Integral straight from quantum physics

My Gavanan folks. Today we have an integral from the quantum realm and by that I mean this is an actual integral we find out of solving a problem in quantum mechanics. For the interested viewers this is actually an exercise in Sakurai's brilliant text on quantum mechanics. So just some more background again for the interested viewers. This actual problem is about calculating a sort of expectation value for an operator. That is to say we're interested in inner product of zero e to the i k x cap zero. So this would be the expectation value of the operator e to the i k x cap in the ground state. That's what the zero ket means. The zero ket actually represents the ground state for the simple harmonic oscillator or the state of minimum energy. An x cap here is the position operator. That's the quantum mechanics way of introducing the observable or physical quantity that is position. And psi nod here is actually the wave function of the harmonic oscillator in this ground state. More specifically it's the position space wave function which means it equals the inner product of x and zero. Where x ket here represents a position eigenstate and this leftward ket is the bra. So that's the position eigenbra. God damn it I absolutely love the Dirac-Dirac-ket notation. It is absolutely brilliant. Now to get from here to here all we need to do is make use of the fact that the position eigenstates form a complete set. And if you have an analytic function of some observable or physical quantity a and you apply that to an eigenstate of that observable then you just get another eigenvalue equation where the eigenvalue is the function applied to the eigenvalue corresponding to that eigenstate. It's relatively simple and I will skip that bit for now but don't worry I will provide that bit of the solution in an Instagram post. So do follow me over there as well. Anyway so we do know what the ground state wave function for the harmonic oscillator is. It is in fact 1 over root x nought times root pi times exp of negative one half x squared over x nought squared where x nought is a constant that is root h cut over m for the mass of the particle and omega for its frequency. Now because this thing is real valued its conjugate equals well itself which implies that the absolute value of phi nought of x squared is well just 1 over x nought root pi e of negative x squared over x nought squared. Now returning back to the target integral we realize that it equals 1 over x nought root pi integral negative to positive infinity e to the ikx times e to the negative x squared over x nought squared dx. Of course we can multiply out the two exponentials and get 1 over x nought root pi integral negative to positive infinity e to the negative sign factored out we have x squared over x nought squared minus ikx dx. And now for a simple completing square algorithm. So we have x squared over x nought squared minus ikx which can be written out as x over x nought whole thing squared minus 2 x over x nought. And what number would I need to get ikx over here? Well for that I would just need ikx nought over 2. Okay cool. And now we need a version of zero and the very special version of zero we need in this case is of course plus ikx nought over 2 squared minus ikx nought over 2 squared. Now all of this is just a perfect square that is of course x over x nought minus ikx nought over 2 whole thing squared minus i squared k squared x nought squared over 4. I squared is of course equal to negative 1 so we have a cancellation of negative signs and we have k squared x nought squared over 4. So all of this implies that the target integral i is 1 over x nought root pi integral negative to positive infinity. What exactly? Oh yeah we have e to the negative x over x nought minus ikx over 2 whole thing squared minus k squared x nought squared over 4. Of course again using properties of the exponential function we'll write this as another exponential being multiplied. e to the negative k squared x nought squared over 4 dx. And we realize that this thing is just e to some constant which of course means it is a constant. So we have 1 over x nought root pi terribly sorry about that e to the negative k squared x nought squared over 4 times the integral from negative to positive infinity of e to the negative x over x nought minus ikx over 2 whole thing squared dx. So this is actually interesting. This is a Gaussian integral with an imaginary shift and just for convenience purposes I'm going to transform x over x nought into t which implies that dx is just x nought dt or wait even better yet we do have a dx over x nought situation going on. So dx over x nought is just dt and we'll write the integral in a slightly different manner. It's 1 over root pi e to the negative k squared x nought squared over 4 integral. The limits are clearly not bothered and we have e to the negative t minus ikx over 2 whole thing squared dt. So again Gaussian integral with a complex shift and we're about to see that the complex that the complex shift simply does not matter. Now that seems like a bold statement to make but hey this is quantum mechanics after all. So how are we going to prove it? Well I'll define an integral function i of a parameter alpha as the integral from negative to positive infinity of e to the negative x plus i alpha squared dx where alpha is a real number and I plan to differentiate this thing with respect to alpha yielding i prime of alpha on the left and on the right after switching up the order of integration and differentiation operators. We have the integral of the partial derivative with respect to alpha of e to the negative x plus i alpha squared dx. So differentiating with respect to alpha now yields well the exponential function repeated and then because of the chain rule I have to differentiate the argument. So that is a negative 2 outside and differentiating with the power rule means I still have this factor here of x plus i times alpha and now differentiating this further because of the chain rule I have an extra factor of i. Okay cool but now notice that because we're integrating with respect to x we actually have this really nice anti-derivative scenario for which I actually need the negative 2 so on integration we'll get i times e to the negative x plus i alpha squared with the limits being negative and positive infinity and if you expand the square then we do get a factor of e to the negative x squared times another exponential function and this term goes to zero as x approaches positive or negative infinity so this means that the integral just crashes out to zero in other words i prime of alpha is zero but that means i of alpha is a constant with respect to alpha which means if you know one value of i of alpha you know them all and we do know one value of i of alpha that is i of zero which is just the integral from negative to positive infinity of e to the negative x squared dx which of course is the classic Gaussian integral which evaluates out to root pi which is absolutely beautiful and that means we have the result for our target integral and let me just copy and paste this down here so this is the target integral and we see that the one over root pi cancels out with the result of the integral so this means that i here equals e to the negative k squared x naught squared over four but we do know what x naught and hence x naught square sorts out to so that's h cut over m times omega which further implies that i here equals exp of negative k squared terribly sorry about that negative k squared h cut over 4 m omega which is a pretty good looking result indeed i hope you enjoyed the video be sure to like and subscribe do drop your follow on instagram as well share the video to you know likes on your friends that hey i just checked out this really awesome quantum integral i don't know thank you see you next time