Energy measurement from position eigenstate
$begingroup$
Given that the eigenstates of the position operator can be written as $delta(x-x')$, and suppose we measure a particle in an infinite potential with walls at $x=0$ and $x=L$. I measure the particle to be in the position $x=L/2$, so the particle is in the eigenstate $ |x rangle = delta(x-L/2)$. Suppose now that I want to measure the energy of the particle. The eigenstates of the energy operator are given by:
$$ |psi_nrangle = sqrt{frac{2}{L}}sin left( frac{npi x}{L} right) $$
In order to measure energy I understand that I have to expand the original eigenstate in terms of the new energy eigenstates:
$$
|xrangle = sum|psi_nranglelanglepsi_n|xrangle
$$
where the probability of collapse into the eigenstate is given by:
$$
P_n = |langlepsi_n|xrangle|^2
$$
But now I sort of run into an issue. Sure then, I can say that:
$$
langlepsi_n|xrangle = int sqrt{frac{2}{L}}sin left( frac{npi x}{L} right)delta(x-L/2)dx
$$
and since
$$
int delta(x-x')f(x)dx = f(x')
$$
I can say
$$
langlepsi_n|xrangle = sqrt{frac{2}{L}}sin left( frac{npi }{2} right)
$$
and,
$$
P_n=|langlepsi_n|xrangle|^2 = frac{2}{L}sin^2 left( frac{npi}{2} right)
$$
I know that this means that all odd values of n are equally probably and all even values are not possible, but probability is supposed to be dimensionless, so what's happened here? What rookie error have I made?
quantum-mechanics homework-and-exercises energy potential quantum-measurement
edited 6 hours ago
Aaron Stevens
13.5k42250
asked 9 hours ago
monkeyofsciencemonkeyofscience
625
$endgroup$
|
show 6 more comments
$begingroup$
Given that the eigenstates of the position operator can be written as $delta(x-x')$, and suppose we measure a particle in an infinite potential with walls at $x=0$ and $x=L$. I measure the particle to be in the position $x=L/2$, so the particle is in the eigenstate $ |x rangle = delta(x-L/2)$. Suppose now that I want to measure the energy of the particle. The eigenstates of the energy operator are given by:
$$ |psi_nrangle = sqrt{frac{2}{L}}sin left( frac{npi x}{L} right) $$
In order to measure energy I understand that I have to expand the original eigenstate in terms of the new energy eigenstates:
$$
|xrangle = sum|psi_nranglelanglepsi_n|xrangle
$$
where the probability of collapse into the eigenstate is given by:
$$
P_n = |langlepsi_n|xrangle|^2
$$
But now I sort of run into an issue. Sure then, I can say that:
$$
langlepsi_n|xrangle = int sqrt{frac{2}{L}}sin left( frac{npi x}{L} right)delta(x-L/2)dx
$$
and since
$$
int delta(x-x')f(x)dx = f(x')
$$
I can say
$$
langlepsi_n|xrangle = sqrt{frac{2}{L}}sin left( frac{npi }{2} right)
$$
and,
$$
P_n=|langlepsi_n|xrangle|^2 = frac{2}{L}sin^2 left( frac{npi}{2} right)
$$
I know that this means that all odd values of n are equally probably and all even values are not possible, but probability is supposed to be dimensionless, so what's happened here? What rookie error have I made?
quantum-mechanics homework-and-exercises energy potential quantum-measurement
edited 6 hours ago
Aaron Stevens
13.5k42250
asked 9 hours ago
monkeyofsciencemonkeyofscience
625
$endgroup$
$begingroup$
Rookie. $|psirangle$ is dimensionless, but $|xrangle$ has dimension $L^{-1/2}$, as you may check from $langle x| x'rangle=delta(x-x')$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
That is... confusing kets with wave functions as you do makes the first few formulas you write magnificently meaningless... Recall $psi(x)=langle x| psirangle$, with dimension $L^{-1/2}$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
Gotcha. So would it be correct to say that $psi (x) = langle x|L/2 rangle$? And $|psirangle = |L/2rangle$? So then if my energy eigenstates are represented by $|phirangle$ then $P_n = |langle phi| L/2 rangle|^2$?
$endgroup$
– monkeyofscience
7 hours ago
1
$begingroup$
No, normalized w.f.s have dimension $L^{-1/2}$, but what you wrote has dimensions of 1/L, like a delta function. So that is your dimensional mismatch (your "And ...?"). You may try a truly normalized Gaussian w.f. with labile width, and take the width to zero in a limit, but the expressions you write in this comment are visibly inconsistent.
$endgroup$
– Cosmas Zachos
7 hours ago
$begingroup$
Dang it. So how on earth do I get the position eigenfunctions to a point where I can talk meaningfully about the probability of an energy eigenfunction?
$endgroup$
– monkeyofscience
7 hours ago
|
show 6 more comments
$begingroup$
Given that the eigenstates of the position operator can be written as $delta(x-x')$, and suppose we measure a particle in an infinite potential with walls at $x=0$ and $x=L$. I measure the particle to be in the position $x=L/2$, so the particle is in the eigenstate $ |x rangle = delta(x-L/2)$. Suppose now that I want to measure the energy of the particle. The eigenstates of the energy operator are given by:
$$ |psi_nrangle = sqrt{frac{2}{L}}sin left( frac{npi x}{L} right) $$
In order to measure energy I understand that I have to expand the original eigenstate in terms of the new energy eigenstates:
$$
|xrangle = sum|psi_nranglelanglepsi_n|xrangle
$$
where the probability of collapse into the eigenstate is given by:
$$
P_n = |langlepsi_n|xrangle|^2
$$
But now I sort of run into an issue. Sure then, I can say that:
$$
langlepsi_n|xrangle = int sqrt{frac{2}{L}}sin left( frac{npi x}{L} right)delta(x-L/2)dx
$$
and since
$$
int delta(x-x')f(x)dx = f(x')
$$
I can say
$$
langlepsi_n|xrangle = sqrt{frac{2}{L}}sin left( frac{npi }{2} right)
$$
and,
$$
P_n=|langlepsi_n|xrangle|^2 = frac{2}{L}sin^2 left( frac{npi}{2} right)
$$
I know that this means that all odd values of n are equally probably and all even values are not possible, but probability is supposed to be dimensionless, so what's happened here? What rookie error have I made?
quantum-mechanics homework-and-exercises energy potential quantum-measurement
edited 6 hours ago
Aaron Stevens
13.5k42250
asked 9 hours ago
monkeyofsciencemonkeyofscience
625
$endgroup$
Given that the eigenstates of the position operator can be written as $delta(x-x')$, and suppose we measure a particle in an infinite potential with walls at $x=0$ and $x=L$. I measure the particle to be in the position $x=L/2$, so the particle is in the eigenstate $ |x rangle = delta(x-L/2)$. Suppose now that I want to measure the energy of the particle. The eigenstates of the energy operator are given by:
$$ |psi_nrangle = sqrt{frac{2}{L}}sin left( frac{npi x}{L} right) $$
In order to measure energy I understand that I have to expand the original eigenstate in terms of the new energy eigenstates:
$$
|xrangle = sum|psi_nranglelanglepsi_n|xrangle
$$
where the probability of collapse into the eigenstate is given by:
$$
P_n = |langlepsi_n|xrangle|^2
$$
But now I sort of run into an issue. Sure then, I can say that:
$$
langlepsi_n|xrangle = int sqrt{frac{2}{L}}sin left( frac{npi x}{L} right)delta(x-L/2)dx
$$
and since
$$
int delta(x-x')f(x)dx = f(x')
$$
I can say
$$
langlepsi_n|xrangle = sqrt{frac{2}{L}}sin left( frac{npi }{2} right)
$$
and,
$$
P_n=|langlepsi_n|xrangle|^2 = frac{2}{L}sin^2 left( frac{npi}{2} right)
$$
I know that this means that all odd values of n are equally probably and all even values are not possible, but probability is supposed to be dimensionless, so what's happened here? What rookie error have I made?
quantum-mechanics homework-and-exercises energy potential quantum-measurement
quantum-mechanics homework-and-exercises energy potential quantum-measurement
edited 6 hours ago
Aaron Stevens
13.5k42250
asked 9 hours ago
monkeyofsciencemonkeyofscience
625
edited 6 hours ago
Aaron Stevens
13.5k42250
asked 9 hours ago
monkeyofsciencemonkeyofscience
625
edited 6 hours ago
Aaron Stevens
13.5k42250
edited 6 hours ago
Aaron Stevens
13.5k42250
edited 6 hours ago
Aaron Stevens
13.5k42250
13.5k42250
asked 9 hours ago
monkeyofsciencemonkeyofscience
625
asked 9 hours ago
monkeyofsciencemonkeyofscience
625
asked 9 hours ago
monkeyofsciencemonkeyofscience
625
625
$begingroup$
Rookie. $|psirangle$ is dimensionless, but $|xrangle$ has dimension $L^{-1/2}$, as you may check from $langle x| x'rangle=delta(x-x')$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
That is... confusing kets with wave functions as you do makes the first few formulas you write magnificently meaningless... Recall $psi(x)=langle x| psirangle$, with dimension $L^{-1/2}$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
Gotcha. So would it be correct to say that $psi (x) = langle x|L/2 rangle$? And $|psirangle = |L/2rangle$? So then if my energy eigenstates are represented by $|phirangle$ then $P_n = |langle phi| L/2 rangle|^2$?
$endgroup$
– monkeyofscience
7 hours ago
1
$begingroup$
No, normalized w.f.s have dimension $L^{-1/2}$, but what you wrote has dimensions of 1/L, like a delta function. So that is your dimensional mismatch (your "And ...?"). You may try a truly normalized Gaussian w.f. with labile width, and take the width to zero in a limit, but the expressions you write in this comment are visibly inconsistent.
$endgroup$
– Cosmas Zachos
7 hours ago
$begingroup$
Dang it. So how on earth do I get the position eigenfunctions to a point where I can talk meaningfully about the probability of an energy eigenfunction?
$endgroup$
– monkeyofscience
7 hours ago
|
show 6 more comments
$begingroup$
Rookie. $|psirangle$ is dimensionless, but $|xrangle$ has dimension $L^{-1/2}$, as you may check from $langle x| x'rangle=delta(x-x')$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
That is... confusing kets with wave functions as you do makes the first few formulas you write magnificently meaningless... Recall $psi(x)=langle x| psirangle$, with dimension $L^{-1/2}$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
Gotcha. So would it be correct to say that $psi (x) = langle x|L/2 rangle$? And $|psirangle = |L/2rangle$? So then if my energy eigenstates are represented by $|phirangle$ then $P_n = |langle phi| L/2 rangle|^2$?
$endgroup$
– monkeyofscience
7 hours ago
1
$begingroup$
No, normalized w.f.s have dimension $L^{-1/2}$, but what you wrote has dimensions of 1/L, like a delta function. So that is your dimensional mismatch (your "And ...?"). You may try a truly normalized Gaussian w.f. with labile width, and take the width to zero in a limit, but the expressions you write in this comment are visibly inconsistent.
$endgroup$
– Cosmas Zachos
7 hours ago
$begingroup$
Dang it. So how on earth do I get the position eigenfunctions to a point where I can talk meaningfully about the probability of an energy eigenfunction?
$endgroup$
– monkeyofscience
7 hours ago
$begingroup$
Rookie. $|psirangle$ is dimensionless, but $|xrangle$ has dimension $L^{-1/2}$, as you may check from $langle x| x'rangle=delta(x-x')$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
Rookie. $|psirangle$ is dimensionless, but $|xrangle$ has dimension $L^{-1/2}$, as you may check from $langle x| x'rangle=delta(x-x')$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
That is... confusing kets with wave functions as you do makes the first few formulas you write magnificently meaningless... Recall $psi(x)=langle x| psirangle$, with dimension $L^{-1/2}$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
That is... confusing kets with wave functions as you do makes the first few formulas you write magnificently meaningless... Recall $psi(x)=langle x| psirangle$, with dimension $L^{-1/2}$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
Gotcha. So would it be correct to say that $psi (x) = langle x|L/2 rangle$? And $|psirangle = |L/2rangle$? So then if my energy eigenstates are represented by $|phirangle$ then $P_n = |langle phi| L/2 rangle|^2$?
$endgroup$
– monkeyofscience
7 hours ago
$begingroup$
Gotcha. So would it be correct to say that $psi (x) = langle x|L/2 rangle$? And $|psirangle = |L/2rangle$? So then if my energy eigenstates are represented by $|phirangle$ then $P_n = |langle phi| L/2 rangle|^2$?
$endgroup$
– monkeyofscience
7 hours ago
1
1
$begingroup$
No, normalized w.f.s have dimension $L^{-1/2}$, but what you wrote has dimensions of 1/L, like a delta function. So that is your dimensional mismatch (your "And ...?"). You may try a truly normalized Gaussian w.f. with labile width, and take the width to zero in a limit, but the expressions you write in this comment are visibly inconsistent.
$endgroup$
– Cosmas Zachos
7 hours ago
$begingroup$
No, normalized w.f.s have dimension $L^{-1/2}$, but what you wrote has dimensions of 1/L, like a delta function. So that is your dimensional mismatch (your "And ...?"). You may try a truly normalized Gaussian w.f. with labile width, and take the width to zero in a limit, but the expressions you write in this comment are visibly inconsistent.
$endgroup$
– Cosmas Zachos
7 hours ago
$begingroup$
Dang it. So how on earth do I get the position eigenfunctions to a point where I can talk meaningfully about the probability of an energy eigenfunction?
$endgroup$
– monkeyofscience
7 hours ago
$begingroup$
Dang it. So how on earth do I get the position eigenfunctions to a point where I can talk meaningfully about the probability of an energy eigenfunction?
$endgroup$
– monkeyofscience
7 hours ago
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
The formula
$$
P_n = |langlepsi_n|psirangle|^2
$$
assumes that the pre-measurement state $|psirangle$ and the observable's eigenstates $|psi_nrangle$ are both normalized to be unit state-vectors. In other words, the formula for arbitrary non-zero state-vectors is
$$
P_n = frac{|langlepsi_n|psirangle|^2}{
langlepsi_n|psi_nrangle,langlepsi|psirangle}.
$$
Notice that this expression for $P_n$ is dimensionless by construction.
The problem with the case described in the OP is that $|xrangle=delta(x-L/2)$ is not normalizable: it does not belong to the Hilbert space, so it can't be used for the pre-measurement state $|psirangle$.
That's not a problem in principle, because real position-measurments don't have infinite precision, so the state after a real position-measurement would not be $|xrangle$. It would be some normalizable state-vector $|psirangle$ whose corresponding wavefunction is sharply concentrated near a particular position, but with a non-zero width that makes it normalizable.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21541
$endgroup$
1
$begingroup$
How does your equation for $P$ work if you switched the two vectors? i.e. for the position probability density we usually see of $|langle x|psirangle|^2$ ?
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens That's a great question. (You always ask great questions!) For the position probability density, we would use $P(x)=|langle x|psirangle|^2/langlepsi|psirangle$. The other factor in the denominator isn't included in this case, and the normalization is instead absorbed into the meaning of "density."
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
@AaronStevens Another way to think of it is that the possible outcomes of a measurement are represented by projection operators that project the originally-normalizable state onto a new normalizable state. In the case of a position measurement, we can project into a finite interval, but not to a point; but we can define a density that can be integrated over any finite interval to give the same result. Thanks for asking that question -- it needed to be addressed.
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
That all makes sense. It just seems odd to me that one is doable and one is not because $|langle x|psirangle|^2=langle psi|xranglelangle x |psirangle=langle x|psiranglelanglepsi|xrangle=|langlepsi|xrangle|^2$ But perhaps the issue is in the interpretations of how these values relate to measurements. Certainly the probability of finding the particle at any specified location is $0$, so the problem assumes an impossible thing already (and we don't have to think about normalization for that).
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens Yeah, that's a quirk of the notation. The order in which the inner product is written is not what matters; what matters is the roles of the two state-vectors. One of them is the pre-measurement state (which must be normalizable), and one of them is the post-measurement state (which must be normalizable if we want a probability, but can be non-normalizable if we're content with a density). The asymmetry in the meaning is not reflected in the notation. That's one advantage of expressing things in terms of projection operators; then the notation is more clearly asymmetric.
$endgroup$
– Chiral Anomaly
4 hours ago
|
show 4 more comments
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1 Answer
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1 Answer
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oldest
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oldest
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active
oldest
votes
$begingroup$
The formula
$$
P_n = |langlepsi_n|psirangle|^2
$$
assumes that the pre-measurement state $|psirangle$ and the observable's eigenstates $|psi_nrangle$ are both normalized to be unit state-vectors. In other words, the formula for arbitrary non-zero state-vectors is
$$
P_n = frac{|langlepsi_n|psirangle|^2}{
langlepsi_n|psi_nrangle,langlepsi|psirangle}.
$$
Notice that this expression for $P_n$ is dimensionless by construction.
The problem with the case described in the OP is that $|xrangle=delta(x-L/2)$ is not normalizable: it does not belong to the Hilbert space, so it can't be used for the pre-measurement state $|psirangle$.
That's not a problem in principle, because real position-measurments don't have infinite precision, so the state after a real position-measurement would not be $|xrangle$. It would be some normalizable state-vector $|psirangle$ whose corresponding wavefunction is sharply concentrated near a particular position, but with a non-zero width that makes it normalizable.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21541
$endgroup$
1
$begingroup$
How does your equation for $P$ work if you switched the two vectors? i.e. for the position probability density we usually see of $|langle x|psirangle|^2$ ?
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens That's a great question. (You always ask great questions!) For the position probability density, we would use $P(x)=|langle x|psirangle|^2/langlepsi|psirangle$. The other factor in the denominator isn't included in this case, and the normalization is instead absorbed into the meaning of "density."
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
@AaronStevens Another way to think of it is that the possible outcomes of a measurement are represented by projection operators that project the originally-normalizable state onto a new normalizable state. In the case of a position measurement, we can project into a finite interval, but not to a point; but we can define a density that can be integrated over any finite interval to give the same result. Thanks for asking that question -- it needed to be addressed.
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
That all makes sense. It just seems odd to me that one is doable and one is not because $|langle x|psirangle|^2=langle psi|xranglelangle x |psirangle=langle x|psiranglelanglepsi|xrangle=|langlepsi|xrangle|^2$ But perhaps the issue is in the interpretations of how these values relate to measurements. Certainly the probability of finding the particle at any specified location is $0$, so the problem assumes an impossible thing already (and we don't have to think about normalization for that).
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens Yeah, that's a quirk of the notation. The order in which the inner product is written is not what matters; what matters is the roles of the two state-vectors. One of them is the pre-measurement state (which must be normalizable), and one of them is the post-measurement state (which must be normalizable if we want a probability, but can be non-normalizable if we're content with a density). The asymmetry in the meaning is not reflected in the notation. That's one advantage of expressing things in terms of projection operators; then the notation is more clearly asymmetric.
$endgroup$
– Chiral Anomaly
4 hours ago
|
show 4 more comments
$begingroup$
The formula
$$
P_n = |langlepsi_n|psirangle|^2
$$
assumes that the pre-measurement state $|psirangle$ and the observable's eigenstates $|psi_nrangle$ are both normalized to be unit state-vectors. In other words, the formula for arbitrary non-zero state-vectors is
$$
P_n = frac{|langlepsi_n|psirangle|^2}{
langlepsi_n|psi_nrangle,langlepsi|psirangle}.
$$
Notice that this expression for $P_n$ is dimensionless by construction.
The problem with the case described in the OP is that $|xrangle=delta(x-L/2)$ is not normalizable: it does not belong to the Hilbert space, so it can't be used for the pre-measurement state $|psirangle$.
That's not a problem in principle, because real position-measurments don't have infinite precision, so the state after a real position-measurement would not be $|xrangle$. It would be some normalizable state-vector $|psirangle$ whose corresponding wavefunction is sharply concentrated near a particular position, but with a non-zero width that makes it normalizable.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21541
$endgroup$
1
$begingroup$
How does your equation for $P$ work if you switched the two vectors? i.e. for the position probability density we usually see of $|langle x|psirangle|^2$ ?
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens That's a great question. (You always ask great questions!) For the position probability density, we would use $P(x)=|langle x|psirangle|^2/langlepsi|psirangle$. The other factor in the denominator isn't included in this case, and the normalization is instead absorbed into the meaning of "density."
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
@AaronStevens Another way to think of it is that the possible outcomes of a measurement are represented by projection operators that project the originally-normalizable state onto a new normalizable state. In the case of a position measurement, we can project into a finite interval, but not to a point; but we can define a density that can be integrated over any finite interval to give the same result. Thanks for asking that question -- it needed to be addressed.
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
That all makes sense. It just seems odd to me that one is doable and one is not because $|langle x|psirangle|^2=langle psi|xranglelangle x |psirangle=langle x|psiranglelanglepsi|xrangle=|langlepsi|xrangle|^2$ But perhaps the issue is in the interpretations of how these values relate to measurements. Certainly the probability of finding the particle at any specified location is $0$, so the problem assumes an impossible thing already (and we don't have to think about normalization for that).
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens Yeah, that's a quirk of the notation. The order in which the inner product is written is not what matters; what matters is the roles of the two state-vectors. One of them is the pre-measurement state (which must be normalizable), and one of them is the post-measurement state (which must be normalizable if we want a probability, but can be non-normalizable if we're content with a density). The asymmetry in the meaning is not reflected in the notation. That's one advantage of expressing things in terms of projection operators; then the notation is more clearly asymmetric.
$endgroup$
– Chiral Anomaly
4 hours ago
|
show 4 more comments
$begingroup$
The formula
$$
P_n = |langlepsi_n|psirangle|^2
$$
assumes that the pre-measurement state $|psirangle$ and the observable's eigenstates $|psi_nrangle$ are both normalized to be unit state-vectors. In other words, the formula for arbitrary non-zero state-vectors is
$$
P_n = frac{|langlepsi_n|psirangle|^2}{
langlepsi_n|psi_nrangle,langlepsi|psirangle}.
$$
Notice that this expression for $P_n$ is dimensionless by construction.
The problem with the case described in the OP is that $|xrangle=delta(x-L/2)$ is not normalizable: it does not belong to the Hilbert space, so it can't be used for the pre-measurement state $|psirangle$.
That's not a problem in principle, because real position-measurments don't have infinite precision, so the state after a real position-measurement would not be $|xrangle$. It would be some normalizable state-vector $|psirangle$ whose corresponding wavefunction is sharply concentrated near a particular position, but with a non-zero width that makes it normalizable.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21541
$endgroup$
The formula
$$
P_n = |langlepsi_n|psirangle|^2
$$
assumes that the pre-measurement state $|psirangle$ and the observable's eigenstates $|psi_nrangle$ are both normalized to be unit state-vectors. In other words, the formula for arbitrary non-zero state-vectors is
$$
P_n = frac{|langlepsi_n|psirangle|^2}{
langlepsi_n|psi_nrangle,langlepsi|psirangle}.
$$
Notice that this expression for $P_n$ is dimensionless by construction.
The problem with the case described in the OP is that $|xrangle=delta(x-L/2)$ is not normalizable: it does not belong to the Hilbert space, so it can't be used for the pre-measurement state $|psirangle$.
That's not a problem in principle, because real position-measurments don't have infinite precision, so the state after a real position-measurement would not be $|xrangle$. It would be some normalizable state-vector $|psirangle$ whose corresponding wavefunction is sharply concentrated near a particular position, but with a non-zero width that makes it normalizable.
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21541
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21541
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21541
answered 8 hours ago
Chiral AnomalyChiral Anomaly
12.4k21541
12.4k21541
1
$begingroup$
How does your equation for $P$ work if you switched the two vectors? i.e. for the position probability density we usually see of $|langle x|psirangle|^2$ ?
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens That's a great question. (You always ask great questions!) For the position probability density, we would use $P(x)=|langle x|psirangle|^2/langlepsi|psirangle$. The other factor in the denominator isn't included in this case, and the normalization is instead absorbed into the meaning of "density."
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
@AaronStevens Another way to think of it is that the possible outcomes of a measurement are represented by projection operators that project the originally-normalizable state onto a new normalizable state. In the case of a position measurement, we can project into a finite interval, but not to a point; but we can define a density that can be integrated over any finite interval to give the same result. Thanks for asking that question -- it needed to be addressed.
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
That all makes sense. It just seems odd to me that one is doable and one is not because $|langle x|psirangle|^2=langle psi|xranglelangle x |psirangle=langle x|psiranglelanglepsi|xrangle=|langlepsi|xrangle|^2$ But perhaps the issue is in the interpretations of how these values relate to measurements. Certainly the probability of finding the particle at any specified location is $0$, so the problem assumes an impossible thing already (and we don't have to think about normalization for that).
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens Yeah, that's a quirk of the notation. The order in which the inner product is written is not what matters; what matters is the roles of the two state-vectors. One of them is the pre-measurement state (which must be normalizable), and one of them is the post-measurement state (which must be normalizable if we want a probability, but can be non-normalizable if we're content with a density). The asymmetry in the meaning is not reflected in the notation. That's one advantage of expressing things in terms of projection operators; then the notation is more clearly asymmetric.
$endgroup$
– Chiral Anomaly
4 hours ago
|
show 4 more comments
1
$begingroup$
How does your equation for $P$ work if you switched the two vectors? i.e. for the position probability density we usually see of $|langle x|psirangle|^2$ ?
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens That's a great question. (You always ask great questions!) For the position probability density, we would use $P(x)=|langle x|psirangle|^2/langlepsi|psirangle$. The other factor in the denominator isn't included in this case, and the normalization is instead absorbed into the meaning of "density."
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
@AaronStevens Another way to think of it is that the possible outcomes of a measurement are represented by projection operators that project the originally-normalizable state onto a new normalizable state. In the case of a position measurement, we can project into a finite interval, but not to a point; but we can define a density that can be integrated over any finite interval to give the same result. Thanks for asking that question -- it needed to be addressed.
$endgroup$
– Chiral Anomaly
4 hours ago
1
$begingroup$
That all makes sense. It just seems odd to me that one is doable and one is not because $|langle x|psirangle|^2=langle psi|xranglelangle x |psirangle=langle x|psiranglelanglepsi|xrangle=|langlepsi|xrangle|^2$ But perhaps the issue is in the interpretations of how these values relate to measurements. Certainly the probability of finding the particle at any specified location is $0$, so the problem assumes an impossible thing already (and we don't have to think about normalization for that).
$endgroup$
– Aaron Stevens
4 hours ago
1
$begingroup$
@AaronStevens Yeah, that's a quirk of the notation. The order in which the inner product is written is not what matters; what matters is the roles of the two state-vectors. One of them is the pre-measurement state (which must be normalizable), and one of them is the post-measurement state (which must be normalizable if we want a probability, but can be non-normalizable if we're content with a density). The asymmetry in the meaning is not reflected in the notation. That's one advantage of expressing things in terms of projection operators; then the notation is more clearly asymmetric.
$endgroup$
– Chiral Anomaly
4 hours ago
1
1
$begingroup$
How does your equation for $P$ work if you switched the two vectors? i.e. for the position probability density we usually see of $|langle x|psirangle|^2$ ?
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
How does your equation for $P$ work if you switched the two vectors? i.e. for the position probability density we usually see of $|langle x|psirangle|^2$ ?
$endgroup$
– Aaron Stevens
4 hours ago
1
1
$begingroup$
@AaronStevens That's a great question. (You always ask great questions!) For the position probability density, we would use $P(x)=|langle x|psirangle|^2/langlepsi|psirangle$. The other factor in the denominator isn't included in this case, and the normalization is instead absorbed into the meaning of "density."
$endgroup$
– Chiral Anomaly
4 hours ago
$begingroup$
@AaronStevens That's a great question. (You always ask great questions!) For the position probability density, we would use $P(x)=|langle x|psirangle|^2/langlepsi|psirangle$. The other factor in the denominator isn't included in this case, and the normalization is instead absorbed into the meaning of "density."
$endgroup$
– Chiral Anomaly
4 hours ago
1
1
$begingroup$
@AaronStevens Another way to think of it is that the possible outcomes of a measurement are represented by projection operators that project the originally-normalizable state onto a new normalizable state. In the case of a position measurement, we can project into a finite interval, but not to a point; but we can define a density that can be integrated over any finite interval to give the same result. Thanks for asking that question -- it needed to be addressed.
$endgroup$
– Chiral Anomaly
4 hours ago
$begingroup$
@AaronStevens Another way to think of it is that the possible outcomes of a measurement are represented by projection operators that project the originally-normalizable state onto a new normalizable state. In the case of a position measurement, we can project into a finite interval, but not to a point; but we can define a density that can be integrated over any finite interval to give the same result. Thanks for asking that question -- it needed to be addressed.
$endgroup$
– Chiral Anomaly
4 hours ago
1
1
$begingroup$
That all makes sense. It just seems odd to me that one is doable and one is not because $|langle x|psirangle|^2=langle psi|xranglelangle x |psirangle=langle x|psiranglelanglepsi|xrangle=|langlepsi|xrangle|^2$ But perhaps the issue is in the interpretations of how these values relate to measurements. Certainly the probability of finding the particle at any specified location is $0$, so the problem assumes an impossible thing already (and we don't have to think about normalization for that).
$endgroup$
– Aaron Stevens
4 hours ago
$begingroup$
That all makes sense. It just seems odd to me that one is doable and one is not because $|langle x|psirangle|^2=langle psi|xranglelangle x |psirangle=langle x|psiranglelanglepsi|xrangle=|langlepsi|xrangle|^2$ But perhaps the issue is in the interpretations of how these values relate to measurements. Certainly the probability of finding the particle at any specified location is $0$, so the problem assumes an impossible thing already (and we don't have to think about normalization for that).
$endgroup$
– Aaron Stevens
4 hours ago
1
1
$begingroup$
@AaronStevens Yeah, that's a quirk of the notation. The order in which the inner product is written is not what matters; what matters is the roles of the two state-vectors. One of them is the pre-measurement state (which must be normalizable), and one of them is the post-measurement state (which must be normalizable if we want a probability, but can be non-normalizable if we're content with a density). The asymmetry in the meaning is not reflected in the notation. That's one advantage of expressing things in terms of projection operators; then the notation is more clearly asymmetric.
$endgroup$
– Chiral Anomaly
4 hours ago
$begingroup$
@AaronStevens Yeah, that's a quirk of the notation. The order in which the inner product is written is not what matters; what matters is the roles of the two state-vectors. One of them is the pre-measurement state (which must be normalizable), and one of them is the post-measurement state (which must be normalizable if we want a probability, but can be non-normalizable if we're content with a density). The asymmetry in the meaning is not reflected in the notation. That's one advantage of expressing things in terms of projection operators; then the notation is more clearly asymmetric.
$endgroup$
– Chiral Anomaly
4 hours ago
|
show 4 more comments
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$begingroup$
Rookie. $|psirangle$ is dimensionless, but $|xrangle$ has dimension $L^{-1/2}$, as you may check from $langle x| x'rangle=delta(x-x')$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
That is... confusing kets with wave functions as you do makes the first few formulas you write magnificently meaningless... Recall $psi(x)=langle x| psirangle$, with dimension $L^{-1/2}$.
$endgroup$
– Cosmas Zachos
8 hours ago
$begingroup$
Gotcha. So would it be correct to say that $psi (x) = langle x|L/2 rangle$? And $|psirangle = |L/2rangle$? So then if my energy eigenstates are represented by $|phirangle$ then $P_n = |langle phi| L/2 rangle|^2$?
$endgroup$
– monkeyofscience
7 hours ago
1
$begingroup$
No, normalized w.f.s have dimension $L^{-1/2}$, but what you wrote has dimensions of 1/L, like a delta function. So that is your dimensional mismatch (your "And ...?"). You may try a truly normalized Gaussian w.f. with labile width, and take the width to zero in a limit, but the expressions you write in this comment are visibly inconsistent.
$endgroup$
– Cosmas Zachos
7 hours ago
$begingroup$
Dang it. So how on earth do I get the position eigenfunctions to a point where I can talk meaningfully about the probability of an energy eigenfunction?
$endgroup$
– monkeyofscience
7 hours ago