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The location of an object is gauge dependent. Therefore, it's not measurable?
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The location of an object $x$ depends on how we choose our coordinate system. If we move the zero point, $x$ also changes. However, since we have translational invariance, we can always do such shifts without changing anything.
Now, in quantum theories, there is usually a lot of emphasis on quantities that are gauge independent. For example, the phase of a wave function in quantum mechanics can be changed using global $U(1)$ transformations and is therefore gauge dependent. So completely analogous to how we can shift the position of an object using translations, we can here shift here the phase of the wave function.
How are these two situations different? Since both, the location and the phase of the wave function, depend on how we choose our coordinate systems, they both shouldn't be measurable?!
gauge-theory gauge-invariance gauge
asked 2 hours ago
JakobH
3,01611245
add a comment |Â
up vote
1
down vote
favorite
The location of an object $x$ depends on how we choose our coordinate system. If we move the zero point, $x$ also changes. However, since we have translational invariance, we can always do such shifts without changing anything.
Now, in quantum theories, there is usually a lot of emphasis on quantities that are gauge independent. For example, the phase of a wave function in quantum mechanics can be changed using global $U(1)$ transformations and is therefore gauge dependent. So completely analogous to how we can shift the position of an object using translations, we can here shift here the phase of the wave function.
How are these two situations different? Since both, the location and the phase of the wave function, depend on how we choose our coordinate systems, they both shouldn't be measurable?!
gauge-theory gauge-invariance gauge
asked 2 hours ago
JakobH
3,01611245
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The location of an object $x$ depends on how we choose our coordinate system. If we move the zero point, $x$ also changes. However, since we have translational invariance, we can always do such shifts without changing anything.
Now, in quantum theories, there is usually a lot of emphasis on quantities that are gauge independent. For example, the phase of a wave function in quantum mechanics can be changed using global $U(1)$ transformations and is therefore gauge dependent. So completely analogous to how we can shift the position of an object using translations, we can here shift here the phase of the wave function.
How are these two situations different? Since both, the location and the phase of the wave function, depend on how we choose our coordinate systems, they both shouldn't be measurable?!
gauge-theory gauge-invariance gauge
asked 2 hours ago
JakobH
3,01611245
The location of an object $x$ depends on how we choose our coordinate system. If we move the zero point, $x$ also changes. However, since we have translational invariance, we can always do such shifts without changing anything.
Now, in quantum theories, there is usually a lot of emphasis on quantities that are gauge independent. For example, the phase of a wave function in quantum mechanics can be changed using global $U(1)$ transformations and is therefore gauge dependent. So completely analogous to how we can shift the position of an object using translations, we can here shift here the phase of the wave function.
How are these two situations different? Since both, the location and the phase of the wave function, depend on how we choose our coordinate systems, they both shouldn't be measurable?!
gauge-theory gauge-invariance gauge
gauge-theory gauge-invariance gauge
asked 2 hours ago
JakobH
3,01611245
asked 2 hours ago
JakobH
3,01611245
asked 2 hours ago
JakobH
3,01611245
asked 2 hours ago
JakobH
3,01611245
asked 2 hours ago
JakobH
3,01611245
3,01611245
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
The location of an object x depends on how we choose our coordinate system.
This is not true. The position of an object is invariant; it is the label which we assign to that position which depends on our coordinate system.
Take out a blank sheet of paper and draw a dot on it. Where is the dot? It seems like a bit of a non-answer, but a reasonable response would be "it is where it is."
This isn't particularly useful for computing things, so we typically decide to assign the point a numerical label. We can do this in a number of ways - we could choose a rectilinear grid, or we could choose a polar grid, or something more exotic; we could choose the coordinate origin to be in the center of the page, or we could choose the bottom left instead; and given any valid coordinate system, we can stretch or compress it to get a different one.
However, the dot that you drew on the paper hasn't moved - swapping labels around, which physicists do in our imaginations, has no effect on the world, or on any measurements we could perform in it. A coordinate system is simply a choice of labels for the observable quantity position.
How are these two situations different?
At this level, they aren't. Saying that the coordinates of a particular point are $(2,3)$ means absolutely nothing unless I specify a coordinate system, which essentially amounts to a choice of gauge. Similarly, saying that the global phase of a wavefunction is $pi/4$ is meaningless unless I establish a reference point of some kind. This wouldn't be impossible - I could demand that the wave function evaluated at $x=0$ be purely real at $t=0$, and then calculate the global phase of the wave function at any other time based on this reference point.
The difference lies in the fact that position is an observable quantity while the wave function is not. If it were somehow possible to ascertain the precise value of $psi(x)$, then we could set a reference point as mentioned above and define a meaningful notion of observable global phase. However, as we can only actually measure $|psi|^2$, we can't go backward to unambiguously determine $psi$, and so the global phase of a wave function does not have measurable physical content.
answered 1 hour ago
J. Murray
5,7282519
Thanks! What confuses me is that it’s usually argued that the phase of the wave function (or the wave function itself) are not observable BECAUSE they can be changed arbitrarily by U(1) transformations. However this does not seem to be true, given the above analogy with the location of an object.
– JakobH
16 mins ago
In some sense, I would even argue that the situation is even more analogous. We can measure the phase of a wave function if we provide a point of reference. In other words, we can measure the phase relative to the phase of another wave function. Analogously, we can only measure the location relative to the location of other objects.
– JakobH
14 mins ago
@JakobH From a top-down theory perspective, the (pure) states of a quantum system are elements of a projective Hilbert space, which means that changing the global phase corresponds to precisely the same physical state - this is why the phase is not measurable.
– J. Murray
2 mins ago
The U(1) invariance refers to global phase - multiplying the entire state by some $e^iphi$. Relative phases, e.g. between different points in space, are indeed measurable.
– J. Murray
41 secs ago
add a comment |Â
up vote
0
down vote
An experimentalist's answer:
One can always define an (x,y,z,t) for a particle measured in the lab, a point on a screen as a simple example and the start of a clock. Then one has to make sure that the same coordinates will be used for the next particle measured.
One can never define a phase for a single wavefunction and measure it. Measurement means to find a point in (x,y,z,t) where that wavefunction has a given phase and measure it: to get a probability distribution, which is the only measurable quantity for defining a wavefunction, one needs many measurements, there is no way to connect a single wavefunction by measurement to a single (x,y,z,t). One can measure the phase in connection with a second wavefunction and see the difference in the probability distribution of the superpositions.
answered 2 hours ago
anna v
152k7146436
Thanks! But Isn’t the fact that we can measure the „phase in connection with a second wavefunction“, in some sense, analogous to how we can only measure the location relative to the location of other objects?
– JakobH
11 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The location of an object x depends on how we choose our coordinate system.
This is not true. The position of an object is invariant; it is the label which we assign to that position which depends on our coordinate system.
Take out a blank sheet of paper and draw a dot on it. Where is the dot? It seems like a bit of a non-answer, but a reasonable response would be "it is where it is."
This isn't particularly useful for computing things, so we typically decide to assign the point a numerical label. We can do this in a number of ways - we could choose a rectilinear grid, or we could choose a polar grid, or something more exotic; we could choose the coordinate origin to be in the center of the page, or we could choose the bottom left instead; and given any valid coordinate system, we can stretch or compress it to get a different one.
However, the dot that you drew on the paper hasn't moved - swapping labels around, which physicists do in our imaginations, has no effect on the world, or on any measurements we could perform in it. A coordinate system is simply a choice of labels for the observable quantity position.
How are these two situations different?
At this level, they aren't. Saying that the coordinates of a particular point are $(2,3)$ means absolutely nothing unless I specify a coordinate system, which essentially amounts to a choice of gauge. Similarly, saying that the global phase of a wavefunction is $pi/4$ is meaningless unless I establish a reference point of some kind. This wouldn't be impossible - I could demand that the wave function evaluated at $x=0$ be purely real at $t=0$, and then calculate the global phase of the wave function at any other time based on this reference point.
The difference lies in the fact that position is an observable quantity while the wave function is not. If it were somehow possible to ascertain the precise value of $psi(x)$, then we could set a reference point as mentioned above and define a meaningful notion of observable global phase. However, as we can only actually measure $|psi|^2$, we can't go backward to unambiguously determine $psi$, and so the global phase of a wave function does not have measurable physical content.
answered 1 hour ago
J. Murray
5,7282519
Thanks! What confuses me is that it’s usually argued that the phase of the wave function (or the wave function itself) are not observable BECAUSE they can be changed arbitrarily by U(1) transformations. However this does not seem to be true, given the above analogy with the location of an object.
– JakobH
16 mins ago
In some sense, I would even argue that the situation is even more analogous. We can measure the phase of a wave function if we provide a point of reference. In other words, we can measure the phase relative to the phase of another wave function. Analogously, we can only measure the location relative to the location of other objects.
– JakobH
14 mins ago
@JakobH From a top-down theory perspective, the (pure) states of a quantum system are elements of a projective Hilbert space, which means that changing the global phase corresponds to precisely the same physical state - this is why the phase is not measurable.
– J. Murray
2 mins ago
The U(1) invariance refers to global phase - multiplying the entire state by some $e^iphi$. Relative phases, e.g. between different points in space, are indeed measurable.
– J. Murray
41 secs ago
add a comment |Â
up vote
3
down vote
The location of an object x depends on how we choose our coordinate system.
This is not true. The position of an object is invariant; it is the label which we assign to that position which depends on our coordinate system.
Take out a blank sheet of paper and draw a dot on it. Where is the dot? It seems like a bit of a non-answer, but a reasonable response would be "it is where it is."
This isn't particularly useful for computing things, so we typically decide to assign the point a numerical label. We can do this in a number of ways - we could choose a rectilinear grid, or we could choose a polar grid, or something more exotic; we could choose the coordinate origin to be in the center of the page, or we could choose the bottom left instead; and given any valid coordinate system, we can stretch or compress it to get a different one.
However, the dot that you drew on the paper hasn't moved - swapping labels around, which physicists do in our imaginations, has no effect on the world, or on any measurements we could perform in it. A coordinate system is simply a choice of labels for the observable quantity position.
How are these two situations different?
At this level, they aren't. Saying that the coordinates of a particular point are $(2,3)$ means absolutely nothing unless I specify a coordinate system, which essentially amounts to a choice of gauge. Similarly, saying that the global phase of a wavefunction is $pi/4$ is meaningless unless I establish a reference point of some kind. This wouldn't be impossible - I could demand that the wave function evaluated at $x=0$ be purely real at $t=0$, and then calculate the global phase of the wave function at any other time based on this reference point.
The difference lies in the fact that position is an observable quantity while the wave function is not. If it were somehow possible to ascertain the precise value of $psi(x)$, then we could set a reference point as mentioned above and define a meaningful notion of observable global phase. However, as we can only actually measure $|psi|^2$, we can't go backward to unambiguously determine $psi$, and so the global phase of a wave function does not have measurable physical content.
answered 1 hour ago
J. Murray
5,7282519
Thanks! What confuses me is that it’s usually argued that the phase of the wave function (or the wave function itself) are not observable BECAUSE they can be changed arbitrarily by U(1) transformations. However this does not seem to be true, given the above analogy with the location of an object.
– JakobH
16 mins ago
In some sense, I would even argue that the situation is even more analogous. We can measure the phase of a wave function if we provide a point of reference. In other words, we can measure the phase relative to the phase of another wave function. Analogously, we can only measure the location relative to the location of other objects.
– JakobH
14 mins ago
@JakobH From a top-down theory perspective, the (pure) states of a quantum system are elements of a projective Hilbert space, which means that changing the global phase corresponds to precisely the same physical state - this is why the phase is not measurable.
– J. Murray
2 mins ago
The U(1) invariance refers to global phase - multiplying the entire state by some $e^iphi$. Relative phases, e.g. between different points in space, are indeed measurable.
– J. Murray
41 secs ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The location of an object x depends on how we choose our coordinate system.
This is not true. The position of an object is invariant; it is the label which we assign to that position which depends on our coordinate system.
Take out a blank sheet of paper and draw a dot on it. Where is the dot? It seems like a bit of a non-answer, but a reasonable response would be "it is where it is."
This isn't particularly useful for computing things, so we typically decide to assign the point a numerical label. We can do this in a number of ways - we could choose a rectilinear grid, or we could choose a polar grid, or something more exotic; we could choose the coordinate origin to be in the center of the page, or we could choose the bottom left instead; and given any valid coordinate system, we can stretch or compress it to get a different one.
However, the dot that you drew on the paper hasn't moved - swapping labels around, which physicists do in our imaginations, has no effect on the world, or on any measurements we could perform in it. A coordinate system is simply a choice of labels for the observable quantity position.
How are these two situations different?
At this level, they aren't. Saying that the coordinates of a particular point are $(2,3)$ means absolutely nothing unless I specify a coordinate system, which essentially amounts to a choice of gauge. Similarly, saying that the global phase of a wavefunction is $pi/4$ is meaningless unless I establish a reference point of some kind. This wouldn't be impossible - I could demand that the wave function evaluated at $x=0$ be purely real at $t=0$, and then calculate the global phase of the wave function at any other time based on this reference point.
The difference lies in the fact that position is an observable quantity while the wave function is not. If it were somehow possible to ascertain the precise value of $psi(x)$, then we could set a reference point as mentioned above and define a meaningful notion of observable global phase. However, as we can only actually measure $|psi|^2$, we can't go backward to unambiguously determine $psi$, and so the global phase of a wave function does not have measurable physical content.
answered 1 hour ago
J. Murray
5,7282519
The location of an object x depends on how we choose our coordinate system.
This is not true. The position of an object is invariant; it is the label which we assign to that position which depends on our coordinate system.
Take out a blank sheet of paper and draw a dot on it. Where is the dot? It seems like a bit of a non-answer, but a reasonable response would be "it is where it is."
This isn't particularly useful for computing things, so we typically decide to assign the point a numerical label. We can do this in a number of ways - we could choose a rectilinear grid, or we could choose a polar grid, or something more exotic; we could choose the coordinate origin to be in the center of the page, or we could choose the bottom left instead; and given any valid coordinate system, we can stretch or compress it to get a different one.
However, the dot that you drew on the paper hasn't moved - swapping labels around, which physicists do in our imaginations, has no effect on the world, or on any measurements we could perform in it. A coordinate system is simply a choice of labels for the observable quantity position.
How are these two situations different?
At this level, they aren't. Saying that the coordinates of a particular point are $(2,3)$ means absolutely nothing unless I specify a coordinate system, which essentially amounts to a choice of gauge. Similarly, saying that the global phase of a wavefunction is $pi/4$ is meaningless unless I establish a reference point of some kind. This wouldn't be impossible - I could demand that the wave function evaluated at $x=0$ be purely real at $t=0$, and then calculate the global phase of the wave function at any other time based on this reference point.
The difference lies in the fact that position is an observable quantity while the wave function is not. If it were somehow possible to ascertain the precise value of $psi(x)$, then we could set a reference point as mentioned above and define a meaningful notion of observable global phase. However, as we can only actually measure $|psi|^2$, we can't go backward to unambiguously determine $psi$, and so the global phase of a wave function does not have measurable physical content.
answered 1 hour ago
J. Murray
5,7282519
answered 1 hour ago
J. Murray
5,7282519
answered 1 hour ago
J. Murray
5,7282519
answered 1 hour ago
J. Murray
5,7282519
5,7282519
Thanks! What confuses me is that it’s usually argued that the phase of the wave function (or the wave function itself) are not observable BECAUSE they can be changed arbitrarily by U(1) transformations. However this does not seem to be true, given the above analogy with the location of an object.
– JakobH
16 mins ago
In some sense, I would even argue that the situation is even more analogous. We can measure the phase of a wave function if we provide a point of reference. In other words, we can measure the phase relative to the phase of another wave function. Analogously, we can only measure the location relative to the location of other objects.
– JakobH
14 mins ago
@JakobH From a top-down theory perspective, the (pure) states of a quantum system are elements of a projective Hilbert space, which means that changing the global phase corresponds to precisely the same physical state - this is why the phase is not measurable.
– J. Murray
2 mins ago
The U(1) invariance refers to global phase - multiplying the entire state by some $e^iphi$. Relative phases, e.g. between different points in space, are indeed measurable.
– J. Murray
41 secs ago
add a comment |Â
Thanks! What confuses me is that it’s usually argued that the phase of the wave function (or the wave function itself) are not observable BECAUSE they can be changed arbitrarily by U(1) transformations. However this does not seem to be true, given the above analogy with the location of an object.
– JakobH
16 mins ago
In some sense, I would even argue that the situation is even more analogous. We can measure the phase of a wave function if we provide a point of reference. In other words, we can measure the phase relative to the phase of another wave function. Analogously, we can only measure the location relative to the location of other objects.
– JakobH
14 mins ago
@JakobH From a top-down theory perspective, the (pure) states of a quantum system are elements of a projective Hilbert space, which means that changing the global phase corresponds to precisely the same physical state - this is why the phase is not measurable.
– J. Murray
2 mins ago
The U(1) invariance refers to global phase - multiplying the entire state by some $e^iphi$. Relative phases, e.g. between different points in space, are indeed measurable.
– J. Murray
41 secs ago
Thanks! What confuses me is that it’s usually argued that the phase of the wave function (or the wave function itself) are not observable BECAUSE they can be changed arbitrarily by U(1) transformations. However this does not seem to be true, given the above analogy with the location of an object.
– JakobH
16 mins ago
Thanks! What confuses me is that it’s usually argued that the phase of the wave function (or the wave function itself) are not observable BECAUSE they can be changed arbitrarily by U(1) transformations. However this does not seem to be true, given the above analogy with the location of an object.
– JakobH
16 mins ago
In some sense, I would even argue that the situation is even more analogous. We can measure the phase of a wave function if we provide a point of reference. In other words, we can measure the phase relative to the phase of another wave function. Analogously, we can only measure the location relative to the location of other objects.
– JakobH
14 mins ago
In some sense, I would even argue that the situation is even more analogous. We can measure the phase of a wave function if we provide a point of reference. In other words, we can measure the phase relative to the phase of another wave function. Analogously, we can only measure the location relative to the location of other objects.
– JakobH
14 mins ago
@JakobH From a top-down theory perspective, the (pure) states of a quantum system are elements of a projective Hilbert space, which means that changing the global phase corresponds to precisely the same physical state - this is why the phase is not measurable.
– J. Murray
2 mins ago
@JakobH From a top-down theory perspective, the (pure) states of a quantum system are elements of a projective Hilbert space, which means that changing the global phase corresponds to precisely the same physical state - this is why the phase is not measurable.
– J. Murray
2 mins ago
The U(1) invariance refers to global phase - multiplying the entire state by some $e^iphi$. Relative phases, e.g. between different points in space, are indeed measurable.
– J. Murray
41 secs ago
The U(1) invariance refers to global phase - multiplying the entire state by some $e^iphi$. Relative phases, e.g. between different points in space, are indeed measurable.
– J. Murray
41 secs ago
add a comment |Â
up vote
0
down vote
An experimentalist's answer:
One can always define an (x,y,z,t) for a particle measured in the lab, a point on a screen as a simple example and the start of a clock. Then one has to make sure that the same coordinates will be used for the next particle measured.
One can never define a phase for a single wavefunction and measure it. Measurement means to find a point in (x,y,z,t) where that wavefunction has a given phase and measure it: to get a probability distribution, which is the only measurable quantity for defining a wavefunction, one needs many measurements, there is no way to connect a single wavefunction by measurement to a single (x,y,z,t). One can measure the phase in connection with a second wavefunction and see the difference in the probability distribution of the superpositions.
answered 2 hours ago
anna v
152k7146436
Thanks! But Isn’t the fact that we can measure the „phase in connection with a second wavefunction“, in some sense, analogous to how we can only measure the location relative to the location of other objects?
– JakobH
11 mins ago
add a comment |Â
up vote
0
down vote
An experimentalist's answer:
One can always define an (x,y,z,t) for a particle measured in the lab, a point on a screen as a simple example and the start of a clock. Then one has to make sure that the same coordinates will be used for the next particle measured.
One can never define a phase for a single wavefunction and measure it. Measurement means to find a point in (x,y,z,t) where that wavefunction has a given phase and measure it: to get a probability distribution, which is the only measurable quantity for defining a wavefunction, one needs many measurements, there is no way to connect a single wavefunction by measurement to a single (x,y,z,t). One can measure the phase in connection with a second wavefunction and see the difference in the probability distribution of the superpositions.
answered 2 hours ago
anna v
152k7146436
Thanks! But Isn’t the fact that we can measure the „phase in connection with a second wavefunction“, in some sense, analogous to how we can only measure the location relative to the location of other objects?
– JakobH
11 mins ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
An experimentalist's answer:
One can always define an (x,y,z,t) for a particle measured in the lab, a point on a screen as a simple example and the start of a clock. Then one has to make sure that the same coordinates will be used for the next particle measured.
One can never define a phase for a single wavefunction and measure it. Measurement means to find a point in (x,y,z,t) where that wavefunction has a given phase and measure it: to get a probability distribution, which is the only measurable quantity for defining a wavefunction, one needs many measurements, there is no way to connect a single wavefunction by measurement to a single (x,y,z,t). One can measure the phase in connection with a second wavefunction and see the difference in the probability distribution of the superpositions.
answered 2 hours ago
anna v
152k7146436
An experimentalist's answer:
One can always define an (x,y,z,t) for a particle measured in the lab, a point on a screen as a simple example and the start of a clock. Then one has to make sure that the same coordinates will be used for the next particle measured.
One can never define a phase for a single wavefunction and measure it. Measurement means to find a point in (x,y,z,t) where that wavefunction has a given phase and measure it: to get a probability distribution, which is the only measurable quantity for defining a wavefunction, one needs many measurements, there is no way to connect a single wavefunction by measurement to a single (x,y,z,t). One can measure the phase in connection with a second wavefunction and see the difference in the probability distribution of the superpositions.
answered 2 hours ago
anna v
152k7146436
answered 2 hours ago
anna v
152k7146436
answered 2 hours ago
anna v
152k7146436
answered 2 hours ago
anna v
152k7146436
152k7146436
Thanks! But Isn’t the fact that we can measure the „phase in connection with a second wavefunction“, in some sense, analogous to how we can only measure the location relative to the location of other objects?
– JakobH
11 mins ago
add a comment |Â
Thanks! But Isn’t the fact that we can measure the „phase in connection with a second wavefunction“, in some sense, analogous to how we can only measure the location relative to the location of other objects?
– JakobH
11 mins ago
Thanks! But Isn’t the fact that we can measure the „phase in connection with a second wavefunction“, in some sense, analogous to how we can only measure the location relative to the location of other objects?
– JakobH
11 mins ago
Thanks! But Isn’t the fact that we can measure the „phase in connection with a second wavefunction“, in some sense, analogous to how we can only measure the location relative to the location of other objects?
– JakobH
11 mins ago
add a comment |Â
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