Mixing
Delta distribution and the Schrödinger equation
$begingroup$
While studying the lecture notes of my quantum mechanics course I came across something that seemed a bit odd. There we want to solve the Schrödinger equation for the potential $V(x)=V_0 delta(x)$, where $delta(x)$ represents the "delta-function". We therefore get the equation
$$-frac{hbar^2}{2m}psi''(x)+V_0delta(x)psi(x)=Epsi(x),$$
where the solutions are of the form $psi_{rm left}(x)=Aexp(ikx/hbar)+Bexp(-ikx/hbar)$ with the energy $E= k^2/2m$ for $x<0$ and analogous for $x>0$ we find $psi_{rm right}$. We know that
$$limlimits_{xto 0^-} psi_{rm left} = limlimits_{xto 0^+}psi_{rm right}$$
should hold but also need a second condition. The professor then proceeded to write the following.
We integrate eq. (1) [the first equation in this post] form
$-epsilon$ to $epsilon$ and get
$$-frac{hbar^2}{2m}(psi'(epsilon)-psi'(-epsilon))+V_0psi(0)=Eint^epsilon_{-epsilon}
psi(x) dx.$$ For $epsilon to 0$ we get $$psi ^ { prime } left( 0
^ { + } right) - psi ^ { prime } left( 0 ^ { - } right) = frac {
2 m V } { hbar ^ { 2 } } psi ( 0 ),$$ where $$ f left( 0 ^ { + }
right) = lim _ { x rightarrow 0 atop x > 0 } f ( x ) quad f
left( 0 ^ { - } right) = lim _ { x rightarrow 0 atop x < 0 } f (
x ).$$
My main confusion here comes from the fact that apparently
$$int_{-epsilon}^{epsilon} delta(x)psi(x)dx = psi(0).$$
Could someone explain this?
Notes on my background in math: In a course on mathematical physics we talked briefly about (tempered) distributions and there we saw the definition
$$delta[varphi] =int_mathbb{R^n}varphi(x)delta(x)dx = varphi(0),$$
for $varphi in mathcal{S}(mathbb{R}^n)$ and
$$mathcal { S } left( mathbb { R } ^ { n } right) : = left{ phi in C ^ { infty } left( mathbb { R } ^ { n } right) Big| forall alpha , beta in mathbb { N } _ { 0 } ^ { n } : sup _ { x in mathbb { R } ^ { n } } | x ^ { alpha } D ^ { beta } phi ( x ) | < infty right}.$$
I'm aware of the fact that in physics we often use the $delta$-"function" rather sloppy and tend to brush of problematic arguments away with some kind of intuition (at least that's how it was presented to me up until this point) but I fail to see how it can be considered the same to integrate over an interval $[-epsilon, epsilon]$ and over $mathbb{R}$ in this case.
physics mathematical-physics dirac-delta quantum-mechanics
asked Jan 4 at 19:30
SitoSito
1706
$endgroup$
add a comment |
$begingroup$
While studying the lecture notes of my quantum mechanics course I came across something that seemed a bit odd. There we want to solve the Schrödinger equation for the potential $V(x)=V_0 delta(x)$, where $delta(x)$ represents the "delta-function". We therefore get the equation
$$-frac{hbar^2}{2m}psi''(x)+V_0delta(x)psi(x)=Epsi(x),$$
where the solutions are of the form $psi_{rm left}(x)=Aexp(ikx/hbar)+Bexp(-ikx/hbar)$ with the energy $E= k^2/2m$ for $x<0$ and analogous for $x>0$ we find $psi_{rm right}$. We know that
$$limlimits_{xto 0^-} psi_{rm left} = limlimits_{xto 0^+}psi_{rm right}$$
should hold but also need a second condition. The professor then proceeded to write the following.
We integrate eq. (1) [the first equation in this post] form
$-epsilon$ to $epsilon$ and get
$$-frac{hbar^2}{2m}(psi'(epsilon)-psi'(-epsilon))+V_0psi(0)=Eint^epsilon_{-epsilon}
psi(x) dx.$$ For $epsilon to 0$ we get $$psi ^ { prime } left( 0
^ { + } right) - psi ^ { prime } left( 0 ^ { - } right) = frac {
2 m V } { hbar ^ { 2 } } psi ( 0 ),$$ where $$ f left( 0 ^ { + }
right) = lim _ { x rightarrow 0 atop x > 0 } f ( x ) quad f
left( 0 ^ { - } right) = lim _ { x rightarrow 0 atop x < 0 } f (
x ).$$
My main confusion here comes from the fact that apparently
$$int_{-epsilon}^{epsilon} delta(x)psi(x)dx = psi(0).$$
Could someone explain this?
Notes on my background in math: In a course on mathematical physics we talked briefly about (tempered) distributions and there we saw the definition
$$delta[varphi] =int_mathbb{R^n}varphi(x)delta(x)dx = varphi(0),$$
for $varphi in mathcal{S}(mathbb{R}^n)$ and
$$mathcal { S } left( mathbb { R } ^ { n } right) : = left{ phi in C ^ { infty } left( mathbb { R } ^ { n } right) Big| forall alpha , beta in mathbb { N } _ { 0 } ^ { n } : sup _ { x in mathbb { R } ^ { n } } | x ^ { alpha } D ^ { beta } phi ( x ) | < infty right}.$$
I'm aware of the fact that in physics we often use the $delta$-"function" rather sloppy and tend to brush of problematic arguments away with some kind of intuition (at least that's how it was presented to me up until this point) but I fail to see how it can be considered the same to integrate over an interval $[-epsilon, epsilon]$ and over $mathbb{R}$ in this case.
physics mathematical-physics dirac-delta quantum-mechanics
asked Jan 4 at 19:30
SitoSito
1706
$endgroup$
$begingroup$
For intuition, you can think of the delta function as being an ordinary nonnegative smooth function which has a very sharp peak near $0$, and which is zero outside of a tiny neighborhood of $0$, and such that the area under the curve is $1$. So if $Psi$ is continuous at $0$ then $int_{-epsilon}^epsilon delta(x) Psi(x) , dx approx int_{-epsilon}^epsilon delta(x) Psi(0) , dx =Psi(0) int_{-epsilon}^epsilon delta(x) , dx = Psi(0) cdot 1$.
$endgroup$
– littleO
Jan 4 at 19:59
$begingroup$
In addition to @littleO 's comment, note that the delta function is zero outside the $[-epsilon,epsilon]$ interval, so integrating over $mathbb R$ you can split into three parts: one less than $-epsilon$ (integral $0$), one between $-epsilon$ and $epsilon$, and one interval from$epsilon$ to $infty$. The last one is also zero.
$endgroup$
– Andrei
Jan 4 at 20:52
1
$begingroup$
@FrederikvomEnde I like your comment a lot! Would mark it as answer if you post it as one!
$endgroup$
– Sito
Jan 5 at 16:24
$begingroup$
Alright, just shifted my comment to the answer section!
$endgroup$
– Frederik vom Ende
Jan 5 at 16:55
add a comment |
$begingroup$
While studying the lecture notes of my quantum mechanics course I came across something that seemed a bit odd. There we want to solve the Schrödinger equation for the potential $V(x)=V_0 delta(x)$, where $delta(x)$ represents the "delta-function". We therefore get the equation
$$-frac{hbar^2}{2m}psi''(x)+V_0delta(x)psi(x)=Epsi(x),$$
where the solutions are of the form $psi_{rm left}(x)=Aexp(ikx/hbar)+Bexp(-ikx/hbar)$ with the energy $E= k^2/2m$ for $x<0$ and analogous for $x>0$ we find $psi_{rm right}$. We know that
$$limlimits_{xto 0^-} psi_{rm left} = limlimits_{xto 0^+}psi_{rm right}$$
should hold but also need a second condition. The professor then proceeded to write the following.
We integrate eq. (1) [the first equation in this post] form
$-epsilon$ to $epsilon$ and get
$$-frac{hbar^2}{2m}(psi'(epsilon)-psi'(-epsilon))+V_0psi(0)=Eint^epsilon_{-epsilon}
psi(x) dx.$$ For $epsilon to 0$ we get $$psi ^ { prime } left( 0
^ { + } right) - psi ^ { prime } left( 0 ^ { - } right) = frac {
2 m V } { hbar ^ { 2 } } psi ( 0 ),$$ where $$ f left( 0 ^ { + }
right) = lim _ { x rightarrow 0 atop x > 0 } f ( x ) quad f
left( 0 ^ { - } right) = lim _ { x rightarrow 0 atop x < 0 } f (
x ).$$
My main confusion here comes from the fact that apparently
$$int_{-epsilon}^{epsilon} delta(x)psi(x)dx = psi(0).$$
Could someone explain this?
Notes on my background in math: In a course on mathematical physics we talked briefly about (tempered) distributions and there we saw the definition
$$delta[varphi] =int_mathbb{R^n}varphi(x)delta(x)dx = varphi(0),$$
for $varphi in mathcal{S}(mathbb{R}^n)$ and
$$mathcal { S } left( mathbb { R } ^ { n } right) : = left{ phi in C ^ { infty } left( mathbb { R } ^ { n } right) Big| forall alpha , beta in mathbb { N } _ { 0 } ^ { n } : sup _ { x in mathbb { R } ^ { n } } | x ^ { alpha } D ^ { beta } phi ( x ) | < infty right}.$$
I'm aware of the fact that in physics we often use the $delta$-"function" rather sloppy and tend to brush of problematic arguments away with some kind of intuition (at least that's how it was presented to me up until this point) but I fail to see how it can be considered the same to integrate over an interval $[-epsilon, epsilon]$ and over $mathbb{R}$ in this case.
physics mathematical-physics dirac-delta quantum-mechanics
asked Jan 4 at 19:30
SitoSito
1706
$endgroup$
While studying the lecture notes of my quantum mechanics course I came across something that seemed a bit odd. There we want to solve the Schrödinger equation for the potential $V(x)=V_0 delta(x)$, where $delta(x)$ represents the "delta-function". We therefore get the equation
$$-frac{hbar^2}{2m}psi''(x)+V_0delta(x)psi(x)=Epsi(x),$$
where the solutions are of the form $psi_{rm left}(x)=Aexp(ikx/hbar)+Bexp(-ikx/hbar)$ with the energy $E= k^2/2m$ for $x<0$ and analogous for $x>0$ we find $psi_{rm right}$. We know that
$$limlimits_{xto 0^-} psi_{rm left} = limlimits_{xto 0^+}psi_{rm right}$$
should hold but also need a second condition. The professor then proceeded to write the following.
We integrate eq. (1) [the first equation in this post] form
$-epsilon$ to $epsilon$ and get
$$-frac{hbar^2}{2m}(psi'(epsilon)-psi'(-epsilon))+V_0psi(0)=Eint^epsilon_{-epsilon}
psi(x) dx.$$ For $epsilon to 0$ we get $$psi ^ { prime } left( 0
^ { + } right) - psi ^ { prime } left( 0 ^ { - } right) = frac {
2 m V } { hbar ^ { 2 } } psi ( 0 ),$$ where $$ f left( 0 ^ { + }
right) = lim _ { x rightarrow 0 atop x > 0 } f ( x ) quad f
left( 0 ^ { - } right) = lim _ { x rightarrow 0 atop x < 0 } f (
x ).$$
My main confusion here comes from the fact that apparently
$$int_{-epsilon}^{epsilon} delta(x)psi(x)dx = psi(0).$$
Could someone explain this?
Notes on my background in math: In a course on mathematical physics we talked briefly about (tempered) distributions and there we saw the definition
$$delta[varphi] =int_mathbb{R^n}varphi(x)delta(x)dx = varphi(0),$$
for $varphi in mathcal{S}(mathbb{R}^n)$ and
$$mathcal { S } left( mathbb { R } ^ { n } right) : = left{ phi in C ^ { infty } left( mathbb { R } ^ { n } right) Big| forall alpha , beta in mathbb { N } _ { 0 } ^ { n } : sup _ { x in mathbb { R } ^ { n } } | x ^ { alpha } D ^ { beta } phi ( x ) | < infty right}.$$
I'm aware of the fact that in physics we often use the $delta$-"function" rather sloppy and tend to brush of problematic arguments away with some kind of intuition (at least that's how it was presented to me up until this point) but I fail to see how it can be considered the same to integrate over an interval $[-epsilon, epsilon]$ and over $mathbb{R}$ in this case.
physics mathematical-physics dirac-delta quantum-mechanics
physics mathematical-physics dirac-delta quantum-mechanics
asked Jan 4 at 19:30
SitoSito
1706
asked Jan 4 at 19:30
SitoSito
1706
asked Jan 4 at 19:30
SitoSito
1706
asked Jan 4 at 19:30
SitoSito
1706
asked Jan 4 at 19:30
SitoSito
1706
1706
$begingroup$
For intuition, you can think of the delta function as being an ordinary nonnegative smooth function which has a very sharp peak near $0$, and which is zero outside of a tiny neighborhood of $0$, and such that the area under the curve is $1$. So if $Psi$ is continuous at $0$ then $int_{-epsilon}^epsilon delta(x) Psi(x) , dx approx int_{-epsilon}^epsilon delta(x) Psi(0) , dx =Psi(0) int_{-epsilon}^epsilon delta(x) , dx = Psi(0) cdot 1$.
$endgroup$
– littleO
Jan 4 at 19:59
$begingroup$
In addition to @littleO 's comment, note that the delta function is zero outside the $[-epsilon,epsilon]$ interval, so integrating over $mathbb R$ you can split into three parts: one less than $-epsilon$ (integral $0$), one between $-epsilon$ and $epsilon$, and one interval from$epsilon$ to $infty$. The last one is also zero.
$endgroup$
– Andrei
Jan 4 at 20:52
1
$begingroup$
@FrederikvomEnde I like your comment a lot! Would mark it as answer if you post it as one!
$endgroup$
– Sito
Jan 5 at 16:24
$begingroup$
Alright, just shifted my comment to the answer section!
$endgroup$
– Frederik vom Ende
Jan 5 at 16:55
add a comment |
$begingroup$
For intuition, you can think of the delta function as being an ordinary nonnegative smooth function which has a very sharp peak near $0$, and which is zero outside of a tiny neighborhood of $0$, and such that the area under the curve is $1$. So if $Psi$ is continuous at $0$ then $int_{-epsilon}^epsilon delta(x) Psi(x) , dx approx int_{-epsilon}^epsilon delta(x) Psi(0) , dx =Psi(0) int_{-epsilon}^epsilon delta(x) , dx = Psi(0) cdot 1$.
$endgroup$
– littleO
Jan 4 at 19:59
$begingroup$
In addition to @littleO 's comment, note that the delta function is zero outside the $[-epsilon,epsilon]$ interval, so integrating over $mathbb R$ you can split into three parts: one less than $-epsilon$ (integral $0$), one between $-epsilon$ and $epsilon$, and one interval from$epsilon$ to $infty$. The last one is also zero.
$endgroup$
– Andrei
Jan 4 at 20:52
1
$begingroup$
@FrederikvomEnde I like your comment a lot! Would mark it as answer if you post it as one!
$endgroup$
– Sito
Jan 5 at 16:24
$begingroup$
Alright, just shifted my comment to the answer section!
$endgroup$
– Frederik vom Ende
Jan 5 at 16:55
$begingroup$
For intuition, you can think of the delta function as being an ordinary nonnegative smooth function which has a very sharp peak near $0$, and which is zero outside of a tiny neighborhood of $0$, and such that the area under the curve is $1$. So if $Psi$ is continuous at $0$ then $int_{-epsilon}^epsilon delta(x) Psi(x) , dx approx int_{-epsilon}^epsilon delta(x) Psi(0) , dx =Psi(0) int_{-epsilon}^epsilon delta(x) , dx = Psi(0) cdot 1$.
$endgroup$
– littleO
Jan 4 at 19:59
$begingroup$
For intuition, you can think of the delta function as being an ordinary nonnegative smooth function which has a very sharp peak near $0$, and which is zero outside of a tiny neighborhood of $0$, and such that the area under the curve is $1$. So if $Psi$ is continuous at $0$ then $int_{-epsilon}^epsilon delta(x) Psi(x) , dx approx int_{-epsilon}^epsilon delta(x) Psi(0) , dx =Psi(0) int_{-epsilon}^epsilon delta(x) , dx = Psi(0) cdot 1$.
$endgroup$
– littleO
Jan 4 at 19:59
$begingroup$
In addition to @littleO 's comment, note that the delta function is zero outside the $[-epsilon,epsilon]$ interval, so integrating over $mathbb R$ you can split into three parts: one less than $-epsilon$ (integral $0$), one between $-epsilon$ and $epsilon$, and one interval from$epsilon$ to $infty$. The last one is also zero.
$endgroup$
– Andrei
Jan 4 at 20:52
$begingroup$
In addition to @littleO 's comment, note that the delta function is zero outside the $[-epsilon,epsilon]$ interval, so integrating over $mathbb R$ you can split into three parts: one less than $-epsilon$ (integral $0$), one between $-epsilon$ and $epsilon$, and one interval from$epsilon$ to $infty$. The last one is also zero.
$endgroup$
– Andrei
Jan 4 at 20:52
1
1
$begingroup$
@FrederikvomEnde I like your comment a lot! Would mark it as answer if you post it as one!
$endgroup$
– Sito
Jan 5 at 16:24
$begingroup$
@FrederikvomEnde I like your comment a lot! Would mark it as answer if you post it as one!
$endgroup$
– Sito
Jan 5 at 16:24
$begingroup$
Alright, just shifted my comment to the answer section!
$endgroup$
– Frederik vom Ende
Jan 5 at 16:55
$begingroup$
Alright, just shifted my comment to the answer section!
$endgroup$
– Frederik vom Ende
Jan 5 at 16:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
One may define $delta$ as a measure on $mathbb R$ which on some $Asubseteqmathbb R$ evaluates to $delta(A)=1$ if $0in A$ and $delta(A)=0$ else. Thus assuming $0in A$, one gets
$$
int_{mathbb R}psi(x)delta(x),dx=int_{mathbb R}psi(x),ddelta(x)=∫_Apsi(x),ddelta(x)+∫_{mathbb Rsetminus A}psi(x),ddelta(x)
$$
where the second integral vanishes as $mathbb Rsetminus A$ has $delta$-measure zero. By choosing $A=(-varepsilon,varepsilon)$ we obtain $$psi(0)=int_{mathbb R}psi(x)delta(x),dx=int_{-varepsilon}^varepsilon psi(x)delta(x),dx$$ for all $varepsilon>0$.
answered Jan 5 at 16:53
Frederik vom EndeFrederik vom Ende
6621321
$endgroup$
add a comment |
$begingroup$
Define $phi(x):=psi(x)[|x|leepsilon]$, where the square bracket is an Iverson bracket. Then $int_{-epsilon}^epsilondelta(x)psi(x)dx=int_{Bbb R}phi(x)delta(x)dx=phi(0)=psi(0)$.
answered Jan 4 at 19:43
J.G.J.G.
24.5k22539
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
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votes
$begingroup$
One may define $delta$ as a measure on $mathbb R$ which on some $Asubseteqmathbb R$ evaluates to $delta(A)=1$ if $0in A$ and $delta(A)=0$ else. Thus assuming $0in A$, one gets
$$
int_{mathbb R}psi(x)delta(x),dx=int_{mathbb R}psi(x),ddelta(x)=∫_Apsi(x),ddelta(x)+∫_{mathbb Rsetminus A}psi(x),ddelta(x)
$$
where the second integral vanishes as $mathbb Rsetminus A$ has $delta$-measure zero. By choosing $A=(-varepsilon,varepsilon)$ we obtain $$psi(0)=int_{mathbb R}psi(x)delta(x),dx=int_{-varepsilon}^varepsilon psi(x)delta(x),dx$$ for all $varepsilon>0$.
answered Jan 5 at 16:53
Frederik vom EndeFrederik vom Ende
6621321
$endgroup$
add a comment |
$begingroup$
One may define $delta$ as a measure on $mathbb R$ which on some $Asubseteqmathbb R$ evaluates to $delta(A)=1$ if $0in A$ and $delta(A)=0$ else. Thus assuming $0in A$, one gets
$$
int_{mathbb R}psi(x)delta(x),dx=int_{mathbb R}psi(x),ddelta(x)=∫_Apsi(x),ddelta(x)+∫_{mathbb Rsetminus A}psi(x),ddelta(x)
$$
where the second integral vanishes as $mathbb Rsetminus A$ has $delta$-measure zero. By choosing $A=(-varepsilon,varepsilon)$ we obtain $$psi(0)=int_{mathbb R}psi(x)delta(x),dx=int_{-varepsilon}^varepsilon psi(x)delta(x),dx$$ for all $varepsilon>0$.
answered Jan 5 at 16:53
Frederik vom EndeFrederik vom Ende
6621321
$endgroup$
add a comment |
$begingroup$
One may define $delta$ as a measure on $mathbb R$ which on some $Asubseteqmathbb R$ evaluates to $delta(A)=1$ if $0in A$ and $delta(A)=0$ else. Thus assuming $0in A$, one gets
$$
int_{mathbb R}psi(x)delta(x),dx=int_{mathbb R}psi(x),ddelta(x)=∫_Apsi(x),ddelta(x)+∫_{mathbb Rsetminus A}psi(x),ddelta(x)
$$
where the second integral vanishes as $mathbb Rsetminus A$ has $delta$-measure zero. By choosing $A=(-varepsilon,varepsilon)$ we obtain $$psi(0)=int_{mathbb R}psi(x)delta(x),dx=int_{-varepsilon}^varepsilon psi(x)delta(x),dx$$ for all $varepsilon>0$.
answered Jan 5 at 16:53
Frederik vom EndeFrederik vom Ende
6621321
$endgroup$
One may define $delta$ as a measure on $mathbb R$ which on some $Asubseteqmathbb R$ evaluates to $delta(A)=1$ if $0in A$ and $delta(A)=0$ else. Thus assuming $0in A$, one gets
$$
int_{mathbb R}psi(x)delta(x),dx=int_{mathbb R}psi(x),ddelta(x)=∫_Apsi(x),ddelta(x)+∫_{mathbb Rsetminus A}psi(x),ddelta(x)
$$
where the second integral vanishes as $mathbb Rsetminus A$ has $delta$-measure zero. By choosing $A=(-varepsilon,varepsilon)$ we obtain $$psi(0)=int_{mathbb R}psi(x)delta(x),dx=int_{-varepsilon}^varepsilon psi(x)delta(x),dx$$ for all $varepsilon>0$.
answered Jan 5 at 16:53
Frederik vom EndeFrederik vom Ende
6621321
edited Jan 5 at 16:58
answered Jan 5 at 16:53
Frederik vom EndeFrederik vom Ende
6621321
answered Jan 5 at 16:53
Frederik vom EndeFrederik vom Ende
6621321
answered Jan 5 at 16:53
Frederik vom EndeFrederik vom Ende
6621321
6621321
add a comment |
add a comment |
$begingroup$
Define $phi(x):=psi(x)[|x|leepsilon]$, where the square bracket is an Iverson bracket. Then $int_{-epsilon}^epsilondelta(x)psi(x)dx=int_{Bbb R}phi(x)delta(x)dx=phi(0)=psi(0)$.
answered Jan 4 at 19:43
J.G.J.G.
24.5k22539
$endgroup$
add a comment |
$begingroup$
Define $phi(x):=psi(x)[|x|leepsilon]$, where the square bracket is an Iverson bracket. Then $int_{-epsilon}^epsilondelta(x)psi(x)dx=int_{Bbb R}phi(x)delta(x)dx=phi(0)=psi(0)$.
answered Jan 4 at 19:43
J.G.J.G.
24.5k22539
$endgroup$
add a comment |
$begingroup$
Define $phi(x):=psi(x)[|x|leepsilon]$, where the square bracket is an Iverson bracket. Then $int_{-epsilon}^epsilondelta(x)psi(x)dx=int_{Bbb R}phi(x)delta(x)dx=phi(0)=psi(0)$.
answered Jan 4 at 19:43
J.G.J.G.
24.5k22539
$endgroup$
Define $phi(x):=psi(x)[|x|leepsilon]$, where the square bracket is an Iverson bracket. Then $int_{-epsilon}^epsilondelta(x)psi(x)dx=int_{Bbb R}phi(x)delta(x)dx=phi(0)=psi(0)$.
answered Jan 4 at 19:43
J.G.J.G.
24.5k22539
answered Jan 4 at 19:43
J.G.J.G.
24.5k22539
answered Jan 4 at 19:43
J.G.J.G.
24.5k22539
answered Jan 4 at 19:43
J.G.J.G.
24.5k22539
24.5k22539
add a comment |
add a comment |
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For intuition, you can think of the delta function as being an ordinary nonnegative smooth function which has a very sharp peak near $0$, and which is zero outside of a tiny neighborhood of $0$, and such that the area under the curve is $1$. So if $Psi$ is continuous at $0$ then $int_{-epsilon}^epsilon delta(x) Psi(x) , dx approx int_{-epsilon}^epsilon delta(x) Psi(0) , dx =Psi(0) int_{-epsilon}^epsilon delta(x) , dx = Psi(0) cdot 1$.
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– littleO
Jan 4 at 19:59
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In addition to @littleO 's comment, note that the delta function is zero outside the $[-epsilon,epsilon]$ interval, so integrating over $mathbb R$ you can split into three parts: one less than $-epsilon$ (integral $0$), one between $-epsilon$ and $epsilon$, and one interval from$epsilon$ to $infty$. The last one is also zero.
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– Andrei
Jan 4 at 20:52
1
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@FrederikvomEnde I like your comment a lot! Would mark it as answer if you post it as one!
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– Sito
Jan 5 at 16:24
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Alright, just shifted my comment to the answer section!
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– Frederik vom Ende
Jan 5 at 16:55