Consider a free-particle wave packet in one dimension. At...

Consider a free-particle wave packet in one dimension. At $t=0$ it satisfies the minimum uncertainty relation \[ \left\langle(\Delta x)^{2}\right\rangle\left\langle(\Delta p)^{2}\right\rangle=\frac{\hbar^{2}}{4} \quad(t=0) \] In addition, we know \[ \langle x\rangle=\langle p\rangle=0 \quad(t=0) \] Using the Heisenberg picture, obtain $\left\langle(\Delta x)^{2}\right\rangle_{t}$ as a function of $t(t \geq 0)$ when $\left\langle(\Delta x)^{2}\right\rangle_{t=0}$ is given. (Hint: Take advantage of the property of the minimum uncertainty wave packet you worked out in Chapter $1,$ Problem $1.18 .$ )

Transcript

00:01 We are given here free particle wave packet in a in a one dimension so we are given here free particle and free particle wave packet in a one time one dimension so here the system is one dimension here at time t which is equal to zero that is the minimum uncertainty so we can write here delta here we can delta x square summation of delta x squared multiply we can write a delta p square so summation of delta x squared so summation of delta p this square which is equal we can write here h by devoid over 4 at t which is equal to 0 so we are given this formula now in addition we can write we can we are give we have the summation of x summation of p which is equal to 0 so this t which is equal to here we want to obtain here the delta x square at summation of delta x square at t which is equal to t we want to find this how to find this and we are given here and also when delta x squared summation of delta here summation of delta x squared t which is equal to 0 is greater than that so let's solve this question so we are using the heismangan set in the principle so we can write here the formula here we know that delta x summation of average here we can accept acceptation value of x which is equal acceptation value of p which is equal to 0 so here we can write delta x square and expectation value of delta x square of t which is equal to we can write x squared expectation value of x squared expectation value of x squared now here we know the formula this formula so we are using this formula here and we know that expectation value of x which is equal to 0 so we can write here here we know that that is zero value here so here we can write x of t is a expectation value of x of t which is equal we can write t which is equal to x and here we can write t which is equal to zero so xvitation value of t x t which is equal to zero now here same for the y so we can write here same for the v of t for the y which is equal to zero also we know that expectation value of x x x of not x of zero which is zero and expectation value of p of zero which is equal to now we are using the heismore uncertainty principle so here we can write heisenberg unsurtenay principle so principle is d by d h of a to edge which is equal to we can write 1d by i h bar multiply by we can write here a to h and h so commutator of a to h and h so this is a of equation 1 and we know the formula of h so h which is equal to we know that p squared divided by 2m plus v now for three particle we know the value of v which is v which is equal 0 and value of h so we can write here value of edge is p square divided by 2m so this is the value of this is the value of h now let's equation equation one we come so we just substituting this h value so we can write equation 1 is d by d t of x which is equal we can write 1 divided by ih bar multiple we can write here value commuter can write of x and p squared divided by 2m now we just simplify all this value and we can write here d divided by d h of x of t which is equal to we can write here answer is p divided by m so this is the answer of d d d d d d d d d, d, d, d, t of x.

04:00 Now, here, we just taking integration, so we can write here, integration of d, d, x, which is equal to, we can write p divided by m of integration of dt, plus we can write x of d2...

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