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Prove that the form of the Schrödinger equation ( $11.17$ ) is unchanged under the gauge transformation (11.19)....

Question

Prove that the form of the Schrödinger equation ( $11.17$ ) is unchanged under the gauge transformation (11.19).

Prove that the form of the Schrödinger equation ( $11.17$ ) is unchanged under the gauge transformation (11.19).



Answers

PROVE: Algebraic Invariants A quantity is invariant under rotation if it does not change when the axes are rotated. It was stated in the text that for the general equation of a conic the quantity $B^{2}-4 A C$ is invariant under rotation.
(a) Use the formulas for $A^{\prime}, B^{\prime},$ and $C^{\prime}$ on page 862 to prove that the quantity $B^{2}-4 A C$ is invariant under rotation; that is, show that
$$B^{2}-4 A C=B^{\prime 2}-4 A^{\prime} C^{\prime}$$
(b) Prove that $A+C$ is invariant under rotation.
(c) Is the quantity $F$ invariant under rotation?

We need to show that, as is a linear transformation and were given the condition that, um, t and s satisfy the credibility creations and tease and enough inspiration. Okay. How? Just write down our goal here. You near transformation. Okay. All right. He ordered to do that. We take arbitrary, take any you and B from our in. So we pluck. Are you gonna be to transferred to the ingenious information as we will have X? And why accident. Why, Okay, now we consider t off acts. Now, remember, acts can be represented as ass off you. So plugging as self you now, since ass is it, it's the number. As an S and T said it's by the embers in credibility questions. So this is this keeps us directly back to you. So similarly, we have, um, t apply is factor VII. Okay, so now we consider ask you plus. Yeah. So since you plus V can be represented as a transformation of x and y under T, So that will be t ext plus t y. Now remember the inner transformation by our assumption, the transformation transformation T is a leaner transformation so we can apply the property. Help me near says Formacion here at his t x s t y t X plus y okay. And by our convertibility equations, these keeps us express What? And recall that X is s you And why is SP so that ask you Class sp and our first condition It said his buy. So the next condition we need to check it's whether s off the escape, their off vector under the interview and then you're under the transpiration Information off ass will be the same as the skater times times the transformation off s so that is directly to show that c o s u because s self see you as I see you, that's it. And to do that first consider considered TLC X because he's a linear transformation. So it's just a tee off a sea of t x c times t x and T acts by our previous by our previous reasoning here, T X will be you. It's C times. You okay? So s off. See, Time's view can be written as as soft tee time. See x right, which is a substitute c times you by. I don't trust for Michel t so That's, uh, tee times tfc times eight times X. Now we can we can apply the convertibility equations, so this gives us see X. Now, remember, ax by over by this equation here, acts can be represented as ask you so at a same as C times Ask you so as I see you is C times as here. So we're done. That means ass, he said.

Okay, This question wants us to prove that projection onto a line is a linear transformation. So we can actually do this using our formula. So the projection onto a line of X is equal to the dot product of X and the vector you, which will say is on the line over the magnitude of you squared all times you so that we have an explicit formula for this projection. Now we can actually use our linearity condition. So for a linear transformation, t of a X plus B y is equal to a times TF X plus be times t of Why? So we're just gonna plug plug in a X plus B y into this projection formula. So if we plug in a X plus B y, we would get a X plus b y dotted with you all over the magnitude of you squared times you and we know the dot product itself is linear. So we can distribute this a x and B y so we'd get and a X dotted with you, plus B, why dotted with you all over the magnitude of you squared times you Then we can split up these fractions into a X started with you over the magnitude of you squared plus b y dotted with you over the magnitude of you squared times you or factoring out the constants in front of the dot product Because we can pull scaler multiples in front of the dot product would give us a Times X started with you over the magnitude of you squared times you plus pulling the be out of the second dot product. Why doubted with you over the magnitude of you squared times you. And now if we look at these things in parentheses, Well, this first bracket is the projection of Exxon to you. And the second bracket is the projection of why under you So therefore, since T of a X plus B y is equal to 80 of x plus b t l y, we have a linear transformation.

Okay, This question wants us to prove that the reflection is also a linear transformation. So again, to prove a transformation is a linear, you need to show that t of a X plus b y is equal to a times TX plus b times T y. And again, this is just our linearity condition that we've been applying. So it would make sense that if we're trying to test this, we plug in t of a x plus B y into our reflection. So we want the reflection over a line of a X plus B y, and that would be equal to two times the projection on to this line of a exposed B Y and then minus a X plus B y. And due to our work on a previous problem we've seen the projection itself is linear, so we can actually split this up so we can write. This projection argument has two times the quantity a times the projection on to El of X plus be times the projection on to El of why and then still minus a X plus B y. And now let's group are like terms together. So we get to a projection on the l of X minus a X and then plus to be projection on the l of why minus B y. And now we can factor out and a from the first set. So we get a times two times the projection on the AL of X minus X plus the times two times the projection on the l of y bite us. Why and now we see that this thing in parentheses is what we said. There are a reflection over a line on the X. We're sorry reflection of acts over the line. And then similarly, our second set of parentheses is the reflection of why over the line. And now, since we have a times t of X plus B times TV, why is equal to to you they exposed b y. That is our linearity condition. So we're done. Which tells us that the reflection is indeed linear

So starting with the Lorentz transformation, which gives the prime coordinates in terms of the UN prime coordinates, we go ahead and write ex prime as equaling gamma times X minus U T as well as t prime equaling gamma times t minus you x over C squared as you see here. Also, we have y prime equaling y and Z Prime equaling Z. But those two equations will not change. So we work with these equations, and the first thing we're gonna do is we're gonna add this combination together. So we're going to take the equation for Ex Prime, which is again shown here. We're going to add to it the speed you multiplied by the equation for t primed t prime. So we do that and we now have this combination. So what I have done here is let me go back one step once again, I have taken the equation for t prime which you see right here, and I have multiplied through by the velocity you the speed you So this tea is multiplied by you and you hear becomes you squared since we're all flying by you. So now we combine terms between this quantity and this quantity here. And actually you'll see that we have two terms that canceled. So we have gamma mine times negative ut right here and we have which is right here. And we have gamma ut right here, which is positive. So those will cancel in our next step. We have gamma X times, one which we obtained from right here. And we also have gamma X times Negative. You squared over C squared, which is from this term here. So what I've done is I've factored out Gamma X. So we now have gamma X times one minus. You squared over C squared. But remember that gamma is equal to one over the square root of one minus, you squared over C squared. So with that in mind, we can now write this as one divided by the square root of one minus. You squared over C squared times, X times one minus. You squared over C squared. And this then will simplify thio X times the square root of one minus. You square over C squared, which is then equal to X over gamma. And this then gives us if we multiply both sides by gamma we then have X equals gamma times, ex prime plus ut prime. So this is one of the equations for the inverse of the Lawrence transformation that we started with. So this is analogous to this first equation that you see right here Onley. The sign of the speed is the reverse. Instead of having a minus sign right here we have a plus sign and you can see that. Ah, if we scroll back down, you see that we have a plus sign right here on the speed. So with a similar process, we can get an inverse transformation for the time Variable the time coordinate. So we add this combination. Together we have t prime plus, you oversee square times, ex prime. So we insert the equation for T prime and the equation for ex prime and then we combined terms together. So we've taken the equation for T prime that you see here and the equation for ex prime that you see right here. And we will be inserting these now. And that's where what you see on the right hand side right here and we once again, we'll combine like terms and see what cancels and we do have, um right here we have minus you X over C squared and we have here you x over c squared. And of course, they're both multiplied by gamma. But those will cancel. And that's what you see going on right here. These two terms cancel. We have gamaty, which is right here. And we have ah minus gamma ut right here that is multiplied by this You oversee squared. And that gives us this term minus gamma. You squared t oversee squared. So all that we have left are these first two terms which we can then factor as's follows to get gamaty times one minus you square over c squared. But gamma is one divided by the square root of one minus. You squared over c squared and so we write it this way We get t times the square root of one minus. You squared over C squared, which is then t over gamma. And again we can then right t is Equalling gamma times t prime. Plus, you oversee square times ex prime. And that is the transformation equation, the inverse transformation equation that gives us t in terms of the prime variables. So that is the other result that we obtained. And once again you see that that we have a plus sign instead of a minus. Sign on you.


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