Assume the one-dimensional particle wave function is ? = ...

Assume the one-dimensional particle wave function is ? = \begin{cases} A(1+x)e^{-ax} & \text{for } x>0\\ A(1-x)e^{ax} & \text{for } x<0 \end{cases}, where a is constant. (a) If the potential field is continuous, please calculate the normalized wave function. (5 points) (b) If at x = ?, V(x)=0, please calculate the total energy E and the potential energy V(x). (10 points) (Hint: By using the continuity conditions of ? and ?' at x = 0, we can determine a, and then use the normalization of the wave function to find A. Finally, we can solve for E and V(x) using the stationary Schrödinger equation.)

Transcript

00:01 Now basically in this question you are giving a wave function, right? of a particle, you know, you can into a potential way of certain lengths.

00:08 So you have infinite potential wear, right? with some length air and the particle has a mass m.

00:16 And the wave function you are giving.

00:20 So it's actually basically is a x not initially, okay? at the initial moment, the wave functions, just two over square of seven, eight, and the sign.

00:30 Something in pi x a for a.

00:32 And the next one is six over seven a, sign two pi x over a, and then plus two over screwed of seven a, sign three pi x over a, right? so let's read, we can rewrite this in a bit.

00:55 And actually this is just the ground, this is one, this i would call this, you know, the egg and state of this particle are denoted by a sine or x, right? so you can rewrite this as, you can rewrite this as 2 over square of 7a, and then, you know, squalote of 2 over a, right? and then sign, you know, i don't want to write a sign, actually.

01:21 I want to just write down using, just rewrite it using squarelyt of a over 2, actually.

01:31 And then times sine 1 x, okay, and then plus, similarly, square of 6 over 7a and square of a over 2, sine 2, x basically, and 3rd, you have plus 2 over 7a, so a over 2, and side 3 x, okay.

01:55 So that will give us, you rule that things a little bit, then we find this to be actually 1 over square root 14, right? si 2 and plus square root of 3 over 7, side 2 and plus similarly, 1 over square 14, 7 size 3, right? if we square the sense, if we square this, this will get 1 over 14, this will get 3 over 7 and this will get 1 over 14.

02:29 Right, you put them together, you get actually five...

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