Beginner QM Problems

[xcthulhu]

Hi,

 

I am an amateur trying to teach myself physics by trying to do at least one problem every day.

While these are textbook exercises, nobody has assigned them to me... so helping me isn't cheating unless you think I should be able to handle these problems myself.

 

These are some beginner problems from Griffiths' Introduction to Quantum Mechanics

 

a) Let [math]\psi (x) \varphi (t)[/math] be a seperable solution to Schrodinger's equation corresponding to an allowed energy E. Show that from the assumption of normalizability that E must be real.

 

b) Show that every solution to the time independent solution to Schrodinger's equation can be taken as real (not that they are all real, but rather any non-real solution is decomposable into a linear combination of real solutions).

 

c) If the potential [math]V[/math] is an even function then the time independent solution [math]\psi[/math] can be taken to be even or odd.

 

d) Show that E must exceed the minimum value V(x), for every normalizable solution to the time-independent Schrodinger equation. What is the classical analog to this statement?

 

I will post my answers later today. I love feedback and further reading if anybody has any suggestions.

[xcthulhu]

a) Schrodinger's equation (in one dimension) asserts:

[math]i \hbar \frac{\partial \Psi}{\partial t} = -\frac{\hbar^2}{2 m} \frac{\partial^2 \Psi}{\partial x^2} + V\Psi[/math]

 

If we assume seperability, then [math]\Psi(x,t) = \psi(x)\phi(t)[/math].

 

We can then rewrite the above differential equation as follows:

[math]i \hbar \frac{\phi'(t)}{\phi(t)} = -\frac{\hbar^2 \psi''(x)}{2m \psi(x)} + V(x)[/math]

 

This implies both sides are constant, which is the allowable energy, which we will label [math]E[/math].

 

Now suppose that E were complex; such that [math]E = K + i J[/math]. Then [math]\phi(t) = A e^{-J t} e^{K i t}[/math] according to the differential equation.

From normality we have that [math]\int_{-\infty}^{\infty} |\Psi|^2 d x= 1[/math], but since [math]\Psi(x,t) = \psi(x)\phi(t) = A \psi(x) e^{-J t} e^{K i t}[/math], then

[math]\int_{-\infty}^{\infty} |\Psi|^2 d x = \int_{-\infty}^{\infty} \Psi^\ast\Psi d x = A^2 e^{-2 J t} \int_{-\infty}^{\infty} \psi^\ast(x)\psi(x) dx[/math]

Hence:

[math]A^2 e^{-2 J t} = \int_{-\infty}^{\infty} \psi^\ast(x)\psi(x) dx[/math] (a constant)

Thus it must be that [math]J[/math] is zero.

 

b) We have the following rules for conjugation:

[math](f + g)^\ast = f^ast + g^ast[/math]

[math](f \cdot g)^\ast = f^\ast \cdot g^\ast[/math]

[math](a f)^\ast = a f^\ast[/math] if [math]a[/math] is a real scalar

[math](1/f)^\ast = 1/f^\ast[/math]

[math](f')^\ast = (f^\ast)'[/math]

 

Now add in the observation that from separability we have that [math]E = -\frac{\hbar^2 \psi''(t)}{2m \psi(t)} + V[/math]. Hence the conjugates of both sides of this equation must be equal. But E is its own conjugate since it is real, as well as V (since the potential energy is also defined to be real). With this and using the rules we can push conjugation through to arrive at:

 

[math]E = -\frac{\hbar^2 (\psi^\ast)''(t)}{2m \psi^\ast(t)} + V[/math]

 

(a "stationary solution")

 

Which means that if [math]\psi[/math] indeed satisfies the above so too does [math]\psi^ast[/math].

 

Now it's easy to see that we can rewrite the above as:

[math]E\psi = \hat{H} \psi[/math] and [math]E\psi^\ast = \hat{H} \psi^\ast[/math]

 

Note that [math]\hat{H}[/math] is a linear operator. We add these two equations or subtract them from one another and multiply by i, giving us:

[math]E(\psi+\psi^\ast) = \hat{H} (\psi+\psi^\ast)[/math]

and

[math]E i (\psi-\psi^\ast) = \hat{H} i (\psi-\psi^\ast)[/math]

 

(we know that [math]\hat{H}[/math] distributes because it's a linear operator)

 

Either way, we can see that we have obtained a new stationary solution [math]\psi[/math] which is completely real.

 

c) We can rewrite the stationary solution equation as:

 

[math]V(x) - E = \frac{\hbar^2 \psi''(x)}{m^2 \psi(x)}[/math]

 

Assuming that [math]V(x)[/math] is even then we have that

[math]V(x) - E = V(-x) - E = \frac{\hbar^2 \psi''(-x)}{m^2 \psi(-x)}[/math]

 

Now note that [math](\psi \circ (\lambda x. -x))''(x) = \psi''(-x)[/math]. From this we can deduce that it must be that if [math]\psi(x)[/math] is a solution then [math]\psi(-x)[/math] is a solution, too.

Thus we have [math]\psi(x) + \psi(-x)[/math] (an even function) and [math]\psi(x) - \psi(-x)[/math] (an odd function) are both stationary solutions.

 

I ran out of time on (d), but I'll tackle it later....

Edited July 15, 2011 by xcthulhu

[xcthulhu]

I suppose I'll tackle (d) today.

 

We first take [math]\psi[/math] to be real based on yesterday's results. Today's problem is just a bit of analysis.

 

We can rewrite the equation for the stationary solution as:

[math](V(x) - E) \psi(x) = \frac{\hbar^2}{2m}\psi''(x)[/math]

 

If [math]E < V_{min}[/math], then it must be that [math]\psi(x)[/math] and [math]\psi''(x)[/math] have the same sign.

 

From this we can deduce the following 4 lemmas:

(1) If [math]\psi(a) > 0[/math] and [math]\psi'(a) > 0[/math], then [math]\psi[/math] is increasing on [math][a,\infty)[/math] and [math]\lim_{x \to \infty} \psi(x) = \infty[/math]

(2) If [math]\psi(a) > 0[/math] and [math]\psi'(a) < 0[/math], then [math]\psi[/math] is decreasing on [math](-\infty,a][/math] and [math]\lim_{x \to -\infty} \psi(x) = \infty[/math]

(3) If [math]\psi(a) < 0[/math] and [math]\psi'(a) < 0[/math], then [math]\psi[/math] is decreasing on [math][a,\infty)[/math] and [math]\lim_{x \to \infty} \psi(x) = -\infty[/math]

(4) If [math]\psi(a) < 0[/math] and [math]\psi'(a) > 0[/math], then [math]\psi[/math] is increasing on [math](-\infty,a][/math] and [math]\lim_{x \to -\infty} \psi(x) = -\infty[/math]

 

All of the proofs of these are rather similar. I'll do (4). Assume [math]\psi(a) < 0[/math] and [math]\psi'(a) < 0[/math] and consider any interval [b,a] for which [math]\psi[/math] is negative. I argue that [math]\psi[/math] must be decreasing on the entire interval. For consider any two points [math]c,d\in [b,a][/math] where c < d, then we know by [math]L^2[/math] continuity and the mean value theorem that there's some point e such that c < e < d and [math]\psi''(e) = \frac{\psi'© - \psi'(d)}{c - d}[/math] But evidently [math]\psi(e) < 0[/math] hence [math]\psi''(e) < 0[/math], hence [math]0 > \psi'© - \psi'(d)[/math]. But then that just means that [math]\psi'© < \psi'(d)[/math]. Since by assumption [math]\psi'(a) < 0[/math] then it must be that [math]\psi'(x) < 0 \forall x \in [b,a][/math], which is to say the stationary solution is decreasing on the whole interval [b,a].

 

We can see from the above that it must be that [math]\psi(x) < \psi(a)[/math] for all [math]x < a[/math]. For if there was some [math]\psi© > 0[/math] for [math]c < a[/math] then we find some [math]c < b < a[/math] where

[math]\psi(b) = 0[/math] (by continuity and the intermediate value theorem) and for all [math]\epsilon > 0[/math] we have [math]\psi[/math] is negative on [math][b+\epsilon,a][/math]. Then that means that [math]\psi(b) = \lim_{x\to b} \psi(x) < \psi(a)[/math] (by continuity), which is a contradiction.

 

The proof of the blowup is tricky and I will have to deal with it later.

 

~XCT

Edited July 16, 2011 by xcthulhu

[xcthulhu]

It occurs to me that because I already showed that if [math]\psi[/math] is negative on [b,a] then [math]\psi'©<\psi'(d)[/math] for all [math]c,d\in[b,a][/math]. But since I also showed that for all [math]x < a[/math] it must be taht [math]\psi(x) < \psi(a)[/math], we know that [math]\psi[/math] is negative on the interval [math](-\infty,a][/math], whence we know that [math]\psi(x) << \psi(a)x[/math] on that interval. Hence it must go to negative infinity. That wasn't so hard I just ran out of time last time.

 

Anyway, so either a function is constant or else it is not, in which case one of the four lemmas apply. In each case the function evidently blows up somehow, which is prohibited by assumption.