WKB Approximation to Schrödinger Equation
$begingroup$
Consider the Schrödinger equation: $$y''(x)+EQ(x)y(x)=0,tag{1}$$
where $E>0, Q(x)>0, y(0)=y(pi)=0. $
Use WKB approximation to obtain $$y(x) sim CQ^frac{-1}{4}(x)sinBig(sqrt(E) int_{0}^{x} sqrt(Q(t) dt Big), E to infty. $$
What I did:
Let $phi(x;epsilon)=log (y(x;epsilon)).$ So, $y(x;epsilon)=exp(phi(x;epsilon).$
Substitute this to the main equation to get:
$$phi''(x;epsilon)+phi'(x;epsilon)^2+EQ(x)=0. tag{2}$$
Suppose $$phi(x;epsilon)=frac{1}{epsilon}sum_{n=0}^{infty} phi_n(x)epsilon^{n}, epsilon to 0. $$
Substitute this to $(2)$ using $phi(x;epsilon)=phi_0(x) frac{1}{epsilon}+O(1)$ we get,
$$frac{1}{epsilon}phi''_0(x)+frac{1}{epsilon^2}phi'_0(x)^2+EQ(x)+O(1)=0.$$
This implies $$epsilon phi''_0(x)+phi'_0(x)^2+epsilon^2EQ(x)+O(epsilon^2)=0.$$
So, at $O(1)$, $$phi'_0(x)^2=0.$$
This means $phi_0(x)=C$ where $C$ is constant.
We then continue to the next order by substituting $phi(x;epsilon)=phi_0(x) frac{1}{epsilon}+phi_1(x)+O(epsilon)=frac{C}{epsilon}+phi_1(x)+O(epsilon)$ to $(2)$.
We get $$phi''_1(x)+phi'_1(x)^2+EQ(x)=0.$$
This is where I'm stuck, any idea how to solve this ODE to get $phi_1(x)$?
ordinary-differential-equations numerical-methods asymptotics wave-equation perturbation-theory
asked Dec 12 '18 at 15:41
XIIIXXIIIX
678
$endgroup$
add a comment |
$begingroup$
Consider the Schrödinger equation: $$y''(x)+EQ(x)y(x)=0,tag{1}$$
where $E>0, Q(x)>0, y(0)=y(pi)=0. $
Use WKB approximation to obtain $$y(x) sim CQ^frac{-1}{4}(x)sinBig(sqrt(E) int_{0}^{x} sqrt(Q(t) dt Big), E to infty. $$
What I did:
Let $phi(x;epsilon)=log (y(x;epsilon)).$ So, $y(x;epsilon)=exp(phi(x;epsilon).$
Substitute this to the main equation to get:
$$phi''(x;epsilon)+phi'(x;epsilon)^2+EQ(x)=0. tag{2}$$
Suppose $$phi(x;epsilon)=frac{1}{epsilon}sum_{n=0}^{infty} phi_n(x)epsilon^{n}, epsilon to 0. $$
Substitute this to $(2)$ using $phi(x;epsilon)=phi_0(x) frac{1}{epsilon}+O(1)$ we get,
$$frac{1}{epsilon}phi''_0(x)+frac{1}{epsilon^2}phi'_0(x)^2+EQ(x)+O(1)=0.$$
This implies $$epsilon phi''_0(x)+phi'_0(x)^2+epsilon^2EQ(x)+O(epsilon^2)=0.$$
So, at $O(1)$, $$phi'_0(x)^2=0.$$
This means $phi_0(x)=C$ where $C$ is constant.
We then continue to the next order by substituting $phi(x;epsilon)=phi_0(x) frac{1}{epsilon}+phi_1(x)+O(epsilon)=frac{C}{epsilon}+phi_1(x)+O(epsilon)$ to $(2)$.
We get $$phi''_1(x)+phi'_1(x)^2+EQ(x)=0.$$
This is where I'm stuck, any idea how to solve this ODE to get $phi_1(x)$?
ordinary-differential-equations numerical-methods asymptotics wave-equation perturbation-theory
asked Dec 12 '18 at 15:41
XIIIXXIIIX
678
$endgroup$
add a comment |
$begingroup$
Consider the Schrödinger equation: $$y''(x)+EQ(x)y(x)=0,tag{1}$$
where $E>0, Q(x)>0, y(0)=y(pi)=0. $
Use WKB approximation to obtain $$y(x) sim CQ^frac{-1}{4}(x)sinBig(sqrt(E) int_{0}^{x} sqrt(Q(t) dt Big), E to infty. $$
What I did:
Let $phi(x;epsilon)=log (y(x;epsilon)).$ So, $y(x;epsilon)=exp(phi(x;epsilon).$
Substitute this to the main equation to get:
$$phi''(x;epsilon)+phi'(x;epsilon)^2+EQ(x)=0. tag{2}$$
Suppose $$phi(x;epsilon)=frac{1}{epsilon}sum_{n=0}^{infty} phi_n(x)epsilon^{n}, epsilon to 0. $$
Substitute this to $(2)$ using $phi(x;epsilon)=phi_0(x) frac{1}{epsilon}+O(1)$ we get,
$$frac{1}{epsilon}phi''_0(x)+frac{1}{epsilon^2}phi'_0(x)^2+EQ(x)+O(1)=0.$$
This implies $$epsilon phi''_0(x)+phi'_0(x)^2+epsilon^2EQ(x)+O(epsilon^2)=0.$$
So, at $O(1)$, $$phi'_0(x)^2=0.$$
This means $phi_0(x)=C$ where $C$ is constant.
We then continue to the next order by substituting $phi(x;epsilon)=phi_0(x) frac{1}{epsilon}+phi_1(x)+O(epsilon)=frac{C}{epsilon}+phi_1(x)+O(epsilon)$ to $(2)$.
We get $$phi''_1(x)+phi'_1(x)^2+EQ(x)=0.$$
This is where I'm stuck, any idea how to solve this ODE to get $phi_1(x)$?
ordinary-differential-equations numerical-methods asymptotics wave-equation perturbation-theory
asked Dec 12 '18 at 15:41
XIIIXXIIIX
678
$endgroup$
Consider the Schrödinger equation: $$y''(x)+EQ(x)y(x)=0,tag{1}$$
where $E>0, Q(x)>0, y(0)=y(pi)=0. $
Use WKB approximation to obtain $$y(x) sim CQ^frac{-1}{4}(x)sinBig(sqrt(E) int_{0}^{x} sqrt(Q(t) dt Big), E to infty. $$
What I did:
Let $phi(x;epsilon)=log (y(x;epsilon)).$ So, $y(x;epsilon)=exp(phi(x;epsilon).$
Substitute this to the main equation to get:
$$phi''(x;epsilon)+phi'(x;epsilon)^2+EQ(x)=0. tag{2}$$
Suppose $$phi(x;epsilon)=frac{1}{epsilon}sum_{n=0}^{infty} phi_n(x)epsilon^{n}, epsilon to 0. $$
Substitute this to $(2)$ using $phi(x;epsilon)=phi_0(x) frac{1}{epsilon}+O(1)$ we get,
$$frac{1}{epsilon}phi''_0(x)+frac{1}{epsilon^2}phi'_0(x)^2+EQ(x)+O(1)=0.$$
This implies $$epsilon phi''_0(x)+phi'_0(x)^2+epsilon^2EQ(x)+O(epsilon^2)=0.$$
So, at $O(1)$, $$phi'_0(x)^2=0.$$
This means $phi_0(x)=C$ where $C$ is constant.
We then continue to the next order by substituting $phi(x;epsilon)=phi_0(x) frac{1}{epsilon}+phi_1(x)+O(epsilon)=frac{C}{epsilon}+phi_1(x)+O(epsilon)$ to $(2)$.
We get $$phi''_1(x)+phi'_1(x)^2+EQ(x)=0.$$
This is where I'm stuck, any idea how to solve this ODE to get $phi_1(x)$?
ordinary-differential-equations numerical-methods asymptotics wave-equation perturbation-theory
ordinary-differential-equations numerical-methods asymptotics wave-equation perturbation-theory
asked Dec 12 '18 at 15:41
XIIIXXIIIX
678
asked Dec 12 '18 at 15:41
XIIIXXIIIX
678
asked Dec 12 '18 at 15:41
XIIIXXIIIX
678
asked Dec 12 '18 at 15:41
XIIIXXIIIX
678
asked Dec 12 '18 at 15:41
XIIIXXIIIX
678
678
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, your first balancing decision does not make sense. In the balancing you attempt to get two terms of equal magnitude/scale. You only got one dominating term, which you saw gave a trivial result in $ϕ_0=const.$ and no simplification relative to the original equation in the next equation for $ϕ_1$.
You need to include in the balancing calculations that $E$ is assumed large, and $ϵ$ small so that the balancing of the largest terms among the scale coefficients $ϵ,1,ϵ^2E$ in
$$ϵ[ϵϕ''(x;ϵ)]+[ϵϕ'(x;ϵ)]^2+ϵ^2E[Q(x)]=0$$
gives $ϵ^2E=1$. Then the equation for the first term $ϕ_0$ in the expansion of $ϕ$ is non-trivial
$$
ϕ_0'(x)^2+Q(x)=0.
$$
The next order equation gives
$$
ϕ_0''(x)+2ϕ_0'(x)ϕ_1'(x)=0implies ϕ_1'(x)=-frac{ϕ_0''(x)}{2ϕ_0'(x)}=
-frac{Q'(x)}{4Q(x)}=-frac14(ln|Q(x)|)'
$$
which when integrated gives the cited WKB approximation formula.
The next order term has the equation
$$
ϕ_1''(x)+ϕ_1'(x)^2+2ϕ_0'(x)ϕ_2'(x)=0implies ϕ_2'(x) = frac{ϕ_0'''(x)}{4ϕ_0'(x)^2} - frac{3ϕ_0''(x)^2}{8ϕ_0'(x)^3}=left(frac{ϕ_0''(x)}{4ϕ_0'(x)^2}right)'+frac{ϕ_0''(x)^2}{8ϕ_0'(x)^3}
$$
which does not look like a nice-to-integrate term.
answered Dec 12 '18 at 16:37
LutzLLutzL
60.2k42057
$endgroup$
$begingroup$
So, everything is fine until I said at $O(1)$, $$phi'_0(x)^2=0 ?$$ Also, Can you explain a bit more on this "You need to include in the balancing calculations that E is assumed large, so that the balancing of the largest terms gives $epsilon^2 E=1.$" How do you get $O(epsilon^2 E)=O(1)$ ?
$endgroup$
– XIIIX
Dec 12 '18 at 20:50
1
$begingroup$
The assumptions are that $ϕ_0'$ and $Q$ have middle scale, $ϵ$ is small and $E$ is large. In the set of scale coefficients in the equation for $ϕ_0'$, which are $ϵ,1,ϵ^2E$, you want the largest two to coincide to have a "balance", that is, to avoid that one dominates. The first will always be the smallest, so the second and the third have to be the same scale, for simplicity set them equal.
$endgroup$
– LutzL
Dec 12 '18 at 21:01
add a comment |
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$begingroup$
No, your first balancing decision does not make sense. In the balancing you attempt to get two terms of equal magnitude/scale. You only got one dominating term, which you saw gave a trivial result in $ϕ_0=const.$ and no simplification relative to the original equation in the next equation for $ϕ_1$.
You need to include in the balancing calculations that $E$ is assumed large, and $ϵ$ small so that the balancing of the largest terms among the scale coefficients $ϵ,1,ϵ^2E$ in
$$ϵ[ϵϕ''(x;ϵ)]+[ϵϕ'(x;ϵ)]^2+ϵ^2E[Q(x)]=0$$
gives $ϵ^2E=1$. Then the equation for the first term $ϕ_0$ in the expansion of $ϕ$ is non-trivial
$$
ϕ_0'(x)^2+Q(x)=0.
$$
The next order equation gives
$$
ϕ_0''(x)+2ϕ_0'(x)ϕ_1'(x)=0implies ϕ_1'(x)=-frac{ϕ_0''(x)}{2ϕ_0'(x)}=
-frac{Q'(x)}{4Q(x)}=-frac14(ln|Q(x)|)'
$$
which when integrated gives the cited WKB approximation formula.
The next order term has the equation
$$
ϕ_1''(x)+ϕ_1'(x)^2+2ϕ_0'(x)ϕ_2'(x)=0implies ϕ_2'(x) = frac{ϕ_0'''(x)}{4ϕ_0'(x)^2} - frac{3ϕ_0''(x)^2}{8ϕ_0'(x)^3}=left(frac{ϕ_0''(x)}{4ϕ_0'(x)^2}right)'+frac{ϕ_0''(x)^2}{8ϕ_0'(x)^3}
$$
which does not look like a nice-to-integrate term.
answered Dec 12 '18 at 16:37
LutzLLutzL
60.2k42057
$endgroup$
$begingroup$
So, everything is fine until I said at $O(1)$, $$phi'_0(x)^2=0 ?$$ Also, Can you explain a bit more on this "You need to include in the balancing calculations that E is assumed large, so that the balancing of the largest terms gives $epsilon^2 E=1.$" How do you get $O(epsilon^2 E)=O(1)$ ?
$endgroup$
– XIIIX
Dec 12 '18 at 20:50
1
$begingroup$
The assumptions are that $ϕ_0'$ and $Q$ have middle scale, $ϵ$ is small and $E$ is large. In the set of scale coefficients in the equation for $ϕ_0'$, which are $ϵ,1,ϵ^2E$, you want the largest two to coincide to have a "balance", that is, to avoid that one dominates. The first will always be the smallest, so the second and the third have to be the same scale, for simplicity set them equal.
$endgroup$
– LutzL
Dec 12 '18 at 21:01
add a comment |
$begingroup$
No, your first balancing decision does not make sense. In the balancing you attempt to get two terms of equal magnitude/scale. You only got one dominating term, which you saw gave a trivial result in $ϕ_0=const.$ and no simplification relative to the original equation in the next equation for $ϕ_1$.
You need to include in the balancing calculations that $E$ is assumed large, and $ϵ$ small so that the balancing of the largest terms among the scale coefficients $ϵ,1,ϵ^2E$ in
$$ϵ[ϵϕ''(x;ϵ)]+[ϵϕ'(x;ϵ)]^2+ϵ^2E[Q(x)]=0$$
gives $ϵ^2E=1$. Then the equation for the first term $ϕ_0$ in the expansion of $ϕ$ is non-trivial
$$
ϕ_0'(x)^2+Q(x)=0.
$$
The next order equation gives
$$
ϕ_0''(x)+2ϕ_0'(x)ϕ_1'(x)=0implies ϕ_1'(x)=-frac{ϕ_0''(x)}{2ϕ_0'(x)}=
-frac{Q'(x)}{4Q(x)}=-frac14(ln|Q(x)|)'
$$
which when integrated gives the cited WKB approximation formula.
The next order term has the equation
$$
ϕ_1''(x)+ϕ_1'(x)^2+2ϕ_0'(x)ϕ_2'(x)=0implies ϕ_2'(x) = frac{ϕ_0'''(x)}{4ϕ_0'(x)^2} - frac{3ϕ_0''(x)^2}{8ϕ_0'(x)^3}=left(frac{ϕ_0''(x)}{4ϕ_0'(x)^2}right)'+frac{ϕ_0''(x)^2}{8ϕ_0'(x)^3}
$$
which does not look like a nice-to-integrate term.
answered Dec 12 '18 at 16:37
LutzLLutzL
60.2k42057
$endgroup$
$begingroup$
So, everything is fine until I said at $O(1)$, $$phi'_0(x)^2=0 ?$$ Also, Can you explain a bit more on this "You need to include in the balancing calculations that E is assumed large, so that the balancing of the largest terms gives $epsilon^2 E=1.$" How do you get $O(epsilon^2 E)=O(1)$ ?
$endgroup$
– XIIIX
Dec 12 '18 at 20:50
1
$begingroup$
The assumptions are that $ϕ_0'$ and $Q$ have middle scale, $ϵ$ is small and $E$ is large. In the set of scale coefficients in the equation for $ϕ_0'$, which are $ϵ,1,ϵ^2E$, you want the largest two to coincide to have a "balance", that is, to avoid that one dominates. The first will always be the smallest, so the second and the third have to be the same scale, for simplicity set them equal.
$endgroup$
– LutzL
Dec 12 '18 at 21:01
add a comment |
$begingroup$
No, your first balancing decision does not make sense. In the balancing you attempt to get two terms of equal magnitude/scale. You only got one dominating term, which you saw gave a trivial result in $ϕ_0=const.$ and no simplification relative to the original equation in the next equation for $ϕ_1$.
You need to include in the balancing calculations that $E$ is assumed large, and $ϵ$ small so that the balancing of the largest terms among the scale coefficients $ϵ,1,ϵ^2E$ in
$$ϵ[ϵϕ''(x;ϵ)]+[ϵϕ'(x;ϵ)]^2+ϵ^2E[Q(x)]=0$$
gives $ϵ^2E=1$. Then the equation for the first term $ϕ_0$ in the expansion of $ϕ$ is non-trivial
$$
ϕ_0'(x)^2+Q(x)=0.
$$
The next order equation gives
$$
ϕ_0''(x)+2ϕ_0'(x)ϕ_1'(x)=0implies ϕ_1'(x)=-frac{ϕ_0''(x)}{2ϕ_0'(x)}=
-frac{Q'(x)}{4Q(x)}=-frac14(ln|Q(x)|)'
$$
which when integrated gives the cited WKB approximation formula.
The next order term has the equation
$$
ϕ_1''(x)+ϕ_1'(x)^2+2ϕ_0'(x)ϕ_2'(x)=0implies ϕ_2'(x) = frac{ϕ_0'''(x)}{4ϕ_0'(x)^2} - frac{3ϕ_0''(x)^2}{8ϕ_0'(x)^3}=left(frac{ϕ_0''(x)}{4ϕ_0'(x)^2}right)'+frac{ϕ_0''(x)^2}{8ϕ_0'(x)^3}
$$
which does not look like a nice-to-integrate term.
answered Dec 12 '18 at 16:37
LutzLLutzL
60.2k42057
$endgroup$
No, your first balancing decision does not make sense. In the balancing you attempt to get two terms of equal magnitude/scale. You only got one dominating term, which you saw gave a trivial result in $ϕ_0=const.$ and no simplification relative to the original equation in the next equation for $ϕ_1$.
You need to include in the balancing calculations that $E$ is assumed large, and $ϵ$ small so that the balancing of the largest terms among the scale coefficients $ϵ,1,ϵ^2E$ in
$$ϵ[ϵϕ''(x;ϵ)]+[ϵϕ'(x;ϵ)]^2+ϵ^2E[Q(x)]=0$$
gives $ϵ^2E=1$. Then the equation for the first term $ϕ_0$ in the expansion of $ϕ$ is non-trivial
$$
ϕ_0'(x)^2+Q(x)=0.
$$
The next order equation gives
$$
ϕ_0''(x)+2ϕ_0'(x)ϕ_1'(x)=0implies ϕ_1'(x)=-frac{ϕ_0''(x)}{2ϕ_0'(x)}=
-frac{Q'(x)}{4Q(x)}=-frac14(ln|Q(x)|)'
$$
which when integrated gives the cited WKB approximation formula.
The next order term has the equation
$$
ϕ_1''(x)+ϕ_1'(x)^2+2ϕ_0'(x)ϕ_2'(x)=0implies ϕ_2'(x) = frac{ϕ_0'''(x)}{4ϕ_0'(x)^2} - frac{3ϕ_0''(x)^2}{8ϕ_0'(x)^3}=left(frac{ϕ_0''(x)}{4ϕ_0'(x)^2}right)'+frac{ϕ_0''(x)^2}{8ϕ_0'(x)^3}
$$
which does not look like a nice-to-integrate term.
answered Dec 12 '18 at 16:37
LutzLLutzL
60.2k42057
edited Dec 12 '18 at 21:23
answered Dec 12 '18 at 16:37
LutzLLutzL
60.2k42057
answered Dec 12 '18 at 16:37
LutzLLutzL
60.2k42057
answered Dec 12 '18 at 16:37
LutzLLutzL
60.2k42057
60.2k42057
$begingroup$
So, everything is fine until I said at $O(1)$, $$phi'_0(x)^2=0 ?$$ Also, Can you explain a bit more on this "You need to include in the balancing calculations that E is assumed large, so that the balancing of the largest terms gives $epsilon^2 E=1.$" How do you get $O(epsilon^2 E)=O(1)$ ?
$endgroup$
– XIIIX
Dec 12 '18 at 20:50
1
$begingroup$
The assumptions are that $ϕ_0'$ and $Q$ have middle scale, $ϵ$ is small and $E$ is large. In the set of scale coefficients in the equation for $ϕ_0'$, which are $ϵ,1,ϵ^2E$, you want the largest two to coincide to have a "balance", that is, to avoid that one dominates. The first will always be the smallest, so the second and the third have to be the same scale, for simplicity set them equal.
$endgroup$
– LutzL
Dec 12 '18 at 21:01
add a comment |
$begingroup$
So, everything is fine until I said at $O(1)$, $$phi'_0(x)^2=0 ?$$ Also, Can you explain a bit more on this "You need to include in the balancing calculations that E is assumed large, so that the balancing of the largest terms gives $epsilon^2 E=1.$" How do you get $O(epsilon^2 E)=O(1)$ ?
$endgroup$
– XIIIX
Dec 12 '18 at 20:50
1
$begingroup$
The assumptions are that $ϕ_0'$ and $Q$ have middle scale, $ϵ$ is small and $E$ is large. In the set of scale coefficients in the equation for $ϕ_0'$, which are $ϵ,1,ϵ^2E$, you want the largest two to coincide to have a "balance", that is, to avoid that one dominates. The first will always be the smallest, so the second and the third have to be the same scale, for simplicity set them equal.
$endgroup$
– LutzL
Dec 12 '18 at 21:01
$begingroup$
So, everything is fine until I said at $O(1)$, $$phi'_0(x)^2=0 ?$$ Also, Can you explain a bit more on this "You need to include in the balancing calculations that E is assumed large, so that the balancing of the largest terms gives $epsilon^2 E=1.$" How do you get $O(epsilon^2 E)=O(1)$ ?
$endgroup$
– XIIIX
Dec 12 '18 at 20:50
$begingroup$
So, everything is fine until I said at $O(1)$, $$phi'_0(x)^2=0 ?$$ Also, Can you explain a bit more on this "You need to include in the balancing calculations that E is assumed large, so that the balancing of the largest terms gives $epsilon^2 E=1.$" How do you get $O(epsilon^2 E)=O(1)$ ?
$endgroup$
– XIIIX
Dec 12 '18 at 20:50
1
1
$begingroup$
The assumptions are that $ϕ_0'$ and $Q$ have middle scale, $ϵ$ is small and $E$ is large. In the set of scale coefficients in the equation for $ϕ_0'$, which are $ϵ,1,ϵ^2E$, you want the largest two to coincide to have a "balance", that is, to avoid that one dominates. The first will always be the smallest, so the second and the third have to be the same scale, for simplicity set them equal.
$endgroup$
– LutzL
Dec 12 '18 at 21:01
$begingroup$
The assumptions are that $ϕ_0'$ and $Q$ have middle scale, $ϵ$ is small and $E$ is large. In the set of scale coefficients in the equation for $ϕ_0'$, which are $ϵ,1,ϵ^2E$, you want the largest two to coincide to have a "balance", that is, to avoid that one dominates. The first will always be the smallest, so the second and the third have to be the same scale, for simplicity set them equal.
$endgroup$
– LutzL
Dec 12 '18 at 21:01
add a comment |
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