Normalization of wave function and estimation of values

[Rorshach]

Homework Statement

Hello, I have this problem with seemingly simple process, but there are things I either don't know, or make some stupid mistake on the way over and over. Here's the problem:

At a particular time given by the wave function ψ(x)=N*x*exp(-(x/a)²)
Determine N so that the wave function is normalized. Then determine xψ, p_xψ, p_x²ψ and Hψ. For this case assume that V (x) = 1/2kx². Calculate the <H>.

Homework Equations

p_x=±ihbarred(d/dx)

The Attempt at a Solution

I tried to normalize the function using wolframalpha, but the result came out with unknown to me error function(I don't know this function at all, this is my first time meeting it), hence my problem with normalizing the function. I do not know why they ask for the values in this form (p_x*ψ- why the *ψ?), but according to the textbook I can get p_x from the equation I gave above, just by differentiating the wave function on x values and include the i*hbarred, which I estimated to be (N*exp(-(x/a)²)*(a²-2x²))/a². However in the answers I got totally different result:
-i*hbarred*N*(1-2x²/a²)*exp(-(x/a)²)

 

[DrClaude]

I tried to normalize the function using wolframalpha, but the result came out with unknown to me error function(I don't know this function at all, this is my first time meeting it), hence my problem with normalizing the function.

What exactly did you try to calculate?

I do not know why they ask for the values in this form (p_x*ψ- why the *ψ?),

They're just asking to apply the operator to the wave function.

which I estimated to be (N*exp(-(x/a)²)*(a²-2x²))/a². However in the answers I got totally different result:
-i*hbarred*N*(1-2x²/a²)*exp(-(x/a)²)

Apart from the fact that you forgot the factor $-i \hbar$, I don't see the difference between the two results.

 

[Rorshach]

Gosh, I must be blind:P Of course You are right, no idea how I missed that;)
I tried to calculate the infinite integral for square of wave function ψ(x)=N*x*exp(-(x/a)2) to get N, but then wolframalpha gave the result with error function.
As for the notation- I just haven't encountered this type of notation, and don't really understand its meaning- notation I've been familiar so far is the one with wave function symbol as the lower capital (in this case it could be a bit confusing, standing next to x).

 

[DrClaude]

Gosh, I must be blind:P Of course You are right, no idea how I missed that;)
I tried to calculate the infinite integral for square of wave function ψ(x)=N*x*exp(-(x/a)2) to get N, but then wolframalpha gave the result with error function.

Try this in WolframAlpha:

Integrate[(N x Exp[-(x/a)^2])^2, {x, -\[Infinity], \[Infinity]}]

As for the notation- I just haven't encountered this type of notation, and don't really understand its meaning-

Better get used to it! This is standard operator notation. If you see something like $p_x \psi(x)$, you replace the $p_x$ by its operator value, so you get
$$
p_x \psi(x) = -i \hbar \frac{d}{dx} \psi(x)
$$
$x$ is also an operator: if the wave function is expressed as a function of $x$, $\psi(x)$, then $x \psi(x)$ is the trivial operation "multiply by $x$". Note that you can also represent the wave function in momentum space, $\psi(p)$, in which case the $p$ operator is simple multiplication and $x$ becomes a differential operator.

notation I've been familiar so far is the one with wave function symbol as the lower capital (in this case it could be a bit confusing, standing next to x).

I don't understand what you mean here.

 

[Rorshach]

ψThanks for the wolframalpha formula, it explains a lot:) By the notation I have been familiar so far I mean this:
for example momentum operator of function ψ: p_(x)ψ
I know it may look weird, but my teacher had this manner of writing it in this way.
In the case of x we are looking for <x>, right?

 

[jtbell]

I tried to calculate the infinite integral for square of wave function ψ(x)=N*x*exp(-(x/a)2) to get N, but then wolframalpha gave the result with error function.

These kinds of integrals are called "Gaussian integrals" and are very common in quantum mechanics. Many QM textbooks have little tables of them in the appendices at the end. Or there's good old Wikipedia:

http://en.wikipedia.org/wiki/Gaussian_integral

Usually you can make a substitution to get your integral into the form of one of the standard Gaussian integrals, and then do it by hand.

 

[Rorshach]

Thanks:) I have to refresh some stuff with those integrals:P About xψ- I thought to calculate <x> according to formula ∫ψ*(x,t)xψ(x,t)dx

 

[DrClaude]

Thanks:) I have to refresh some stuff with those integrals:P About xψ- I thought to calculate <x> according to formula ∫ψ*(x,t)xψ(x,t)dx

As stated, the problem asks for $x \psi$, not $\langle \psi | x | \psi \rangle$.

 

[Rorshach]

Wouldn't it then be just multiplication? Or x represents some operator?

 

[DrClaude]

Wouldn't it then be just multiplication? Or x represents some operator?

Yes and yes.

My guess is that this is the point of the exercise. It teaches you about how to think of operators, and that things like "just multiplication" by $x$ is actually an operation.

 

[Rorshach]

so it is going to be xψ=N*x*exp(-(x/a)²)*x=Nx²exp(-(x/a)²)
right?

 

[DrClaude]

so it is going to be xψ=N*x*exp(-(x/a)²)*x=Nx²exp(-(x/a)²)
right?

That's how I would answer it.

 

[Rorshach]

Thank You;)

 

[Rorshach]

One more thing: if I want to obtain <H>, the formula states <H>=<ψ|Hψ>, meaning integral of th conjugate of the wave function times H times wave function? I am not proficient in bra-ket notation.

 

[DrClaude]

One more thing: if I want to obtain <H>, the formula states <H>=<ψ|Hψ>, meaning integral of th conjugate of the wave function times H times wave function? I am not proficient in bra-ket notation.

Correct.