
#1
Jan2613, 02:42 PM

P: 268

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



#2
Jan2613, 07:51 PM

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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
Jan2713, 01:18 AM

P: 268

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
Jan2713, 01:31 AM

P: 987

Quantum state
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
Jan2713, 01:47 AM

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The wavefunction encodes the statistics of whatever is being considered.
As andrien says, it's squaremodulus is a probability density function. From the wavefunction 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
Jan2713, 02:29 AM

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A state [itex]\psi\rangle[/itex] 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.
[tex]\psi(x) = \langle x  \psi \rangle[/tex] 



#7
Jan2713, 04:22 AM

P: 268

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



#8
Jan2713, 04:58 AM

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



#9
Jan2713, 10:03 AM

P: 343

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 δfunctions 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
Jan2713, 10:34 AM

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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 "shutupandcalculate 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 welldefined 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 [itex]\psi \rangle[/itex] 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 selfadjoint operators, defined on a dense subspace of Hilbert space. The possible values an observable can take is given by the spectrum of this selfadjoint 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 [itex]a[/itex] of an observable [itex]A[/itex], represented by the selfadjoint operator [itex]\hat{A}[/itex] is given by [tex]P_{\psi}(a)=\int \mathrm{d} \beta \langle{a,\beta}\psi \rangle^2,[/tex] where [itex]a,\beta \rangle[/itex] is a complete set of orthonormalized (generalized) eigenvectors of [itex]\hat{A}[/itex] for the spectral value (generalized eigenvalue) [itex]a[/itex], and [itex]\beta[/itex] 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 [itex]A_1,\ldots,A_n[/itex] of compatible observables which have only onedimensional common (generalized) eigenvectors [itex]a_1,\ldots,a_n[/itex]. Then the probability to measure a possible set of values [itex](a_1,\ldots,a_n)[/itex] is simply given by [tex]P_{\psi}(a_1,\ldots,a_n)=\langle a_1,\ldots,a_n\psi \rangle^2.[/tex] 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 [itex]A_2,\ldots,A_n[/itex] to get the probability distribution for [itex]A_1[/itex]: [tex]P_{\psi}(a_1)=\int \mathrm{d} a_2 \cdots \mathrm{d} a_{n} P_{\psi}(a_1,a_2,\ldots,a_n).[/tex] 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 nonlocal 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
Jan2713, 05:46 PM

P: 724

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 singleatom electron spin qubit in silicon" http://www.nature.com/nature/journal...ture11449.html Then this: "Landmark in quantum computing" http://www.youtube.com/watch?v=veDS0...ature=youtu.be 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. 



#12
Jan2813, 02:24 AM

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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! 



#13
Jan2813, 04:20 AM

P: 724

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 75year long dream, etc." consipracy interpretations. 



#14
Jan2813, 05:58 AM

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I don't see, where a twolevel quantum system can contradict the Born interpretation. I've just tried to discuss a very nice experiment, concerning the uncertainty relation, done with neutronspin measurements.
http://physicsforums.com/showthread....ty+disturbance 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! Spin1/2 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, AddisonWesley; 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 qubit in detail. I'd like to do that, if I'd only understood what you imply by your statement. Of course such a twolevel system exists and nowadays one can measure many of those (spin1/2 systems in SternGerlachlike settings; polarization of single and multiphoton 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 [itex]s=1/2[/itex] and thus the spinstate space is twodimensional. The standard basis is given by the eigenbasis of [itex]\hat{\sigma}_z[/itex] with the eigenvalues [itex]\pm 1/2[/itex]. 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 [itex]\hat{\sigma}_z[/itex], which is feasible since 1923 by using a SternGerlach apparatus. Then this electron has a definite [itex]\sigma_z[/itex] component, taking the corresponding definitive value, i.e., either [itex]+1/2[/itex] or [itex]1/2[/itex], never both values "at the same time". The corresponding eigenstates are, as it must be for eigenstates of a selfadjoint operator with different eigenvalues, orthogonal to each other, and thus the preparation in one of the spinz eigenstates excludes it to take the other value. (2) The electron is not prepared in a certain eigenstate of [itex]\hat{\sigma}_z[/itex]. Then, in general, it's spin state is described by a statistical operator, i.e., a positive semidefinite hermitian matrix [itex]\hat{R}[/itex] with [itex]\mathrm{Tr} \hat{R}=1.[/itex] It can also be in a pure state, but that's included in the general case, because then you just set [itex]\hat{R}=\psi \rangle \langle \psi.[/itex] Then the electron has no definite [itex]\sigma_z[/itex] component. Measuring [itex]\sigma_z[/itex] has the probability to find either [itex]\sigma_z=+1/2[/itex] or [itex]\sigma_z=1/2[/itex] with the probabilities given by Born's rule: [tex]P_{R}(\sigma_z)=\langle \sigma_z\hat{R}\sigma_z \rangle.[/tex] In neither of the two cases, it makes the slightest sense to claim, the particle's spinzcomponent has "both values a the same time". It's simply contradicting quantum theory, no matter which interpretation you adapt. 



#15
Jan2813, 10:15 AM

P: 268

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
Jan2813, 10:17 AM

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and 2state systems in general have been around since the birth of QM, a good example would be the SternGerlach 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
Jan2813, 12:49 PM

P: 724

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 Breakthrough 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: http://www.youtube.com/watch?v=b0xbyqngSA8 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). 



#18
Jan2813, 01:53 PM

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P: 2,200

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. 


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