Non-Relativistic Limit of Klein-Gordon Probability Density
$begingroup$
In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
begin{eqnarray}
P & = & dfrac{ihbar}{2mc^2}left(Phi^*dfrac{partialPhi}{partial t}-Phidfrac{partialPhi^*}{partial t}right) \
vec{j} &=& dfrac{hbar}{2mi}left(Phi^*vec{nabla}Phi-Phivec{nabla}Phi^*right)
end{eqnarray}
$$
together with the statement that:
One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.
The 'known' expressions are:
$$
begin{eqnarray}
rho &=& Psi^*Psi \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
When taking a 'non-relativistic limit', I am used to taking the limit $c to infty$, which does give the right result for $vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?
quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation
edited Dec 23 '18 at 21:37
Qmechanic♦
103k121861182
asked Dec 23 '18 at 21:21
SimonSimon
771313
$endgroup$
add a comment |
$begingroup$
In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
begin{eqnarray}
P & = & dfrac{ihbar}{2mc^2}left(Phi^*dfrac{partialPhi}{partial t}-Phidfrac{partialPhi^*}{partial t}right) \
vec{j} &=& dfrac{hbar}{2mi}left(Phi^*vec{nabla}Phi-Phivec{nabla}Phi^*right)
end{eqnarray}
$$
together with the statement that:
One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.
The 'known' expressions are:
$$
begin{eqnarray}
rho &=& Psi^*Psi \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
When taking a 'non-relativistic limit', I am used to taking the limit $c to infty$, which does give the right result for $vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?
quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation
edited Dec 23 '18 at 21:37
Qmechanic♦
103k121861182
asked Dec 23 '18 at 21:21
SimonSimon
771313
$endgroup$
1
$begingroup$
For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
$endgroup$
– Qmechanic♦
Dec 23 '18 at 21:39
add a comment |
$begingroup$
In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
begin{eqnarray}
P & = & dfrac{ihbar}{2mc^2}left(Phi^*dfrac{partialPhi}{partial t}-Phidfrac{partialPhi^*}{partial t}right) \
vec{j} &=& dfrac{hbar}{2mi}left(Phi^*vec{nabla}Phi-Phivec{nabla}Phi^*right)
end{eqnarray}
$$
together with the statement that:
One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.
The 'known' expressions are:
$$
begin{eqnarray}
rho &=& Psi^*Psi \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
When taking a 'non-relativistic limit', I am used to taking the limit $c to infty$, which does give the right result for $vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?
quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation
edited Dec 23 '18 at 21:37
Qmechanic♦
103k121861182
asked Dec 23 '18 at 21:21
SimonSimon
771313
$endgroup$
In the lecture notes accompanying an introductory course in relativistic quantum mechanics, the Klein-Gordon probability density and current are defined as:
$$
begin{eqnarray}
P & = & dfrac{ihbar}{2mc^2}left(Phi^*dfrac{partialPhi}{partial t}-Phidfrac{partialPhi^*}{partial t}right) \
vec{j} &=& dfrac{hbar}{2mi}left(Phi^*vec{nabla}Phi-Phivec{nabla}Phi^*right)
end{eqnarray}
$$
together with the statement that:
One can show that in the non-relativistic limit, the known expressions for the probability density and current are recovered.
The 'known' expressions are:
$$
begin{eqnarray}
rho &=& Psi^*Psi \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
When taking a 'non-relativistic limit', I am used to taking the limit $c to infty$, which does give the right result for $vec{j}$, but for the density produces $P=0$. How should one then take said limit to recover the non-relativistic equations?
quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation
quantum-mechanics wavefunction speed-of-light schroedinger-equation klein-gordon-equation
edited Dec 23 '18 at 21:37
Qmechanic♦
103k121861182
asked Dec 23 '18 at 21:21
SimonSimon
771313
edited Dec 23 '18 at 21:37
Qmechanic♦
103k121861182
asked Dec 23 '18 at 21:21
SimonSimon
771313
edited Dec 23 '18 at 21:37
Qmechanic♦
103k121861182
edited Dec 23 '18 at 21:37
Qmechanic♦
103k121861182
edited Dec 23 '18 at 21:37
Qmechanic♦
103k121861182
103k121861182
asked Dec 23 '18 at 21:21
SimonSimon
771313
asked Dec 23 '18 at 21:21
SimonSimon
771313
asked Dec 23 '18 at 21:21
SimonSimon
771313
771313
1
$begingroup$
For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
$endgroup$
– Qmechanic♦
Dec 23 '18 at 21:39
add a comment |
1
$begingroup$
For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
$endgroup$
– Qmechanic♦
Dec 23 '18 at 21:39
1
1
$begingroup$
For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
$endgroup$
– Qmechanic♦
Dec 23 '18 at 21:39
$begingroup$
For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
$endgroup$
– Qmechanic♦
Dec 23 '18 at 21:39
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The trick is to make the approach for the relativistic Klein-Gordon wave function
$$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
(The physical reasoning behind this approach is:
The fast oscillating exponential is the solution for the particle at rest.
And compared to that, $Psi$ will give only slow variations.)
From that you find its derivatives
$$
begin{eqnarray}
frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
end{eqnarray}
$$
Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
$$
begin{eqnarray}
P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.
answered Dec 24 '18 at 0:54
Thomas FritschThomas Fritsch
34929
$endgroup$
add a comment |
$begingroup$
You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.
answered Dec 24 '18 at 0:32
my2ctsmy2cts
4,9422618
$endgroup$
add a comment |
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%2f450076%2fnon-relativistic-limit-of-klein-gordon-probability-density%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The trick is to make the approach for the relativistic Klein-Gordon wave function
$$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
(The physical reasoning behind this approach is:
The fast oscillating exponential is the solution for the particle at rest.
And compared to that, $Psi$ will give only slow variations.)
From that you find its derivatives
$$
begin{eqnarray}
frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
end{eqnarray}
$$
Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
$$
begin{eqnarray}
P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.
answered Dec 24 '18 at 0:54
Thomas FritschThomas Fritsch
34929
$endgroup$
add a comment |
$begingroup$
The trick is to make the approach for the relativistic Klein-Gordon wave function
$$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
(The physical reasoning behind this approach is:
The fast oscillating exponential is the solution for the particle at rest.
And compared to that, $Psi$ will give only slow variations.)
From that you find its derivatives
$$
begin{eqnarray}
frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
end{eqnarray}
$$
Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
$$
begin{eqnarray}
P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.
answered Dec 24 '18 at 0:54
Thomas FritschThomas Fritsch
34929
$endgroup$
add a comment |
$begingroup$
The trick is to make the approach for the relativistic Klein-Gordon wave function
$$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
(The physical reasoning behind this approach is:
The fast oscillating exponential is the solution for the particle at rest.
And compared to that, $Psi$ will give only slow variations.)
From that you find its derivatives
$$
begin{eqnarray}
frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
end{eqnarray}
$$
Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
$$
begin{eqnarray}
P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.
answered Dec 24 '18 at 0:54
Thomas FritschThomas Fritsch
34929
$endgroup$
The trick is to make the approach for the relativistic Klein-Gordon wave function
$$ Phi(vec{r},t) = Psi(vec{r},t) e^{-imc^2t/hbar} $$
(The physical reasoning behind this approach is:
The fast oscillating exponential is the solution for the particle at rest.
And compared to that, $Psi$ will give only slow variations.)
From that you find its derivatives
$$
begin{eqnarray}
frac{partialPhi}{partial t} &=& left( frac{partialPsi}{partial t} - frac{imc^2}{hbar}Psi right) e^{-imc^2t/hbar} \
vec{nabla}Phi &=& vec{nabla}Psi e^{-imc^2t/hbar}
end{eqnarray}
$$
Plug these into the definitions of the Klein-Gordon probability density and current ($P$ and $vec{j}$) and you get
$$
begin{eqnarray}
P &=& Psi^* Psi + frac{ihbar}{2mc^2} left( Psi^*frac{partialPsi}{partial t} - Psi frac{partialPsi^*}{partial t} right) \
vec{j} &=& dfrac{hbar}{2mi}left(Psi^*vec{nabla}Psi-Psivec{nabla}Psi^*right)
end{eqnarray}
$$
Now you can do the limit $c to infty$ and get the 'known' expressions of the non-relativistic Schrödinger probability density and current.
answered Dec 24 '18 at 0:54
Thomas FritschThomas Fritsch
34929
edited Jan 7 at 23:38
answered Dec 24 '18 at 0:54
Thomas FritschThomas Fritsch
34929
answered Dec 24 '18 at 0:54
Thomas FritschThomas Fritsch
34929
answered Dec 24 '18 at 0:54
Thomas FritschThomas Fritsch
34929
34929
add a comment |
add a comment |
$begingroup$
You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.
answered Dec 24 '18 at 0:32
my2ctsmy2cts
4,9422618
$endgroup$
add a comment |
$begingroup$
You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.
answered Dec 24 '18 at 0:32
my2ctsmy2cts
4,9422618
$endgroup$
add a comment |
$begingroup$
You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.
answered Dec 24 '18 at 0:32
my2ctsmy2cts
4,9422618
$endgroup$
You can substitute $Phi = e^{-mc^2t/hbar} Psi$ and then neglect the second order time derivative of $Psi$. Drop the constant $mc^2$ and you will have recovered the Schrödinger equation.
answered Dec 24 '18 at 0:32
my2ctsmy2cts
4,9422618
answered Dec 24 '18 at 0:32
my2ctsmy2cts
4,9422618
answered Dec 24 '18 at 0:32
my2ctsmy2cts
4,9422618
answered Dec 24 '18 at 0:32
my2ctsmy2cts
4,9422618
4,9422618
add a comment |
add a comment |
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%2f450076%2fnon-relativistic-limit-of-klein-gordon-probability-density%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
1
$begingroup$
For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein.
$endgroup$
– Qmechanic♦
Dec 23 '18 at 21:39