Entangled protons in a uniform magnetic field?

  • B
  • Thread starter docnet
  • Start date
  • #1
docnet
Gold Member
572
237
Summary:
two protons in a uniform magnetic field exchange a photon. are the two protons entangled?
In NMR, protons that are near each other exchange energy with each other by photon emission and absorption.

Proton A is spin down, and proton B is spin up. proton A emits a photon and becomes spin up, and proton B absorbs this photon and becomes spin down. have the two protons been entangled?
 

Answers and Replies

  • #2
anuttarasammyak
Gold Member
1,344
613
Presence of uniform magnetic field represented by Hamiltonian ##H_M## may disturb forming entanglement. Please do not take this amateur view seriously.
 
  • #3
hutchphd
Science Advisor
Homework Helper
3,969
3,139
Summary:: two protons in a uniform magnetic field exchange a photon. are the two protons entangled?

Proton A is spin down, and proton B is spin up. proton A emits a photon and becomes spin up, and proton B absorbs this photon and becomes spin down. have the two protons been entangled?
How do you isolate this system down to two particles? The radiofrequency "photons" in NMR actually saturate the field making the effective magnetic temperature very high (i.e. up and down proton states equally populated) The signal is emitted by the system to reach magnetic thermal equilibrium. The pairwise exchange you envision is very unlikely. To really consider this one need be much more specific and one ends up very likely with the usual simpler beam experiments (polarizers and Stern -Gerlach) but for protons which are harder to deal with.
 
  • #4
docnet
Gold Member
572
237
Presence of uniform magnetic field represented by Hamiltonian ##H_M## may disturb forming entanglement. Please do not take this amateur view seriously.
Could you please explain like I'm 5 years old how the magnetic field disturbs the entanglement? i thought the magnetic field causes the splitting of energy levels of spin states up and down, and causes spin precession (a degree of freedom unrelated to the two energy states)
 
  • #5
docnet
Gold Member
572
237
How do you isolate this system down to two particles? The radiofrequency "photons" in NMR actually saturate the field making the effective magnetic temperature very high (i.e. up and down proton states equally populated) The signal is emitted by the system to reach magnetic thermal equilibrium. The pairwise exchange you envision is very unlikely. To really consider this one need be much more specific and one ends up very likely with the usual simpler beam experiments (polarizers and Stern -Gerlach) but for protons which are harder to deal with.
It seems like NMR spectroscopists can create situations where over many identical molecules, there is an exchange of spin populations between two unique protons. i think the term "nuclear overhauser" describes this situation of photon exchange between neighboring spins that occupy different energy states. (the energy exchange is nowhere near 100% efficient)
 
  • #6
hutchphd
Science Advisor
Homework Helper
3,969
3,139
How do you know which two protons?
 
  • #7
docnet
Gold Member
572
237
How do you know which two protons?
the protons have characteristic Larmor frequencies that depends on the local magnetic field (unique location within the molecule) and can be identified that way. its worth mentioning that NMR is a "bulk" method (the spectrometer is a classical EM wave detector) and it can't detect a spin transition of one spin. there is no way of telling which proton, in which specific molecule is undergoing spin transition at a given time, but NMR measures the average effect over many molecules as "one" signal.

the spectroscopist selects specific protons by coding a classical radiofrequency pulse. the pulse (selectively) excites one population of protons by using a RF pulse of their Larmor frequency. then they selectively match the pair of nearby proton's characteristic Larmor frequencies (the delta E between spin up and spin down) to maximize the transfer of energy between them. honestly NMR theory seems so complicated.

edit 1: oops, sorry the last thing I mentioned about the selective transfer of energy can only happen between heteronuclei (like H and C, or H and N) because the pulse needs to be selective for the Larmor frequencies of one population. without this detail, my description at the end would not make sense

clarification: to selectively act on one population of nuclei, and leave the other population alone, the Larmor frequencies have to be far apart.
 
Last edited:
  • #10
Could you please explain like I'm 5 years old how the magnetic field disturbs the entanglement? i thought the magnetic field causes the splitting of energy levels of spin states up and down, and causes spin precession (a degree of freedom unrelated to the two energy states)
Greetings,

That is asking a lot, and I do not think it possible. So let us begin at the very beginning. Spin is quantized. Spin does not precess. The semi-classical vector diagrams that are so widely used show macroscopic magnetization, not spin. That might appear to be nitpicking, but it is not--it is basics.

The phenomena that appear to be under discussion can be generally characterized as cross-polarization and cross-relaxation phenomena.

Cross-polarization typically refers to heteronuclear systems, for example the effect on a rare spin population via manipulation of an abundant spin system: ##_{}^{13}\textrm{C}## and ##_{}^{1}\textrm{H}## respectively would be good examples. Alex Pines (University of California, Berkeley) has contributed a great deal to cross-polarization studies.

The Overhauser Effect is a cross-relaxation phenomenon. It is complicated. It can be applied to homonuclear systems as well as heteronuclear systems. It has also been used in two-dimensional NMR experiments. Richard Ernst (ETH Zurich) made pioneering contributions to these fields.

I realize that my reply does not address the original question regarding entanglement. But there were so many misconceptions in the posts above that I felt we needed to start somewhere.

Best regards,
ES
 
  • #11
hutchphd
Science Advisor
Homework Helper
3,969
3,139
This seems very unlikely to me, so I poked around Wikipedia (rather than thinking hard about it). My reservations are well encompassed by the article

https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance_quantum_computer

and in particular from it

More recent work, particularly by Caves and others, shows that all experiments in liquid state bulk ensemble NMR quantum computing to date do not possess quantum entanglement, thought to be required for quantum computation. Hence NMR quantum computing experiments are likely to have been only classical simulations of a quantum computer.[12]
 
  • Like
Likes anuttarasammyak and docnet
  • #12
docnet
Gold Member
572
237
Greetings,

That is asking a lot, and I do not think it possible. So let us begin at the very beginning. Spin is quantized. Spin does not precess. The semi-classical vector diagrams that are so widely used show macroscopic magnetization, not spin. That might appear to be nitpicking, but it is not--it is basics.
yes, spin is quantized but that fact isn't relevant because my question only refers to the spin 1/2 particles. The macroscopic magnetization you speak of (and which is manipulated by RF pulses in NMR) derives from the average expectation value of the ensemble of many identical spins. Can't the precession of spin be derived from the magnetic field hamiltonian in basic quantum mechanics?

The 3 components of spin cannot be measured simultaneously because the Pauli matrices do not commute with each other, so determining the precise precession axis isn't possible, but the classical vector diagram can still be used to visualize spin behavior as the vector represents the simultaneous behavior of many spins precessing in phase with each other. any thoughts about this?
 
  • #13
docnet
Gold Member
572
237
This seems very unlikely to me, so I poked around Wikipedia (rather than thinking hard about it). My reservations are well encompassed by the article

https://en.wikipedia.org/wiki/Nuclear_magnetic_resonance_quantum_computer

and in particular from it

More recent work, particularly by Caves and others, shows that all experiments in liquid state bulk ensemble NMR quantum computing to date do not possess quantum entanglement, thought to be required for quantum computation. Hence NMR quantum computing experiments are likely to have been only classical simulations of a quantum computer.[12]
It is consistent with the things I've learned about NMR (and quantum mechanics) so far. It is true that NMR spectrometers are classical measurement apparati because it detects they detect a classical alternating electric current. In principle, it is too dull a measurement tool for things like entanglement.
 
  • Skeptical
Likes EigenState137
  • #14
yes, spin is quantized but that fact isn't relevant because my question only refers to the spin 1/2 particles. The macroscopic magnetization you speak of (and which is manipulated by RF pulses in NMR) derives from the average expectation value of the ensemble of many identical spins. Can't the precession of spin be derived from the magnetic field hamiltonian in basic quantum mechanics?

The 3 components of spin cannot be measured simultaneously because the Pauli matrices do not commute with each other, so determining the precise precession axis isn't possible, but the classical vector diagram can still be used to visualize spin behavior as the vector represents the simultaneous behavior of many spins precessing in phase with each other. any thoughts about this?
Greetings,

I repeat, spin is quantized, it has no classical analog, it does not precess. There are two observables associated with spin: the magnitude ##S^{2}## and the projection onto a specified axis typically indicated as the z-direction ##S_{z}##. I have used the electron spin notation, but that is totally irrelevant.

The semi-classical vector diagrams are fine in simple cases, but they do not depict the spin, they depict the macroscopic magnetization. But those diagrams utterly fail when attempting to describe more complex situations such as stochastic excitation schemes or multi-dimensional NMR experiments.

In actuality, if you want to describe an ensemble, you should utilize density matrix formalism. But all of this is far beyond what is labelled as a Basic (B) thread.

Best regards,
ES
 
  • #15
hutchphd
Science Advisor
Homework Helper
3,969
3,139
But it is a very good system to study to learn the nuances of coherence. The spin-spin (lateral) coherence is very interesting and good to know. If you understand why it is not really a quantum bit, you will probably know a lot about quantum computing. (Honestly I could give an answer why but I'm not all that sure I understand it completely)
 
  • #16
anuttarasammyak
Gold Member
1,344
613
Revisiting my post #2, silicon qubits which come from spin-spin interaction hate magnetic fields environment.
 
  • #17
docnet
Gold Member
572
237
Greetings,

I repeat, spin is quantized, it has no classical analog, it does not precess. There are two observables associated with spin: the magnitude ##S^{2}## and the projection onto a specified axis typically indicated as the z-direction ##S_{z}##. I have used the electron spin notation, but that is totally irrelevant.

The semi-classical vector diagrams are fine in simple cases, but they do not depict the spin, they depict the macroscopic magnetization. But those diagrams utterly fail when attempting to describe more complex situations such as stochastic excitation schemes or multi-dimensional NMR experiments.

In actuality, if you want to describe an ensemble, you should utilize density matrix formalism. But all of this is far beyond what is labelled as a Basic (B) thread.

Best regards,
ES
I am sorry that this thread has become so confused. .

could you please link a reference that describes a spin does not precess in a magnetic field?

are you sure there are just two observables associated with spin, because you think the x and y components of the spin are not observables? and you say the spins do not behave like little magnetic dipoles?

what I learnd in basic quantum mechanics courses is that ##S## can be measured in any spatial direction. A repeated quantum measurement of of similarly prepared spins along a fixed axis gives a distribution of measurements ##\ket{up}## and ##\ket{down}## along that axis. The orientation of a spin angular momentum in 3D space is given from this distribution.

again, it seems that the description of the ensemble in NMR is obtained averaging the behavior of many spins and in theory removes information about quintessential quantum mechanical phenomena like entanglement and the uncertainty principle. The density matrix formalism used in NMR is no exception for what I described in previous posts because Average Hamiltonian Theory averages the behavior of a large number of spins. So in theory, NMR theory is not fundamental quantum mechanics, but a semi-classical version that is designed to be useful for doing NMR.

edit: i mean "observables" more generally in quantum mechanics. The set of NMR observables are different and not (directly) quantum mechanical, although the observables can be described by an adaptation called NMR theory

edit 2: NMR observables are really just classical EM waves that can be derived from QM.
 
  • #18
could you please link a reference that describes a spin does not precess in a magnetic field?

are you sure there are just two observables associated with spin, because you think the x and y components of the Greetings,spin are not observables? and you say the spins do not behave like little magnetic dipoles?
Greetings,

This is basic quantum mechanics as you can confirm with any decent text on the subject.

And yes i am certain that there are only two observables as stated above. This again can be easily confirmed from any decent text on quantum mechanics. Choose your favorite ones.

##S_{x}## and ##S_{y}## do not commute with ##S_{z}## and ##S^{2}##. The "orientation of a spin angular momentum in 3D space" is a meaningless concept.

NMR may in fact not not provide information on entanglement. To leap to the conclusion that the uncertainty principle does not apply to NMR spectroscopy is simply wrong, as is the conclusion that it is only semi-classical in nature. To provide evidence in support of those statements is far beyond the scope of this forum and a Basic thread in particular. Rather, I suggest that if the topic is of fundamental interest to you that you proceed well beyond introductory quantum mechanics and introductory magnetic resonance spectroscopy lest Bloch, Ernst, and Wüthrich need to surrender their Nobel Prizes.

Best regards,
ES
 
  • #19
36,657
14,482
i am certain that there are only two observables as stated above.
To be more precise, there are only two mutually commuting observables. One is ##S^2##, and the other is ##S_z##, where the ##z## direction is one particular direction in space about which you choose to measure spin. But you could pick that direction to be any direction you like; there are an infinite number of possible directions and each one has its own observable. The key point is that once you've picked one direction, its observable does not commute with the observables for any other direction.

are you sure there are just two observables associated with spin
You can only pick at most two mutually commuting observables, as described above.
 
  • Like
Likes EigenState137 and vanhees71
  • #20
vanhees71
Science Advisor
Insights Author
Gold Member
2021 Award
19,528
10,291
All three components of the spin are observables (within non-relativistic quantum theory, the relativistic case is more complicated). The corresponding self-adjoint operators fulfill the angular-momentum commutation relations (using natural units with ##\hbar=1##),
$$[\hat{s}_j,\hat{s}_k]=\mathrm{i} \epsilon_{jkl} \hat{s}_l.$$
All three components commute with ##\hat{\vec{s}}^2##.

There is a complete set of orthonormal common eigenvectors of any spin-component ##\hat{s}_{\vec{n}}=\vec{n} \cdot \hat{\vec{s}}## and ##\hat{\vec{s}}##, where ##\vec{n} \in \mathbb{R}^3## is an arbitrary unit vector. The possible eigenvalues are for ##\hat{\vec{s}}^2##: ##s(s+1)## with ##s \in \{0,1/2,1,3/2,\ldots \}##. For given ##s## the eigenvalues of the arbitrary component ##s_{\vec{n}}## are ##m_s \in \{-s,-s+1,\ldots,s-1,s \}##.

If the system is prepared in such a common eigenstate ##|s,m_s \rangle## any other spin component is in general indetermined and when measured you get one of the eigenvalues with probabilities given by Born's rule.

The figures drawn in some textbooks have to be interpreted in this and only this way within modern quantum theory. The idea of a "precession" comes from "old quantum mechanics" and is indeed not fully consistent.
 
  • Like
Likes EigenState137
  • #21
Greetings,

@PeterDonis and @vanhees71 Thank you for your elaborations and clarifications. My response was neither sufficiently explicit nor complete.

Best regards,
ES
 

Related Threads on Entangled protons in a uniform magnetic field?

Replies
7
Views
5K
Replies
35
Views
2K
Replies
15
Views
1K
  • Last Post
Replies
3
Views
725
Replies
5
Views
2K
Replies
8
Views
668
Replies
2
Views
3K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
17
Views
9K
  • Last Post
Replies
5
Views
11K
Top