You're wrong. If you won't take my word for it (FWIW, I do have a PhD in quantum optics, although I am not employed as a physicist) go ask at physics.stackexchange.com or any other physics site.
In particular: "Any two electrons would trivially be coherent (in terms of spin) because they differ by a fixed phase angle," demonstrates a misunderstanding of the situation. You are assuming that the electrons share a pure quantum state, but if so, then the electrons are coherent (not just in terms of spin!) by definition. Sharing a pure quantum state is what quantum coherence means.
If the electrons are not coherent, then you cannot define a phase angle between them. (More technically, the phase angle is dependent on the microstate of the environment, which thermodynamics prevents us from knowing.)
Probably not without a better idea of your existing knowledge, but I'll have a go.
Consider a simple point particle in one dimension. The wavefunction for the system is psi(x) where x represents the one-dimensional position. If you have two point particles, and they are coherent, the wavefunction is psi(x,y) where x and y represent the positions of the two particles. (For simplicity I'm assuming the two particles are distinguishable.)
Any particular choice of wavefunction psi(x,y) is a possible quantum state for the entire system. You could, for example, have a wavefunction that is only nonzero when x = y; in this state, whenever you measure the position of the particles you'll always find they are the same.
If the particles are not coherent, then you have one wavefunction psi(x) and another phi(y). (Actually it's more complicated than that, properly speaking you should be using a density matrix at this point, but that's beyond the scope of a slashdot comment.)
In my second example, the even-numbered and odd-numbered electrons are definitely coherent with each other, since otherwise the result of measuring an even-numbered electron could not affect the result of measuring an odd-numbered one.
You don't need coherence for the double-slit experiment, unless you're using different sources for the two slits. Each particle interferes only with itself, so no coherence is necessary.
No, quantum coherence is not about the electrons all having the same quantum state as one another; it's about the system as a whole having a single quantum state.
An example of a coherent system would be one in which the electrons all have the same spin; say a 50% chance that they are all up and a 50% chance that they are all down, but zero chance that some are up and some are down. Another example would be a 50% chance that the odd-numbered electrons are up and the even-numbered onees down, a 50% chance they're the other way around.
If all coffee is hot enough to cause third degree burns (I'm not sure I believe this) then it should not be sold to people in cars. (It probably shouldn't be sold in styrofoam cups, either, even in-shop.) Don't you guys have any sort of OSH legislation?
Assuming for the sake of argument that this *is* true, people wouldn't buy coffee from the drive-through if they knew about it; they do, so they don't. So in the beach analogy it would be as if (almost) nobody had ever heard of seashells, and didn't know that they shouldn't run barefoot. (Still not a great analogy, because if you're on a beach you can wear shoes or walk carefully to eliminate the risk posed by seashells, whereas there's nothing you can do to make a cup of coffee adequately secure when in a car.)
However, I have been told that the McD coffee was (at the time) made deliberately hotter than coffee is normally served at, in order to cater to customers who picked it up at the drive-through but didn't drink it until they reached work. I don't know if this is true.
No reasonable person would expect coffee sold at a drive-through to be hot enough to cause serious burns, and no sensible person would knowingly buy such coffee. Liquids in cars are liable to spill now and then, no matter how much care you take.
By way of analogy, if you paved a footpath with broken glass, and someone fell down and seriously injured themselves, would it be their own fault for falling down or yours for using broken glass? After all, everybody knows that care must be taken to avoid falling down. (What they don't expect is for the consequences to be disproportionate.)
(Except that with physical access you can attack the hardware directly, e.g., rewriting the kernel code by modifying RAM content directly. But that requires somewhat more sophistication than installing a bootkit.)
If the exploit in question really had compromised secure boot, yes, it would be a hack - the whole point of secure boot is that it prevents the boot sector from being used as a way to elevate from administrator access to kernel-mode access. (And yes, in Windows 7 x64 the two are different, at least in theory.)
The primary purpose of secure boot is to prevent malware from installing rootkits. So unless you can change the UEFI configuration from software, it doesn't matter that the OS can't tell for sure whether it was securely booted or not.
However, if you do also want to defend against an attacker who has physical access to the machine, it is my understanding that secure boot can be combined with encryption in such a way that the encryption key is only provided if secure boot was successful. I think that addresses your concern?
If you don't trust your hardware manufacturer, buy from someone else. It should also be noted that the mainstream players have already asserted that their products will allow you to turn secure boot off.
From coreboot.org: "coreboot can scale from specialized applications that run directly from firmware, run operating systems in flash, load custom bootloaders, or implement firmware standards, like PC BIOS services or UEFI."
Thing is, that "small enough" group isn't necessarily going to stay small. Internet TV in particular presents the risk of a scenario where a large fraction of your users' connections are busy simultaneously during peak times.
Granted, for Internet TV you probably only need 2Mbs free-and-clear, not 5Mbs. That makes Bell's cut much more affordable, but I'm still not sure that the overall retail price would be low enough to be feasible.
If I've done my math right, then for Bell-based customers this works out as roughly 14.6GiB per dollar, or seven cents per gigabyte, assuming the network is always congested. The actual cost depends on the peak to off-peak traffic ratio and on how much congestion is considered acceptable, but this provides a minimum.
Folks who want, say, 5Mbs free-and-clear (no congestion and no data cap) would be paying Bell $110.65 per month plus a $14.11 access fee. That's more than I'd prefer to pay myself, but it isn't out of reach.
However, it isn't clear to me exactly what this is buying. I suspect it doesn't include actual internet connectivity, but is just what the retail ISP is paying for Bell to get the traffic from the customer to the ISP. So you need to add the ISPs internal costs, profit margin, any applicable taxes, and whatever wholesale internet rates the ISP pays. I strongly suspect that by the time you've added all this up, 5Mbs free-and-clear is still going to be too expensive for most people.
Slashdot Mirror
You're wrong. If you won't take my word for it (FWIW, I do have a PhD in quantum optics, although I am not employed as a physicist) go ask at physics.stackexchange.com or any other physics site.
In particular: "Any two electrons would trivially be coherent (in terms of spin) because they differ by a fixed phase angle," demonstrates a misunderstanding of the situation. You are assuming that the electrons share a pure quantum state, but if so, then the electrons are coherent (not just in terms of spin!) by definition. Sharing a pure quantum state is what quantum coherence means.
If the electrons are not coherent, then you cannot define a phase angle between them. (More technically, the phase angle is dependent on the microstate of the environment, which thermodynamics prevents us from knowing.)
Probably not without a better idea of your existing knowledge, but I'll have a go.
Consider a simple point particle in one dimension. The wavefunction for the system is psi(x) where x represents the one-dimensional position. If you have two point particles, and they are coherent, the wavefunction is psi(x,y) where x and y represent the positions of the two particles. (For simplicity I'm assuming the two particles are distinguishable.)
Any particular choice of wavefunction psi(x,y) is a possible quantum state for the entire system. You could, for example, have a wavefunction that is only nonzero when x = y; in this state, whenever you measure the position of the particles you'll always find they are the same.
If the particles are not coherent, then you have one wavefunction psi(x) and another phi(y). (Actually it's more complicated than that, properly speaking you should be using a density matrix at this point, but that's beyond the scope of a slashdot comment.)
Coherence is a prerequisite for entanglement.
In my second example, the even-numbered and odd-numbered electrons are definitely coherent with each other, since otherwise the result of measuring an even-numbered electron could not affect the result of measuring an odd-numbered one.
You don't need coherence for the double-slit experiment, unless you're using different sources for the two slits. Each particle interferes only with itself, so no coherence is necessary.
No, quantum coherence is not about the electrons all having the same quantum state as one another; it's about the system as a whole having a single quantum state.
An example of a coherent system would be one in which the electrons all have the same spin; say a 50% chance that they are all up and a 50% chance that they are all down, but zero chance that some are up and some are down. Another example would be a 50% chance that the odd-numbered electrons are up and the even-numbered onees down, a 50% chance they're the other way around.
Never mind, my mistake. I see what you mean now.
But you can't explain chemistry without quantum mechanics.
You hold it in your hand, right, that will work. Because cars never have to brake suddenly or get run into from behind.
If all coffee is hot enough to cause third degree burns (I'm not sure I believe this) then it should not be sold to people in cars. (It probably shouldn't be sold in styrofoam cups, either, even in-shop.) Don't you guys have any sort of OSH legislation?
Assuming for the sake of argument that this *is* true, people wouldn't buy coffee from the drive-through if they knew about it; they do, so they don't. So in the beach analogy it would be as if (almost) nobody had ever heard of seashells, and didn't know that they shouldn't run barefoot. (Still not a great analogy, because if you're on a beach you can wear shoes or walk carefully to eliminate the risk posed by seashells, whereas there's nothing you can do to make a cup of coffee adequately secure when in a car.)
However, I have been told that the McD coffee was (at the time) made deliberately hotter than coffee is normally served at, in order to cater to customers who picked it up at the drive-through but didn't drink it until they reached work. I don't know if this is true.
Out of curiosity, what exactly do you think URL means?
No reasonable person would expect coffee sold at a drive-through to be hot enough to cause serious burns, and no sensible person would knowingly buy such coffee. Liquids in cars are liable to spill now and then, no matter how much care you take.
By way of analogy, if you paved a footpath with broken glass, and someone fell down and seriously injured themselves, would it be their own fault for falling down or yours for using broken glass? After all, everybody knows that care must be taken to avoid falling down. (What they don't expect is for the consequences to be disproportionate.)
(Except that with physical access you can attack the hardware directly, e.g., rewriting the kernel code by modifying RAM content directly. But that requires somewhat more sophistication than installing a bootkit.)
If the exploit in question really had compromised secure boot, yes, it would be a hack - the whole point of secure boot is that it prevents the boot sector from being used as a way to elevate from administrator access to kernel-mode access. (And yes, in Windows 7 x64 the two are different, at least in theory.)
The primary purpose of secure boot is to prevent malware from installing rootkits. So unless you can change the UEFI configuration from software, it doesn't matter that the OS can't tell for sure whether it was securely booted or not.
However, if you do also want to defend against an attacker who has physical access to the machine, it is my understanding that secure boot can be combined with encryption in such a way that the encryption key is only provided if secure boot was successful. I think that addresses your concern?
Do you have a reference for this?
I think the exploit has nothing to do with secure boot at all, and the original reporter just made a mistake.
Does Windows even try to determine whether secure boot is enabled or not? What would be the point?
If you don't trust your hardware manufacturer, buy from someone else. It should also be noted that the mainstream players have already asserted that their products will allow you to turn secure boot off.
From coreboot.org: "coreboot can scale from specialized applications that run directly from firmware, run operating systems in flash, load custom bootloaders, or implement firmware standards, like PC BIOS services or UEFI."
Thing is, that "small enough" group isn't necessarily going to stay small. Internet TV in particular presents the risk of a scenario where a large fraction of your users' connections are busy simultaneously during peak times.
Granted, for Internet TV you probably only need 2Mbs free-and-clear, not 5Mbs. That makes Bell's cut much more affordable, but I'm still not sure that the overall retail price would be low enough to be feasible.
Since the hardware determines the bandwidth, I don't see your point. (Are you confusing bandwidth with data?)
But many people think they should be able to. :-)
If I've done my math right, then for Bell-based customers this works out as roughly 14.6GiB per dollar, or seven cents per gigabyte, assuming the network is always congested. The actual cost depends on the peak to off-peak traffic ratio and on how much congestion is considered acceptable, but this provides a minimum.
Folks who want, say, 5Mbs free-and-clear (no congestion and no data cap) would be paying Bell $110.65 per month plus a $14.11 access fee. That's more than I'd prefer to pay myself, but it isn't out of reach.
However, it isn't clear to me exactly what this is buying. I suspect it doesn't include actual internet connectivity, but is just what the retail ISP is paying for Bell to get the traffic from the customer to the ISP. So you need to add the ISPs internal costs, profit margin, any applicable taxes, and whatever wholesale internet rates the ISP pays. I strongly suspect that by the time you've added all this up, 5Mbs free-and-clear is still going to be too expensive for most people.
You store it up. No doubt there are all sorts of subtleties, but basically all you need is a frickin huge capacitor.