# Mass and vectors in QM

1. Nov 7, 2013

### kye

Before measurements, quantum states are vectors in Hilbert space and there may not even be momentum or position chosen before wave function collapse or measurement. I'd like to understand how mass or the higgs field couple to wave function. Can higgs home in to the specific ray or vector in Hilbert space or does the higgs field only couple to it after collapse of the wave function (or measurement). If it is prior, then higgs has capability to navigate the vectors in Hilbert space and find the specific basis? (perhaps explaining why there is mass in the atoms even without being measured?) I understand that in gauge theory, masses can't be forced directly into the equations of the equation or the symmetry is destroyed. So higgs field were proposed to give masses to the properties, but how they exactly interact in Hilbert configuration space.

2. Nov 7, 2013

### dauto

The Higgs field doesn't couple to the wavefunction. It couples to the other fields - the electron fields for instance - giving the particles their masses. When applying Quantum Mechanics to field theory to obtain Quantum Field Theory, the fields play the role of the observable degrees of freedom (equivalent to the position and/or momenta in QM). In other words, the fields are operators, they are not states. Operators don't collapse - states do.

3. Nov 7, 2013

### kye

If so, why does QM has mass if the higgs field can't couple to the wave function.. unless QM is not complete model to integrate mass and one needs quantum field theory even for a single particle in double slit experiment? If so, then since the observable is not position but fields, then how do you treat a one body problem and mass in ordinary qm and its relationship to the double slit experiment for instance?

So before the collapse of the wave function (or related mechanism like decoherence),

4. Nov 7, 2013

### strangerep

You're confusing a general mathematical framework, i.e., quantum theory, with its application to a particular physical scenario (e.g., Newtonian dynamics obeying Galilean symmetry, or Special Relativity obeying Poincare symmetry).

So... general quantum theory does not "have mass". But quantum theory, applied to model realistic dynamical situations does, because both Newtonian and Einsteinian dynamics have a concept of mass. We're simply representing the dynamical variables as operators on some Hilbert space, following the prescription from general quantum theory.

BTW, which QM book(s) have you studied? Ballentine's textbook develops quite well the relationship I summarized above.

Treating that in ordinary QM (without fields) always involves some hand waving and interpretation. I don't like those treatments.

To do it properly, with fields instead of particles, requires proficiency in quantum field theory. I could tell you it's described in the book by Mandel & Wolf on quantum optics and optical coherence, but I'm reasonably sure you're not yet at that level. The basic idea is that events registered on a spatially extended detector (like a photo-sensitive screen) obey a probabilistic formula. I.e., the math of quantum field theory, applied to this case, predicts a certain probability for a given area $\delta A$ on the screen to register an event within a given time interval $\delta t$.

5. Nov 7, 2013

### kye

I read that mass in QM enters as a parameter that relates the momentum operator to the kinetic energy part of the Hamiltonian. Higgs field coupling to electron field details not necessary. Are you saying that it is not possible to apply QM to an electron in a hydrogen atom without using QFT and higgs field?

I know that QFT is applied when there is no fixed numbers of particles and it is moving close to speed of light, etc.

6. Nov 8, 2013

### strangerep

I have no idea how you got that from what I actually said.

The case of hydrogen (electron in a spherically symmetric potential) is discussed in every QM book I know of (and I note that you ignored my question about which QM books you've read).

QFT is of course necessary to achieve the best accuracy (e.g., Lamb shift).

7. Nov 8, 2013

### kye

I own dozens of pop-sci quantum and qft books and also have access to many qm/qft textbooks. You said that "So... general quantum theory does not "have mass". But quantum theory, applied to model realistic dynamical situations does, because both Newtonian and Einsteinian dynamics have a concept of mass.". I thought you meant that general qm didn't use mass, but dynamic situation like particle creation/annihilation, near light speed, etc. did. So what you meant was that qm theory doesn't have mass in the sense that geometry or algebra doesn't have mass. Of course we knew this. You continued "We're simply representing the dynamical variables as operators on some Hilbert space, following the prescription from general quantum theory." The reason I'm asking all this is because with the discovery of higgs.. it is certainty that mass came from the higgs field so I refocus my attention to how QM interact with it. So it doesn't, only QFT did. Or maybe we just don't focus on the detailed interactions of it or coupling in QM and just make mass as given like time. In other words, QM doesn't handle coupling details or dynamics (or just effective theory).

8. Nov 8, 2013

### BruceW

In simple QED I think the electron actually does have mass. And QED is a field theory. So it's not like an absolute rule that in QM particles have mass, but in QFT's particles don't have their own mass. Anyway, (I think) the problem comes along when we try to make a gauge-invariant, renormalisable electroweak field theory, it is not possible for the electron to have mass. And so we need this Higgs model to explain how the electron acquires its mass.

9. Nov 8, 2013

### kye

I understand the Higgs mechanisms and the doublet thing gauge invariance of it and not questioning it.

Isn't it that over 90% of electron mass came from special relativity (energy-mass equivalence), 10% or fewer involves the higgs. Now I'd like to know is how the higgs interact with an electron in between the emitter and detector in a double slit experiment. Before detection, the wave function as a ray or vector in Hilbert space is not yet collapsed so in conventional formalism, they fundamentally don't have positions (except Many world or bohmian's which let's ignore for now). But we can detect the mass at all times. So even though an electron doesn't have position before measurement yet it has mass, so could the higgs field somehow betray their presence by the screening effect giving mass to them or does the higgs field possess the property of superpositions too or maybe it doesn't obey it hence there is always mass in a quantum system before measurement unlike position. What is your comment.

10. Nov 8, 2013

### BruceW

I'm very new to this whole quantum field theory stuff. So bear with me. But I'm reading an amazing book - Quantum field theory by Mandl and Shaw. It's really good, introduces the subject gently. I'd recommend it if you are looking for good books.

So, anyway. I'm not sure what you mean 90% of electron mass comes from special relativity and 10% comes from the higgs... I'm pretty sure that 100% of the invariant mass of the electron comes from the higgs. But if we're talking about relativistic mass, then that is $p^2/c^2+m^2$ (where m is the invariant mass of the electron, and p is the relativistic momentum). So yes, the relativistic mass of the electron is not all from the higgs, some of it comes from the momentum of the electron.

About the higgs interacting with the electron. There's two separate concepts that are important (as far as I can tell). 1) the fact that there is a higgs field, and that it couples with the lepton field, means that the electrons have a specifically defined mass. 2) perturbations of the higgs field are higgs bosons. These are particles that can collide with the electron, just like how a photon can collide with an electron.

So due to 1) yes, even though the electron does not have a specific position, it does have a specific mass. Also, as the electron travels between emitter and detector, (I think) that it is good to imagine Feynman's sum-over-states, so yes there is some probability that the electron will collide with a higgs particle at some place along its path, but since we don't observe exactly what goes on, we must sum over all possible collisions that could have happened with higgs bosons, along all possible paths that the electron could have taken. So even though there is a higgs field, this does not allow us to know the precise position of the electron as if it was moving through some kind of bubble chamber. Also, important to note that the higgs field couples weakly with the electron (and most other particles I think), which is why it's presence is hard to detect. And for this reason, it does not collide with the electron very often.

11. Nov 9, 2013

### vanhees71

The electron mass is due to the Higgs mechanism. That's not true for hadrons, whose mass is to a large amound dynamically generated by the strong interaction.

Further you should not use the idea of an "relativistic mass". For very good reasons in HEP we use the mass always in the sense of invariant mass. What was known as "relativistic mass" in the very early days of relativity is simply the energy of the particle (divided by $c^2$).

12. Nov 15, 2013

### kye

I have questions about this "yes, even though the electron does not have a specific position, it does have a specific mass". When you weight molecules or atoms.. the electron has no position because they are smeared out probability yet they have mass. Can anyone give me other example where something has no positions yet have mass? And how does this get mapped in spacetime? How can spacetime locate no position but can detect mass? Please share how this is viewed or understood by the mainstream in the conceptual and mathematical aspect. Thanks.

13. Nov 16, 2013

### BruceW

I'm not totally sure what you mean. If we measure the position of 4 identically prepared electrons, we can get an answer like 1cm, 11cm, 7cm, 2cm, e.t.c. So even though each of these electrons was identically prepared, we get a different answer for their position each time. And that is why the electron is said to be spread out over space. But, if we take 4 identically prepared electrons and measure their mass, we will get 0.51MeV/c2 every single time. And this is true for any kind of state we prepare that electron to be in. So therefore, in our theory we specify that the electron has a definite mass, so that our theory matches experiment. For example, in the Dirac equation $(i \gamma^\mu \partial_\mu -m)\psi =0$ Automatically specifies that our electrons and positrons have a definite mass. Or, in the field-theoretic interpretation, that our field is made up of electrons and positrons which each have a definite mass.

14. Nov 16, 2013

### kye

in the dirac equations, mass is a parameter and didn't include interactions with the higgs field, isn't it?

anyway, my main question is that in conventional belief. Electrons don't have trajectories in the atoms, because if they do, they may lose electromagnetic energy and fall down to the nucleus, so the consensus is they are probability clouds and we can only describe them as probabilities. Hence there is no position in principle because no trajectories. This means at certain time, the electron is not a particle in the atom, now I'd like to understand how the mass can still exist in the times it is not a particle.. unless you mean the mass becomes energy (from e=mc^2) and this become part of the Hamiltonian of the wave, is this the explanation, pls. elaborate, thank you.

15. Nov 16, 2013

### atyy

Since mass is a parameter relating "momentum" and "energy", in quantum theory momentum is spatial frequency (1/wavelength) and energy is temporal frequency, so "mass" of a wave is a parameter in the relationship between frequency and wavelength. This is what the mass of a wave is, since in quantum theory a "particle" is a wave.

16. Nov 16, 2013

### kye

But they said you can't force the mass terms in the Schroedinger equations or it can destroy the gauge symmetry, that is the precise reason they proposed the higgs mechanism.. so if mass came from the higgs mechanics.. what is its relationship to the "mass" of a wave as "a parameter in the relationship between frequency and wavelength"? How do you put the two together or relate them?

17. Nov 16, 2013

### atyy

With the Higgs mechanism, you can have mass as a parameter between frequency and wavelength in the wave equation for a gauge field.

18. Nov 16, 2013

### BruceW

I am super-confused. I agree, the electron is spread out over space. but it still has mass. If you measure the position of a ground state electron, then in each experiment you will get a different value of the position. But if you measure the mass of the electron, then you always get the same value. Also I don't know what it means "at a certain time, the electron is not a particle in the atom" why not? for example, an electron in the ground state is just going to stay in that ground state, so it is a particle in the atom for a long time.

edit: yes, I'm saying the electron is a particle, even though it is spread out over space. This terminology makes sense in that you can measure the position of the electron as accurately as you want, and you will find it to be a point particle.

19. Nov 16, 2013

### Staff: Mentor

It means no such thing. The electron is in a bound state and therefore part of the atom even though its position is uncertain (it would be more accurate to say that its position is undefined until we measure it).

The Schrodinger equation that we use to calculate the probability of finding the electron at various locations starts with the electron in a bound state; the probabilities that it spits outs are the probabilities for an electron that is bound into the atom.

The bad news is that there's no way of really seeing this except to set up and solve the equations, and that requires paying some fairly serious mathematical dues - a year or so of study beyond introductory differential and integral calculus. The good news is that it's worth the effort - it's the difference between reading about a delicious meal and cooking and eating a delicious meal.

20. Nov 17, 2013

### kye

If you agree the electron is spread out over space, it doesn't have trajectory from a to b, it's like it's simultaneous in a and b.. a wave... note qm only gives probabilities of its positions (it doesn't show it has position before measurement as Bohm emphasized).. see this thread here