Commutator of the Hamiltonian and Parity Operator - evaluation of derivatives
I was studying the commutator of the Hamiltonian and parity operators in the $L^2$ space from Quantum mechanics and came upon the following:
To show that the two operators commute, assuming we have a symmetric potential $V(x)=V(-x)$, we had that the commutator of a function $psi(x)$ was;
begin{align*}[hat{H},hat{P}]=&-frac{hbar^2}{2m}frac{partial^2}{partial x^2}psi(-x)+V(x)psi(-x)+frac{hbar^2}{2m}frac{partial^2}{partial (-x)^2}psi(-x)-V(-x)psi(-x)\=&-frac{hbar^2}{2m}frac{partial^2}{partial x^2}psi(-x)+frac{hbar^2}{2m}frac{partial^2}{partial (-x)^2}psi(-x),
end{align*}
assuming the symmetry of the potential $V(x)$.
Now I struggled to show that the two second derivative terms are the negative of each other. The only possible way I could deduce the were equal was by using the chain rule for the second partial derivative to deduce that:
$$frac{partial psi(-x)}{partial (-x)}=frac{partial psi(-x)}{partial x}frac{partial x}{partial (-x)}=-frac{partial psi(-x)}{partial x},$$ and hence the partial derivative terms both cancel out to show the operators commute.
Now my main issue is struggling with the derivatives, could this be clarified?
As an aside I also had thought about the Lagrange and Leibniz notation, and was wondering whether the following was incorrect or not?
$$frac{partial psi(-x)}{partial x}=-psi '(-x) ,text{ and }, frac{partial psi(-x)}{partial (-x)}=psi'(-x).$$
derivatives quantum-mechanics
asked 2 days ago
user258521user258521
398312
add a comment |
I was studying the commutator of the Hamiltonian and parity operators in the $L^2$ space from Quantum mechanics and came upon the following:
To show that the two operators commute, assuming we have a symmetric potential $V(x)=V(-x)$, we had that the commutator of a function $psi(x)$ was;
begin{align*}[hat{H},hat{P}]=&-frac{hbar^2}{2m}frac{partial^2}{partial x^2}psi(-x)+V(x)psi(-x)+frac{hbar^2}{2m}frac{partial^2}{partial (-x)^2}psi(-x)-V(-x)psi(-x)\=&-frac{hbar^2}{2m}frac{partial^2}{partial x^2}psi(-x)+frac{hbar^2}{2m}frac{partial^2}{partial (-x)^2}psi(-x),
end{align*}
assuming the symmetry of the potential $V(x)$.
Now I struggled to show that the two second derivative terms are the negative of each other. The only possible way I could deduce the were equal was by using the chain rule for the second partial derivative to deduce that:
$$frac{partial psi(-x)}{partial (-x)}=frac{partial psi(-x)}{partial x}frac{partial x}{partial (-x)}=-frac{partial psi(-x)}{partial x},$$ and hence the partial derivative terms both cancel out to show the operators commute.
Now my main issue is struggling with the derivatives, could this be clarified?
As an aside I also had thought about the Lagrange and Leibniz notation, and was wondering whether the following was incorrect or not?
$$frac{partial psi(-x)}{partial x}=-psi '(-x) ,text{ and }, frac{partial psi(-x)}{partial (-x)}=psi'(-x).$$
derivatives quantum-mechanics
asked 2 days ago
user258521user258521
398312
The chain rule you applied is correct.
– Berci
2 days ago
@Berci do you have any idea on the last part of the post?
– user258521
2 days ago
Yes, that's correct too. Note that for any $u=u(x)$ we have $frac{partial f(u)}{partial u}=f'(u)$.
– Berci
2 days ago
I think I figured out the issue I was having with the derivatives ! If we apply the operators in the necessary order then it all sorts itself out... the main issue I was getting confused with was whether to differentiate the function first or evaluate it at $-x$, but this can be dealt with by carefully applying the operators correctly. As an example if $f(x)=x+x^2$ then differentiating first and evaluating at $-x$ gives $f'(-x)=1-2x$ whereas doing so in the reverse order gives $f'(-x)=-1+2x$ and so the order in which we apply these operations was the matter of my confusion.
– user258521
2 days ago
add a comment |
I was studying the commutator of the Hamiltonian and parity operators in the $L^2$ space from Quantum mechanics and came upon the following:
To show that the two operators commute, assuming we have a symmetric potential $V(x)=V(-x)$, we had that the commutator of a function $psi(x)$ was;
begin{align*}[hat{H},hat{P}]=&-frac{hbar^2}{2m}frac{partial^2}{partial x^2}psi(-x)+V(x)psi(-x)+frac{hbar^2}{2m}frac{partial^2}{partial (-x)^2}psi(-x)-V(-x)psi(-x)\=&-frac{hbar^2}{2m}frac{partial^2}{partial x^2}psi(-x)+frac{hbar^2}{2m}frac{partial^2}{partial (-x)^2}psi(-x),
end{align*}
assuming the symmetry of the potential $V(x)$.
Now I struggled to show that the two second derivative terms are the negative of each other. The only possible way I could deduce the were equal was by using the chain rule for the second partial derivative to deduce that:
$$frac{partial psi(-x)}{partial (-x)}=frac{partial psi(-x)}{partial x}frac{partial x}{partial (-x)}=-frac{partial psi(-x)}{partial x},$$ and hence the partial derivative terms both cancel out to show the operators commute.
Now my main issue is struggling with the derivatives, could this be clarified?
As an aside I also had thought about the Lagrange and Leibniz notation, and was wondering whether the following was incorrect or not?
$$frac{partial psi(-x)}{partial x}=-psi '(-x) ,text{ and }, frac{partial psi(-x)}{partial (-x)}=psi'(-x).$$
derivatives quantum-mechanics
asked 2 days ago
user258521user258521
398312
I was studying the commutator of the Hamiltonian and parity operators in the $L^2$ space from Quantum mechanics and came upon the following:
To show that the two operators commute, assuming we have a symmetric potential $V(x)=V(-x)$, we had that the commutator of a function $psi(x)$ was;
begin{align*}[hat{H},hat{P}]=&-frac{hbar^2}{2m}frac{partial^2}{partial x^2}psi(-x)+V(x)psi(-x)+frac{hbar^2}{2m}frac{partial^2}{partial (-x)^2}psi(-x)-V(-x)psi(-x)\=&-frac{hbar^2}{2m}frac{partial^2}{partial x^2}psi(-x)+frac{hbar^2}{2m}frac{partial^2}{partial (-x)^2}psi(-x),
end{align*}
assuming the symmetry of the potential $V(x)$.
Now I struggled to show that the two second derivative terms are the negative of each other. The only possible way I could deduce the were equal was by using the chain rule for the second partial derivative to deduce that:
$$frac{partial psi(-x)}{partial (-x)}=frac{partial psi(-x)}{partial x}frac{partial x}{partial (-x)}=-frac{partial psi(-x)}{partial x},$$ and hence the partial derivative terms both cancel out to show the operators commute.
Now my main issue is struggling with the derivatives, could this be clarified?
As an aside I also had thought about the Lagrange and Leibniz notation, and was wondering whether the following was incorrect or not?
$$frac{partial psi(-x)}{partial x}=-psi '(-x) ,text{ and }, frac{partial psi(-x)}{partial (-x)}=psi'(-x).$$
derivatives quantum-mechanics
derivatives quantum-mechanics
asked 2 days ago
user258521user258521
398312
asked 2 days ago
user258521user258521
398312
asked 2 days ago
user258521user258521
398312
asked 2 days ago
user258521user258521
398312
asked 2 days ago
user258521user258521
398312
398312
The chain rule you applied is correct.
– Berci
2 days ago
@Berci do you have any idea on the last part of the post?
– user258521
2 days ago
Yes, that's correct too. Note that for any $u=u(x)$ we have $frac{partial f(u)}{partial u}=f'(u)$.
– Berci
2 days ago
I think I figured out the issue I was having with the derivatives ! If we apply the operators in the necessary order then it all sorts itself out... the main issue I was getting confused with was whether to differentiate the function first or evaluate it at $-x$, but this can be dealt with by carefully applying the operators correctly. As an example if $f(x)=x+x^2$ then differentiating first and evaluating at $-x$ gives $f'(-x)=1-2x$ whereas doing so in the reverse order gives $f'(-x)=-1+2x$ and so the order in which we apply these operations was the matter of my confusion.
– user258521
2 days ago
add a comment |
The chain rule you applied is correct.
– Berci
2 days ago
@Berci do you have any idea on the last part of the post?
– user258521
2 days ago
Yes, that's correct too. Note that for any $u=u(x)$ we have $frac{partial f(u)}{partial u}=f'(u)$.
– Berci
2 days ago
I think I figured out the issue I was having with the derivatives ! If we apply the operators in the necessary order then it all sorts itself out... the main issue I was getting confused with was whether to differentiate the function first or evaluate it at $-x$, but this can be dealt with by carefully applying the operators correctly. As an example if $f(x)=x+x^2$ then differentiating first and evaluating at $-x$ gives $f'(-x)=1-2x$ whereas doing so in the reverse order gives $f'(-x)=-1+2x$ and so the order in which we apply these operations was the matter of my confusion.
– user258521
2 days ago
The chain rule you applied is correct.
– Berci
2 days ago
The chain rule you applied is correct.
– Berci
2 days ago
@Berci do you have any idea on the last part of the post?
– user258521
2 days ago
@Berci do you have any idea on the last part of the post?
– user258521
2 days ago
Yes, that's correct too. Note that for any $u=u(x)$ we have $frac{partial f(u)}{partial u}=f'(u)$.
– Berci
2 days ago
Yes, that's correct too. Note that for any $u=u(x)$ we have $frac{partial f(u)}{partial u}=f'(u)$.
– Berci
2 days ago
I think I figured out the issue I was having with the derivatives ! If we apply the operators in the necessary order then it all sorts itself out... the main issue I was getting confused with was whether to differentiate the function first or evaluate it at $-x$, but this can be dealt with by carefully applying the operators correctly. As an example if $f(x)=x+x^2$ then differentiating first and evaluating at $-x$ gives $f'(-x)=1-2x$ whereas doing so in the reverse order gives $f'(-x)=-1+2x$ and so the order in which we apply these operations was the matter of my confusion.
– user258521
2 days ago
I think I figured out the issue I was having with the derivatives ! If we apply the operators in the necessary order then it all sorts itself out... the main issue I was getting confused with was whether to differentiate the function first or evaluate it at $-x$, but this can be dealt with by carefully applying the operators correctly. As an example if $f(x)=x+x^2$ then differentiating first and evaluating at $-x$ gives $f'(-x)=1-2x$ whereas doing so in the reverse order gives $f'(-x)=-1+2x$ and so the order in which we apply these operations was the matter of my confusion.
– user258521
2 days ago
add a comment |
0
active
oldest
votes
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: "69"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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%2fmath.stackexchange.com%2fquestions%2f3063092%2fcommutator-of-the-hamiltonian-and-parity-operator-evaluation-of-derivatives%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics 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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2fmath.stackexchange.com%2fquestions%2f3063092%2fcommutator-of-the-hamiltonian-and-parity-operator-evaluation-of-derivatives%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
The chain rule you applied is correct.
– Berci
2 days ago
@Berci do you have any idea on the last part of the post?
– user258521
2 days ago
Yes, that's correct too. Note that for any $u=u(x)$ we have $frac{partial f(u)}{partial u}=f'(u)$.
– Berci
2 days ago
I think I figured out the issue I was having with the derivatives ! If we apply the operators in the necessary order then it all sorts itself out... the main issue I was getting confused with was whether to differentiate the function first or evaluate it at $-x$, but this can be dealt with by carefully applying the operators correctly. As an example if $f(x)=x+x^2$ then differentiating first and evaluating at $-x$ gives $f'(-x)=1-2x$ whereas doing so in the reverse order gives $f'(-x)=-1+2x$ and so the order in which we apply these operations was the matter of my confusion.
– user258521
2 days ago