I was struggling with the structure of Schrödinger's equation over time when we consider a real physical process. The problem can be stated in two parts as follows,
Is it possible to construct a wave function for a system with a new introduced potential from the previously known wave function solution of the same system before the introduction of the potential. Eg: I want to get the solution of an infinite square well where the dirac delta potential is placed at the center from the usual square well problem without potential. We are not talking about entirely two things. Basically, it is a process of introducing an infinite square well and slowly putting the delta potential at the center over time.
The second part is where we don't introduce any kind of bizarre potentials, instead we change the boundary conditions over time. It is like expanding the square well or making some regions in space as inaccessible to the wave function.
Note:
There is a difference between wave function going to zero at some place and the space itself becomes inaccessible to the wave function.
One should not to be confused these things with the concept of perturbation even though it looks similar in some places. In perturbation, there is no change in boundary conditions or any introduction of completely arbitrary potentials. It is a very small change in the initial potential with same conditions maintained at
all the time.
To understand this, I took a series of problems with different levels in ascending order. First, we try to solve the general schrodinger equation for one dimensional infinite square well with a very little change by moving it to length "a" and "2a". Except for the limits, the solution is written in normal convention of "k" and "E" , we have $$ \psi = A e^{ikx} + B e^{-ikx}$$ and applying the left, right limits, $$A e^{ika} + Be^{-ka} = 0 \,\,\,...(1)\\ A e^{i2ka} + B e^{-i2ka}\,\,\,...(2)$$ (1) gives $$ A e^{i2ka}= -B $$ provided $ e^{ika}$ don't become zero. $$(2) \rightarrow \,\,\, B (e^{-i2ka}-1) = 0 \\ B = 0 \,\,or\,\, e^{-2ika} = 1$$ B = 0 gives A=0 i.e. trivial solution where there is no wave function. It leaves us with $$ e^{-i2ka} = 1 \\ A = -B\\ cos2ka = 1 \,\,and\,\, sin 2ka = 0$$both conditions need to be satisfied if, $$ sin 2ka = 0 \,\,and \,\,cos2ka = 1 \\ 2ka = n\pi \,\,and\,\, 2ka = 2n\pi$$ larger condition is the common one i.e. $$ ka = n\pi$$ which gives us the exact solution $$\psi = A e^{ika} - A e^{-ika} = C sin 2ka $$ where "C" is again an overall constant to be determined by normalization and $ ka = n\pi$ Normalization gives again the same constant [ since, $$ sin^2 2kx = \frac{1 - cos2kx}{2} $$ where integration is from "a" to "2a" i.e. sin 2ka = sin ka = 0 ] i.e. $\sqrt{\frac{2}{a}}$ giving us the complete solution,$$\psi (x) = \sqrt{\frac{2}{a}} sinkx \,\,\,\,and\,\, k = \frac{n\pi}{a}$$
It was quite a bit surprise for me, that there is no any kind of reflection about the shift of the coordinate system or the shift about the coordinate syystem. Even though the wave function depends on the position, it doesn't reflect anything i.e. independent of coordinate system.
But don't think they are exactly the same in every way because now the limits have changed to give negative values for all x values in between "a" and "2a". But, it doesn't matter since the measurable quantities depend on the square of the wave function - saves the day.
With this information, if I take the a box of length "2a" and split it into two parts "0" to "a" and "a" to "2a" with an imaginary line drawn at x = a . Actually, there is no anything physical about this problem, but I just wanted to see how the problem will evolve from one to two.
As we have two parts, the solution is, $$ \psi(x) = Ae^{ikx} + Be^{-ikx} \,\,\,\,0\leq{x}\leq{a} \\ Ce^{ikx} + De^{-ikx} \,\,\,\,{a}\leq{x}\leq{2a}$$ with the boundary condition that the first part should vanish at the left x= 0 and the right should vanish at x=2a, we have, $$ A+B = 0 \\ C e^{2ika} + De^{-i2ka} = 0 \\ \rightarrow \,\,\,\, Ce^{i4ka} = -D $$
If we seek the continuity at x = a, by comparing independent sine and cosine terms $$ A - B = C - D \\ (C+D) coska = 0 $$ Putting all the conditions together and working out gives, $$ A = C \\ B = D \\ A= -B\\ k = \frac{n\pi}{2a}$$ which gives the wave function as, $$ \psi(x) = F sin(kx) $$ After Normalizing, $$\psi(x) = \frac{1}{\sqrt{a}} sin(kx) \,\,\,\,or\\ \frac{1}{\sqrt{a}} i sin(kx) \\ k = \frac{n\pi}{2a} \,\,\,\, n= 0, \pm{1},\pm{2},..$$
I put 'i' because it is the exact solution I got and I just wanted to show "i" doesn't affect our wave function in anyway. The physical results are the same.
Actually we can talk a little about this 'i' . Any wave function, even though it is normalized, there is no unique form for its' structure. It is always possible to multiply it by 'i' or '-i' like we can do it in classical sense, multiplying by '1'. Hence, there is no way of telling whether your wave function solution is real or complex by just looking at it.
This itself is a justification for why the modulus square of the wave function that all matters for any physical result (which is called the probability density) and not the value of the wave function (probability amplitude).
Is it possible to construct a wave function for a system with a new introduced potential from the previously known wave function solution of the same system before the introduction of the potential. Eg: I want to get the solution of an infinite square well where the dirac delta potential is placed at the center from the usual square well problem without potential. We are not talking about entirely two things. Basically, it is a process of introducing an infinite square well and slowly putting the delta potential at the center over time.
The second part is where we don't introduce any kind of bizarre potentials, instead we change the boundary conditions over time. It is like expanding the square well or making some regions in space as inaccessible to the wave function.
Note:
There is a difference between wave function going to zero at some place and the space itself becomes inaccessible to the wave function.
One should not to be confused these things with the concept of perturbation even though it looks similar in some places. In perturbation, there is no change in boundary conditions or any introduction of completely arbitrary potentials. It is a very small change in the initial potential with same conditions maintained at
all the time.
To understand this, I took a series of problems with different levels in ascending order. First, we try to solve the general schrodinger equation for one dimensional infinite square well with a very little change by moving it to length "a" and "2a". Except for the limits, the solution is written in normal convention of "k" and "E" , we have $$ \psi = A e^{ikx} + B e^{-ikx}$$ and applying the left, right limits, $$A e^{ika} + Be^{-ka} = 0 \,\,\,...(1)\\ A e^{i2ka} + B e^{-i2ka}\,\,\,...(2)$$ (1) gives $$ A e^{i2ka}= -B $$ provided $ e^{ika}$ don't become zero. $$(2) \rightarrow \,\,\, B (e^{-i2ka}-1) = 0 \\ B = 0 \,\,or\,\, e^{-2ika} = 1$$ B = 0 gives A=0 i.e. trivial solution where there is no wave function. It leaves us with $$ e^{-i2ka} = 1 \\ A = -B\\ cos2ka = 1 \,\,and\,\, sin 2ka = 0$$both conditions need to be satisfied if, $$ sin 2ka = 0 \,\,and \,\,cos2ka = 1 \\ 2ka = n\pi \,\,and\,\, 2ka = 2n\pi$$ larger condition is the common one i.e. $$ ka = n\pi$$ which gives us the exact solution $$\psi = A e^{ika} - A e^{-ika} = C sin 2ka $$ where "C" is again an overall constant to be determined by normalization and $ ka = n\pi$ Normalization gives again the same constant [ since, $$ sin^2 2kx = \frac{1 - cos2kx}{2} $$ where integration is from "a" to "2a" i.e. sin 2ka = sin ka = 0 ] i.e. $\sqrt{\frac{2}{a}}$ giving us the complete solution,$$\psi (x) = \sqrt{\frac{2}{a}} sinkx \,\,\,\,and\,\, k = \frac{n\pi}{a}$$
It was quite a bit surprise for me, that there is no any kind of reflection about the shift of the coordinate system or the shift about the coordinate syystem. Even though the wave function depends on the position, it doesn't reflect anything i.e. independent of coordinate system.
But don't think they are exactly the same in every way because now the limits have changed to give negative values for all x values in between "a" and "2a". But, it doesn't matter since the measurable quantities depend on the square of the wave function - saves the day.
With this information, if I take the a box of length "2a" and split it into two parts "0" to "a" and "a" to "2a" with an imaginary line drawn at x = a . Actually, there is no anything physical about this problem, but I just wanted to see how the problem will evolve from one to two.
As we have two parts, the solution is, $$ \psi(x) = Ae^{ikx} + Be^{-ikx} \,\,\,\,0\leq{x}\leq{a} \\ Ce^{ikx} + De^{-ikx} \,\,\,\,{a}\leq{x}\leq{2a}$$ with the boundary condition that the first part should vanish at the left x= 0 and the right should vanish at x=2a, we have, $$ A+B = 0 \\ C e^{2ika} + De^{-i2ka} = 0 \\ \rightarrow \,\,\,\, Ce^{i4ka} = -D $$
If we seek the continuity at x = a, by comparing independent sine and cosine terms $$ A - B = C - D \\ (C+D) coska = 0 $$ Putting all the conditions together and working out gives, $$ A = C \\ B = D \\ A= -B\\ k = \frac{n\pi}{2a}$$ which gives the wave function as, $$ \psi(x) = F sin(kx) $$ After Normalizing, $$\psi(x) = \frac{1}{\sqrt{a}} sin(kx) \,\,\,\,or\\ \frac{1}{\sqrt{a}} i sin(kx) \\ k = \frac{n\pi}{2a} \,\,\,\, n= 0, \pm{1},\pm{2},..$$
I put 'i' because it is the exact solution I got and I just wanted to show "i" doesn't affect our wave function in anyway. The physical results are the same.
Actually we can talk a little about this 'i' . Any wave function, even though it is normalized, there is no unique form for its' structure. It is always possible to multiply it by 'i' or '-i' like we can do it in classical sense, multiplying by '1'. Hence, there is no way of telling whether your wave function solution is real or complex by just looking at it.
This itself is a justification for why the modulus square of the wave function that all matters for any physical result (which is called the probability density) and not the value of the wave function (probability amplitude).
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