Why can't a particle penetrate an infinite potential barrier?
Clash Royale CLAN TAG#URR8PPP
I am studying basic quantum theory. My question is:
Why can't a particle penetrate an infinite potential barrier?
The reasoning that I have applied is that particles under consideration have finite energy. So, to cross an infinite potential barrier the particle requires infinite energy. But I cannot think of the mathematical relation between potential and energy so that indeed I am convinced that to cross an infinite potential barrier the particle needs infinite energy.
What is the relation between the potential and energy of quantum mechanical particles?
quantum-mechanics potential probability quantum-tunneling
add a comment |
I am studying basic quantum theory. My question is:
Why can't a particle penetrate an infinite potential barrier?
The reasoning that I have applied is that particles under consideration have finite energy. So, to cross an infinite potential barrier the particle requires infinite energy. But I cannot think of the mathematical relation between potential and energy so that indeed I am convinced that to cross an infinite potential barrier the particle needs infinite energy.
What is the relation between the potential and energy of quantum mechanical particles?
quantum-mechanics potential probability quantum-tunneling
1
You start with the tunneling probability knowing that it is exponentially small with the finite barrier height, therefore if the latter is infinite the former is zero. Once you see this you may use the infinite high potential well as a mathematical model for an impenetrable barrier.
– hyportnex
Dec 19 '18 at 15:53
add a comment |
I am studying basic quantum theory. My question is:
Why can't a particle penetrate an infinite potential barrier?
The reasoning that I have applied is that particles under consideration have finite energy. So, to cross an infinite potential barrier the particle requires infinite energy. But I cannot think of the mathematical relation between potential and energy so that indeed I am convinced that to cross an infinite potential barrier the particle needs infinite energy.
What is the relation between the potential and energy of quantum mechanical particles?
quantum-mechanics potential probability quantum-tunneling
I am studying basic quantum theory. My question is:
Why can't a particle penetrate an infinite potential barrier?
The reasoning that I have applied is that particles under consideration have finite energy. So, to cross an infinite potential barrier the particle requires infinite energy. But I cannot think of the mathematical relation between potential and energy so that indeed I am convinced that to cross an infinite potential barrier the particle needs infinite energy.
What is the relation between the potential and energy of quantum mechanical particles?
quantum-mechanics potential probability quantum-tunneling
quantum-mechanics potential probability quantum-tunneling
edited Dec 19 '18 at 18:51
Qmechanic♦
101k121831151
101k121831151
asked Dec 19 '18 at 15:43
Soumee
1425
1425
1
You start with the tunneling probability knowing that it is exponentially small with the finite barrier height, therefore if the latter is infinite the former is zero. Once you see this you may use the infinite high potential well as a mathematical model for an impenetrable barrier.
– hyportnex
Dec 19 '18 at 15:53
add a comment |
1
You start with the tunneling probability knowing that it is exponentially small with the finite barrier height, therefore if the latter is infinite the former is zero. Once you see this you may use the infinite high potential well as a mathematical model for an impenetrable barrier.
– hyportnex
Dec 19 '18 at 15:53
1
1
You start with the tunneling probability knowing that it is exponentially small with the finite barrier height, therefore if the latter is infinite the former is zero. Once you see this you may use the infinite high potential well as a mathematical model for an impenetrable barrier.
– hyportnex
Dec 19 '18 at 15:53
You start with the tunneling probability knowing that it is exponentially small with the finite barrier height, therefore if the latter is infinite the former is zero. Once you see this you may use the infinite high potential well as a mathematical model for an impenetrable barrier.
– hyportnex
Dec 19 '18 at 15:53
add a comment |
3 Answers
3
active
oldest
votes
The relation between the particle's wave function $psi(x)$, potential $V(x)$ and energy is
$$
E = int dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right) quad
label((*)
$$
Suppose $V(x)$ is bounded from below and is equal to $+infty$ on some interval $[x_1,x_2]$. If $psi(x)neq 0$ for $xin[x_1,x_2]$, then the energy $E$ is infinite. The term containing second derivative is always non-negative, so it can not compensate this infinity.
Update. This relation is well known in the quantum mechanics. I didn't mention that the norm of a wave function is usually taken to be $1$:
$$
int dx psi^*(x)psi(x) = 1
$$
Under this condition the Schrodinger equation
$$
-frachbar^22mpsi''(x) + V(x)psi(x) = Epsi(x)
$$
been multiplied by $psi^*(x)$ and integrated by $x$ gives the relation (*).
The term
$$
-frachbar^22mint dx psi^*(x)psi''(x)
$$
corresponds to the kinetic energy of a particle, so it must be non-negative. Indeed, integration by parts leads to the following manifestly non-negative expression
$$
frachbar^22mint dx psi'^*(x)psi'(x).
$$
By the way, quantity $psi''(x)/psi(x)$ can be either positive or negative.
1
Can I arrive at the same conclusion from the following equation: $-frachbar^22mfrac1psi (x)fracpartial ^2psi (x)partial x^2+V(x)=E$
– Soumee
Dec 19 '18 at 17:25
1
Can you please tell from which equation has the equation that you have mentioned been derived. It seems like it must have been something like: $dE = dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right)$ , which physically translates into small amount of energy in the interval dx. So, can we find out the small amount of energy in the interval dx? If so, from where?
– Soumee
Dec 19 '18 at 17:34
@Soumee I am not sure if you can come to this conclusion from the Schrodinger equation directly. You should consider properties of the $psi''(x)/psi(x)$ term in this case. I'll update my answer.
– Gec
Dec 19 '18 at 17:38
Thirdly, since $psi''(x)$ is non negative, the term $-frachbar^22mpsi''(x)$ as a whole is negative, which brings down the energy by a small amount from infinity, but ultimately $E=infty$
– Soumee
Dec 19 '18 at 17:39
@Soumee I have updated my answer.
– Gec
Dec 19 '18 at 17:59
|
show 15 more comments
Imagine a finite potential well of the form
$$
V(x) = begincases 0 & |x| < L/2 \ V_0 & rm otherwiseendcases
$$
You can solve Schrodinger's equation in the usual way, by splitting the domain in three parts, the resulting wave function will look something like this
$$
psi(x) = begincases psi_1(x) & x < L/2 \ psi_2(x) & |x| leq L/2 \ psi_3(x) & x > L/2endcases
$$
Inside the box $psi_2(x) sim e^pm ikx$, but outside the box you will find
$$
psi_3(x) sim e^-alpha x
$$
where
$$
alpha = fracsqrt2m(V_0 - E)hbar
$$
Now calculate the limit $V_0toinfty$ (infinity potential barrier), and you will see that $psi_3(x)to 0$, same as $psi_1(x)$. So in that sense the particle cannot penetrate the barrier and remains confined in the region $|x| leq L/2$
I think the OP is interested in a barrier where the potential goes back to 0 at some point instead of a well like you have here. Although the conclusion will ultimately look the same either way.
– Aaron Stevens
Dec 19 '18 at 16:10
While this does answer the question for the OPs exact scenario it doesn't put to rest the issue for similar scenarios. You'd need to redo the full calculation each time for any possible V(x) on either side of the barrier (which is obviously not doable).
– jacob1729
Dec 19 '18 at 18:07
add a comment |
Gec's answer is the one I would consider rigorous, but the intuitive answer is this:
Suppose you put a particle detector in the barrier. How often do you expect to measure a particle there? Answer: never, because if you did this then the particle after measurement would be in a position eigenstate that would force you to conclude it had infinite (expectation value of) energy. And we're disallowing that.
The only states where there is no chance to ever measure the particle in the barrier are those with $psi=0$ inside the barrier (or at least over a dense subset).
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
The relation between the particle's wave function $psi(x)$, potential $V(x)$ and energy is
$$
E = int dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right) quad
label((*)
$$
Suppose $V(x)$ is bounded from below and is equal to $+infty$ on some interval $[x_1,x_2]$. If $psi(x)neq 0$ for $xin[x_1,x_2]$, then the energy $E$ is infinite. The term containing second derivative is always non-negative, so it can not compensate this infinity.
Update. This relation is well known in the quantum mechanics. I didn't mention that the norm of a wave function is usually taken to be $1$:
$$
int dx psi^*(x)psi(x) = 1
$$
Under this condition the Schrodinger equation
$$
-frachbar^22mpsi''(x) + V(x)psi(x) = Epsi(x)
$$
been multiplied by $psi^*(x)$ and integrated by $x$ gives the relation (*).
The term
$$
-frachbar^22mint dx psi^*(x)psi''(x)
$$
corresponds to the kinetic energy of a particle, so it must be non-negative. Indeed, integration by parts leads to the following manifestly non-negative expression
$$
frachbar^22mint dx psi'^*(x)psi'(x).
$$
By the way, quantity $psi''(x)/psi(x)$ can be either positive or negative.
1
Can I arrive at the same conclusion from the following equation: $-frachbar^22mfrac1psi (x)fracpartial ^2psi (x)partial x^2+V(x)=E$
– Soumee
Dec 19 '18 at 17:25
1
Can you please tell from which equation has the equation that you have mentioned been derived. It seems like it must have been something like: $dE = dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right)$ , which physically translates into small amount of energy in the interval dx. So, can we find out the small amount of energy in the interval dx? If so, from where?
– Soumee
Dec 19 '18 at 17:34
@Soumee I am not sure if you can come to this conclusion from the Schrodinger equation directly. You should consider properties of the $psi''(x)/psi(x)$ term in this case. I'll update my answer.
– Gec
Dec 19 '18 at 17:38
Thirdly, since $psi''(x)$ is non negative, the term $-frachbar^22mpsi''(x)$ as a whole is negative, which brings down the energy by a small amount from infinity, but ultimately $E=infty$
– Soumee
Dec 19 '18 at 17:39
@Soumee I have updated my answer.
– Gec
Dec 19 '18 at 17:59
|
show 15 more comments
The relation between the particle's wave function $psi(x)$, potential $V(x)$ and energy is
$$
E = int dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right) quad
label((*)
$$
Suppose $V(x)$ is bounded from below and is equal to $+infty$ on some interval $[x_1,x_2]$. If $psi(x)neq 0$ for $xin[x_1,x_2]$, then the energy $E$ is infinite. The term containing second derivative is always non-negative, so it can not compensate this infinity.
Update. This relation is well known in the quantum mechanics. I didn't mention that the norm of a wave function is usually taken to be $1$:
$$
int dx psi^*(x)psi(x) = 1
$$
Under this condition the Schrodinger equation
$$
-frachbar^22mpsi''(x) + V(x)psi(x) = Epsi(x)
$$
been multiplied by $psi^*(x)$ and integrated by $x$ gives the relation (*).
The term
$$
-frachbar^22mint dx psi^*(x)psi''(x)
$$
corresponds to the kinetic energy of a particle, so it must be non-negative. Indeed, integration by parts leads to the following manifestly non-negative expression
$$
frachbar^22mint dx psi'^*(x)psi'(x).
$$
By the way, quantity $psi''(x)/psi(x)$ can be either positive or negative.
1
Can I arrive at the same conclusion from the following equation: $-frachbar^22mfrac1psi (x)fracpartial ^2psi (x)partial x^2+V(x)=E$
– Soumee
Dec 19 '18 at 17:25
1
Can you please tell from which equation has the equation that you have mentioned been derived. It seems like it must have been something like: $dE = dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right)$ , which physically translates into small amount of energy in the interval dx. So, can we find out the small amount of energy in the interval dx? If so, from where?
– Soumee
Dec 19 '18 at 17:34
@Soumee I am not sure if you can come to this conclusion from the Schrodinger equation directly. You should consider properties of the $psi''(x)/psi(x)$ term in this case. I'll update my answer.
– Gec
Dec 19 '18 at 17:38
Thirdly, since $psi''(x)$ is non negative, the term $-frachbar^22mpsi''(x)$ as a whole is negative, which brings down the energy by a small amount from infinity, but ultimately $E=infty$
– Soumee
Dec 19 '18 at 17:39
@Soumee I have updated my answer.
– Gec
Dec 19 '18 at 17:59
|
show 15 more comments
The relation between the particle's wave function $psi(x)$, potential $V(x)$ and energy is
$$
E = int dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right) quad
label((*)
$$
Suppose $V(x)$ is bounded from below and is equal to $+infty$ on some interval $[x_1,x_2]$. If $psi(x)neq 0$ for $xin[x_1,x_2]$, then the energy $E$ is infinite. The term containing second derivative is always non-negative, so it can not compensate this infinity.
Update. This relation is well known in the quantum mechanics. I didn't mention that the norm of a wave function is usually taken to be $1$:
$$
int dx psi^*(x)psi(x) = 1
$$
Under this condition the Schrodinger equation
$$
-frachbar^22mpsi''(x) + V(x)psi(x) = Epsi(x)
$$
been multiplied by $psi^*(x)$ and integrated by $x$ gives the relation (*).
The term
$$
-frachbar^22mint dx psi^*(x)psi''(x)
$$
corresponds to the kinetic energy of a particle, so it must be non-negative. Indeed, integration by parts leads to the following manifestly non-negative expression
$$
frachbar^22mint dx psi'^*(x)psi'(x).
$$
By the way, quantity $psi''(x)/psi(x)$ can be either positive or negative.
The relation between the particle's wave function $psi(x)$, potential $V(x)$ and energy is
$$
E = int dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right) quad
label((*)
$$
Suppose $V(x)$ is bounded from below and is equal to $+infty$ on some interval $[x_1,x_2]$. If $psi(x)neq 0$ for $xin[x_1,x_2]$, then the energy $E$ is infinite. The term containing second derivative is always non-negative, so it can not compensate this infinity.
Update. This relation is well known in the quantum mechanics. I didn't mention that the norm of a wave function is usually taken to be $1$:
$$
int dx psi^*(x)psi(x) = 1
$$
Under this condition the Schrodinger equation
$$
-frachbar^22mpsi''(x) + V(x)psi(x) = Epsi(x)
$$
been multiplied by $psi^*(x)$ and integrated by $x$ gives the relation (*).
The term
$$
-frachbar^22mint dx psi^*(x)psi''(x)
$$
corresponds to the kinetic energy of a particle, so it must be non-negative. Indeed, integration by parts leads to the following manifestly non-negative expression
$$
frachbar^22mint dx psi'^*(x)psi'(x).
$$
By the way, quantity $psi''(x)/psi(x)$ can be either positive or negative.
edited Dec 19 '18 at 17:58
answered Dec 19 '18 at 16:33
Gec
732211
732211
1
Can I arrive at the same conclusion from the following equation: $-frachbar^22mfrac1psi (x)fracpartial ^2psi (x)partial x^2+V(x)=E$
– Soumee
Dec 19 '18 at 17:25
1
Can you please tell from which equation has the equation that you have mentioned been derived. It seems like it must have been something like: $dE = dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right)$ , which physically translates into small amount of energy in the interval dx. So, can we find out the small amount of energy in the interval dx? If so, from where?
– Soumee
Dec 19 '18 at 17:34
@Soumee I am not sure if you can come to this conclusion from the Schrodinger equation directly. You should consider properties of the $psi''(x)/psi(x)$ term in this case. I'll update my answer.
– Gec
Dec 19 '18 at 17:38
Thirdly, since $psi''(x)$ is non negative, the term $-frachbar^22mpsi''(x)$ as a whole is negative, which brings down the energy by a small amount from infinity, but ultimately $E=infty$
– Soumee
Dec 19 '18 at 17:39
@Soumee I have updated my answer.
– Gec
Dec 19 '18 at 17:59
|
show 15 more comments
1
Can I arrive at the same conclusion from the following equation: $-frachbar^22mfrac1psi (x)fracpartial ^2psi (x)partial x^2+V(x)=E$
– Soumee
Dec 19 '18 at 17:25
1
Can you please tell from which equation has the equation that you have mentioned been derived. It seems like it must have been something like: $dE = dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right)$ , which physically translates into small amount of energy in the interval dx. So, can we find out the small amount of energy in the interval dx? If so, from where?
– Soumee
Dec 19 '18 at 17:34
@Soumee I am not sure if you can come to this conclusion from the Schrodinger equation directly. You should consider properties of the $psi''(x)/psi(x)$ term in this case. I'll update my answer.
– Gec
Dec 19 '18 at 17:38
Thirdly, since $psi''(x)$ is non negative, the term $-frachbar^22mpsi''(x)$ as a whole is negative, which brings down the energy by a small amount from infinity, but ultimately $E=infty$
– Soumee
Dec 19 '18 at 17:39
@Soumee I have updated my answer.
– Gec
Dec 19 '18 at 17:59
1
1
Can I arrive at the same conclusion from the following equation: $-frachbar^22mfrac1psi (x)fracpartial ^2psi (x)partial x^2+V(x)=E$
– Soumee
Dec 19 '18 at 17:25
Can I arrive at the same conclusion from the following equation: $-frachbar^22mfrac1psi (x)fracpartial ^2psi (x)partial x^2+V(x)=E$
– Soumee
Dec 19 '18 at 17:25
1
1
Can you please tell from which equation has the equation that you have mentioned been derived. It seems like it must have been something like: $dE = dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right)$ , which physically translates into small amount of energy in the interval dx. So, can we find out the small amount of energy in the interval dx? If so, from where?
– Soumee
Dec 19 '18 at 17:34
Can you please tell from which equation has the equation that you have mentioned been derived. It seems like it must have been something like: $dE = dx psi^*(x)left(-frachbar^22mpsi''(x) + V(x)psi(x)right)$ , which physically translates into small amount of energy in the interval dx. So, can we find out the small amount of energy in the interval dx? If so, from where?
– Soumee
Dec 19 '18 at 17:34
@Soumee I am not sure if you can come to this conclusion from the Schrodinger equation directly. You should consider properties of the $psi''(x)/psi(x)$ term in this case. I'll update my answer.
– Gec
Dec 19 '18 at 17:38
@Soumee I am not sure if you can come to this conclusion from the Schrodinger equation directly. You should consider properties of the $psi''(x)/psi(x)$ term in this case. I'll update my answer.
– Gec
Dec 19 '18 at 17:38
Thirdly, since $psi''(x)$ is non negative, the term $-frachbar^22mpsi''(x)$ as a whole is negative, which brings down the energy by a small amount from infinity, but ultimately $E=infty$
– Soumee
Dec 19 '18 at 17:39
Thirdly, since $psi''(x)$ is non negative, the term $-frachbar^22mpsi''(x)$ as a whole is negative, which brings down the energy by a small amount from infinity, but ultimately $E=infty$
– Soumee
Dec 19 '18 at 17:39
@Soumee I have updated my answer.
– Gec
Dec 19 '18 at 17:59
@Soumee I have updated my answer.
– Gec
Dec 19 '18 at 17:59
|
show 15 more comments
Imagine a finite potential well of the form
$$
V(x) = begincases 0 & |x| < L/2 \ V_0 & rm otherwiseendcases
$$
You can solve Schrodinger's equation in the usual way, by splitting the domain in three parts, the resulting wave function will look something like this
$$
psi(x) = begincases psi_1(x) & x < L/2 \ psi_2(x) & |x| leq L/2 \ psi_3(x) & x > L/2endcases
$$
Inside the box $psi_2(x) sim e^pm ikx$, but outside the box you will find
$$
psi_3(x) sim e^-alpha x
$$
where
$$
alpha = fracsqrt2m(V_0 - E)hbar
$$
Now calculate the limit $V_0toinfty$ (infinity potential barrier), and you will see that $psi_3(x)to 0$, same as $psi_1(x)$. So in that sense the particle cannot penetrate the barrier and remains confined in the region $|x| leq L/2$
I think the OP is interested in a barrier where the potential goes back to 0 at some point instead of a well like you have here. Although the conclusion will ultimately look the same either way.
– Aaron Stevens
Dec 19 '18 at 16:10
While this does answer the question for the OPs exact scenario it doesn't put to rest the issue for similar scenarios. You'd need to redo the full calculation each time for any possible V(x) on either side of the barrier (which is obviously not doable).
– jacob1729
Dec 19 '18 at 18:07
add a comment |
Imagine a finite potential well of the form
$$
V(x) = begincases 0 & |x| < L/2 \ V_0 & rm otherwiseendcases
$$
You can solve Schrodinger's equation in the usual way, by splitting the domain in three parts, the resulting wave function will look something like this
$$
psi(x) = begincases psi_1(x) & x < L/2 \ psi_2(x) & |x| leq L/2 \ psi_3(x) & x > L/2endcases
$$
Inside the box $psi_2(x) sim e^pm ikx$, but outside the box you will find
$$
psi_3(x) sim e^-alpha x
$$
where
$$
alpha = fracsqrt2m(V_0 - E)hbar
$$
Now calculate the limit $V_0toinfty$ (infinity potential barrier), and you will see that $psi_3(x)to 0$, same as $psi_1(x)$. So in that sense the particle cannot penetrate the barrier and remains confined in the region $|x| leq L/2$
I think the OP is interested in a barrier where the potential goes back to 0 at some point instead of a well like you have here. Although the conclusion will ultimately look the same either way.
– Aaron Stevens
Dec 19 '18 at 16:10
While this does answer the question for the OPs exact scenario it doesn't put to rest the issue for similar scenarios. You'd need to redo the full calculation each time for any possible V(x) on either side of the barrier (which is obviously not doable).
– jacob1729
Dec 19 '18 at 18:07
add a comment |
Imagine a finite potential well of the form
$$
V(x) = begincases 0 & |x| < L/2 \ V_0 & rm otherwiseendcases
$$
You can solve Schrodinger's equation in the usual way, by splitting the domain in three parts, the resulting wave function will look something like this
$$
psi(x) = begincases psi_1(x) & x < L/2 \ psi_2(x) & |x| leq L/2 \ psi_3(x) & x > L/2endcases
$$
Inside the box $psi_2(x) sim e^pm ikx$, but outside the box you will find
$$
psi_3(x) sim e^-alpha x
$$
where
$$
alpha = fracsqrt2m(V_0 - E)hbar
$$
Now calculate the limit $V_0toinfty$ (infinity potential barrier), and you will see that $psi_3(x)to 0$, same as $psi_1(x)$. So in that sense the particle cannot penetrate the barrier and remains confined in the region $|x| leq L/2$
Imagine a finite potential well of the form
$$
V(x) = begincases 0 & |x| < L/2 \ V_0 & rm otherwiseendcases
$$
You can solve Schrodinger's equation in the usual way, by splitting the domain in three parts, the resulting wave function will look something like this
$$
psi(x) = begincases psi_1(x) & x < L/2 \ psi_2(x) & |x| leq L/2 \ psi_3(x) & x > L/2endcases
$$
Inside the box $psi_2(x) sim e^pm ikx$, but outside the box you will find
$$
psi_3(x) sim e^-alpha x
$$
where
$$
alpha = fracsqrt2m(V_0 - E)hbar
$$
Now calculate the limit $V_0toinfty$ (infinity potential barrier), and you will see that $psi_3(x)to 0$, same as $psi_1(x)$. So in that sense the particle cannot penetrate the barrier and remains confined in the region $|x| leq L/2$
answered Dec 19 '18 at 15:56
caverac
5,3832923
5,3832923
I think the OP is interested in a barrier where the potential goes back to 0 at some point instead of a well like you have here. Although the conclusion will ultimately look the same either way.
– Aaron Stevens
Dec 19 '18 at 16:10
While this does answer the question for the OPs exact scenario it doesn't put to rest the issue for similar scenarios. You'd need to redo the full calculation each time for any possible V(x) on either side of the barrier (which is obviously not doable).
– jacob1729
Dec 19 '18 at 18:07
add a comment |
I think the OP is interested in a barrier where the potential goes back to 0 at some point instead of a well like you have here. Although the conclusion will ultimately look the same either way.
– Aaron Stevens
Dec 19 '18 at 16:10
While this does answer the question for the OPs exact scenario it doesn't put to rest the issue for similar scenarios. You'd need to redo the full calculation each time for any possible V(x) on either side of the barrier (which is obviously not doable).
– jacob1729
Dec 19 '18 at 18:07
I think the OP is interested in a barrier where the potential goes back to 0 at some point instead of a well like you have here. Although the conclusion will ultimately look the same either way.
– Aaron Stevens
Dec 19 '18 at 16:10
I think the OP is interested in a barrier where the potential goes back to 0 at some point instead of a well like you have here. Although the conclusion will ultimately look the same either way.
– Aaron Stevens
Dec 19 '18 at 16:10
While this does answer the question for the OPs exact scenario it doesn't put to rest the issue for similar scenarios. You'd need to redo the full calculation each time for any possible V(x) on either side of the barrier (which is obviously not doable).
– jacob1729
Dec 19 '18 at 18:07
While this does answer the question for the OPs exact scenario it doesn't put to rest the issue for similar scenarios. You'd need to redo the full calculation each time for any possible V(x) on either side of the barrier (which is obviously not doable).
– jacob1729
Dec 19 '18 at 18:07
add a comment |
Gec's answer is the one I would consider rigorous, but the intuitive answer is this:
Suppose you put a particle detector in the barrier. How often do you expect to measure a particle there? Answer: never, because if you did this then the particle after measurement would be in a position eigenstate that would force you to conclude it had infinite (expectation value of) energy. And we're disallowing that.
The only states where there is no chance to ever measure the particle in the barrier are those with $psi=0$ inside the barrier (or at least over a dense subset).
add a comment |
Gec's answer is the one I would consider rigorous, but the intuitive answer is this:
Suppose you put a particle detector in the barrier. How often do you expect to measure a particle there? Answer: never, because if you did this then the particle after measurement would be in a position eigenstate that would force you to conclude it had infinite (expectation value of) energy. And we're disallowing that.
The only states where there is no chance to ever measure the particle in the barrier are those with $psi=0$ inside the barrier (or at least over a dense subset).
add a comment |
Gec's answer is the one I would consider rigorous, but the intuitive answer is this:
Suppose you put a particle detector in the barrier. How often do you expect to measure a particle there? Answer: never, because if you did this then the particle after measurement would be in a position eigenstate that would force you to conclude it had infinite (expectation value of) energy. And we're disallowing that.
The only states where there is no chance to ever measure the particle in the barrier are those with $psi=0$ inside the barrier (or at least over a dense subset).
Gec's answer is the one I would consider rigorous, but the intuitive answer is this:
Suppose you put a particle detector in the barrier. How often do you expect to measure a particle there? Answer: never, because if you did this then the particle after measurement would be in a position eigenstate that would force you to conclude it had infinite (expectation value of) energy. And we're disallowing that.
The only states where there is no chance to ever measure the particle in the barrier are those with $psi=0$ inside the barrier (or at least over a dense subset).
answered Dec 19 '18 at 18:05
jacob1729
661412
661412
add a comment |
add a comment |
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You start with the tunneling probability knowing that it is exponentially small with the finite barrier height, therefore if the latter is infinite the former is zero. Once you see this you may use the infinite high potential well as a mathematical model for an impenetrable barrier.
– hyportnex
Dec 19 '18 at 15:53