associated with the actual outcome. Because of this in many texts, especially at the introductory level, they basically associate states with elements of a vector space. Strictly speaking it isn't - its a an operator but we all have to start somewhere.

If you want the correct detail see Ballentines excellent book - Quantum Mechanics - A Modern Development:
https://www.amazon.com/Quantum-Mechanics-Development-Leslie-Ballentine/dp/9810241054

Thanks
Bill

Last edited by a moderator: May 6, 2017