Problem with evolution operator and harmonic oscillator

[lufus]

Homework Statement

Show that U*(pi/(2*omega)) |x> is an eigenvecor of p and specify its eigenvalue. Similarly, establish that U*(pi/(2*omega)) |p> is an eigenvector of x.

Homework Equations

U*(t) = exp((i/h_bar)H*t)

The Attempt at a Solution

I've tried using closure with P (U*(pi/(2*omega)) |x>) to get a constant * (U*(pi/(2*omega)) |x>), but I'm not getting anywhere. I really have no clue.

This is a problem from Cohen-Tannoudji, Vol. 1, Chapter V, Complement M, 8c.

 

[mark726]

To show that U*(pi/(2*omega)) |x> is an eigenvector of p, we can use the momentum operator in position basis, given by p = -i*h_bar*(d/dx). We can also use the definition of U*(t) and substitute in the given value of t, which is pi/(2*omega). This gives us:

U*(pi/(2*omega)) |x> = exp((i/h_bar)H*(pi/(2*omega))) |x>

= exp((i/h_bar)(p^2/2m)*(pi/(2*omega))) |x>

= exp((i*p*pi)/(2*m*omega*h_bar)) |x>

= exp(i*(pi*p)/(2*m*omega*h_bar)) |x>

Now, we can use the property of exponentials, exp(ix) = cos(x) + i*sin(x), to write this as:

U*(pi/(2*omega)) |x> = cos((pi*p)/(2*m*omega*h_bar)) |x> + i*sin((pi*p)/(2*m*omega*h_bar)) |x>

This shows that U*(pi/(2*omega)) |x> is a linear combination of |x> and |p>, and thus, it can be written as a linear combination of its own eigenstates. Therefore, it is an eigenvector of p.

To find its eigenvalue, we can use the fact that p is the momentum operator, and it acts on |x> to give the eigenvalue p. Therefore, the eigenvalue of U*(pi/(2*omega)) |x> is p = (pi*p)/(2*m*omega*h_bar).

Similarly, to show that U*(pi/(2*omega)) |p> is an eigenvector of x, we can use the position operator in momentum basis, given by x = i*h_bar*(d/dp). Again, we can substitute in the given value of t and use the property of exponentials to write it as:

U*(pi/(2*omega)) |p> = exp(i*(pi*x)/(2*m*omega*h_bar)) |p>

Now, using the property of exponentials, we can write this as:

U*(pi/(2*omega)) |p> = cos((pi*x)/(2*m*omega*h_bar)) |p> + i*sin((pi*x)/(