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 3 hours ago
Aaron Stevens
13.5k42250
asked 5 hours ago
monkeyofsciencemonkeyofscience
574
$endgroup$
|
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 3 hours ago
Aaron Stevens
13.5k42250
asked 5 hours ago
monkeyofsciencemonkeyofscience
574
$endgroup$
$begingroup$
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
5 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
5 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
4 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
4 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
4 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 3 hours ago
Aaron Stevens
13.5k42250
asked 5 hours ago
monkeyofsciencemonkeyofscience
574
$endgroup$
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 3 hours ago
Aaron Stevens
13.5k42250
asked 5 hours ago
monkeyofsciencemonkeyofscience
574
edited 3 hours ago
Aaron Stevens
13.5k42250
asked 5 hours ago
monkeyofsciencemonkeyofscience
574
edited 3 hours ago
Aaron Stevens
13.5k42250
edited 3 hours ago
Aaron Stevens
13.5k42250
edited 3 hours ago
Aaron Stevens
13.5k42250
13.5k42250
asked 5 hours ago
monkeyofsciencemonkeyofscience
574
asked 5 hours ago
monkeyofsciencemonkeyofscience
574
asked 5 hours ago
monkeyofsciencemonkeyofscience
574
574
$begingroup$
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
5 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
5 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
4 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
4 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
4 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
5 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
5 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
4 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
4 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
4 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
5 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
5 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
5 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
5 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
4 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
4 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
4 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
4 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
4 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
4 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 5 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
$endgroup$
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
|
show 4 more comments
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f468279%2fenergy-measurement-from-position-eigenstate%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
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 5 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
$endgroup$
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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 5 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
$endgroup$
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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 5 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
$endgroup$
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 5 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
answered 5 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
answered 5 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
answered 5 hours ago
Chiral AnomalyChiral Anomaly
12.4k21540
12.4k21540
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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago
|
show 4 more comments
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f468279%2fenergy-measurement-from-position-eigenstate%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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
5 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
5 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
4 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
4 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
4 hours ago