# Can expectation value of observables be imaginary?

1. May 18, 2008

### Raze2dust

I am quite new to Quantum Mechanics and I am studying it from the book by Griffiths, as kind of a self-study..no instructor and all....

For a gaussian wavefunction $$\Psi$$=Aexp(-x$$^{2}$$), I calculated
<p$$^{2}$$> and found it to be equal to ah$$^{2}$$/(1-2aiht/m)

(By h I mean h-bar..not so good at latex yet)

Can it be correct? How can expectation value of p$$^{2}$$ be imaginary?

2. May 18, 2008

### CompuChip

That's not what I got. How did you go about calculating it?

3. May 18, 2008

### Raze2dust

What did you get? <p^2>= ah ?
we know psi(x,0) = Aexp(-x^2)
so i first calulated phi(k) integrating (1/[(2*pi)^(1/2)]) psi(x,0)*exp(-ikx) dx from -inf to inf
then got psi(x,t) (1/[(2*pi)^(1/2)]) phi(k)*exp(i(kx-((h^2)*(k^2)*t)/2m) dk from -inf to inf

till that part im correct, since the answer is given in the book and I verified.

For the next part, I got <x>=<p>=0 and <x^2>=1/(4*w^2) where w=[a/(1+4a2h2t2/m2)] (the 2's are squares)
It's fine till then

Then got <p^2> by doing (-h^2)* integral psi(x,t)* (d^2/dx^2) psi(x,t) dx
the problem is that i get imaginary constant (1-2aiht/m) each time i differentiate psi(x,t)
and this doesn't get cancelled out later.

4. May 18, 2008

### Raze2dust

$$\phi(k)=\int^{-\infty}_{\infty}\psi(x,0)e^{-ikx}dx$$

5. May 18, 2008

### malawi_glenn

but you take the derivative twice, then you should not have any imaginary part left.

6. May 18, 2008

### Raze2dust

(My previous post was a mistake ignore it..)

Here's how I did it..
$$\phi(k)=\stackrel{1}{\overline{\sqrt{2\pi}}}\int^{\infty}_{-\infty}\psi(x,0)e^{-ikx}dx =\stackrel{1}{\overline{\sqrt{2\pi}}}\int^{\infty}_{-\infty}Ae^{ax^{2}+ikx}dx =\stackrel{A}{\overline{\sqrt{2\pi}}}\sqrt{\stackrel{\pi}{\overline{a}}}e^{\stackrel{-k^{2}}{\overline{4a}}}$$
Now we find $$\psi(x,t)$$ as
$$\psi(x,t)=\stackrel{1}{\overline{\sqrt{2\pi}}}\int^{\infty}_{-\infty}\phi(k)e^{i(kx-\stackrel{hk^{2}t}{\overline{2m}})} =A\stackrel{e^{\stackrel{-ax^{2}}{\overline{1+\stackrel{2aiht}{\overline{m}}}}}}{\overline{\sqrt{1+\stackrel{2aiht}{\overline{m}}}}}$$
where A=$$(\stackrel{2a}{\overline{\pi}})^{\stackrel{1}{\overline{4}}}$$
(A should be in the numerator)

So $$\psi(x,t)*=A\stackrel{e^{\stackrel{-ax^{2}}{\overline{1-\stackrel{2aiht}{\overline{m}}}}}}{\overline{\sqrt{1-\stackrel{2aiht}{\overline{m}}}}} =Ce^{-w^{2}x^{2}(1+\stackrel{2aiht}{\overline{m}})}$$
where C is an imaginary factor which gives real factor when combined with the similar factor from $$\psi(x,t)$$

But the imaginary exponential factor comes out in the form of $$(1+\stackrel{2aiht}{\overline{m}})^{2}$$ when we compute <p$$^{2}$$> and doesn't get cancelled by some other real factor. Or am I making some calculation mistake?

7. May 18, 2008

### malawi_glenn

I cant follow what you are doing, it is just a mess..

you want to do <p^2> when you have the wave function in position representation. Then you do:

$$<p^2> = \int_{-\infty}^{\infty} \psi (x)^* (-i\hbar\frac{d}{dx})^2\psi (x) dx$$

since $$\hat{p} = -i\hbar\frac{d}{dx}$$ in position representation.

Where $$\psi (x) = \Psi (x,t=0) = A\exp(-x)$$

That is all you have to do man.

You can take the wave function in momentum representation:

$$\phi (k) = \dfrac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \psi (x) \exp{-ikx} dx$$

Then the expectation value of p^2 is:

$$<p^2> = \int_{-\infty}^{\infty} \phi(p)^* \hat{p} ^2\phi (p) dp$$

where $$\hat{p}$$ is the momentum operator in MOMENTUM representation (i.e $$\hat{p} = p$$), compare with the position operator in position representation: $$\hat{x} = x$$.

compare also with:
$$<x^2> = \int_{-\infty}^{\infty} \psi(x)^* \hat{x}^2\psi (x) dx$$

It seems to me that you have mixed the two representations, you can't do that.

Last edited: May 18, 2008
8. May 18, 2008

### malawi_glenn

and if you want to include time dependence, you apply the time evolution operator, and your position wave function as a function of time, becomes:

$$\Psi (x,t) = A\exp(-x)\exp{(-iEt/\hbar)}$$

Then the expectation value becomes:
$$<p^2>(t) = \int_{-\infty}^{\infty} \Psi (x,t)^* (-i\hbar\frac{d}{dx})^2\Psi (x,t) dx =<p^2>\exp{(Et/\hbar)}$$

9. May 18, 2008

### Raze2dust

yeah thats what I have done..just that $$\psi(x,t)$$ is quite clumsy..
So, whats is the answer u got?

In the book its given as $$<p^{2}>=ah$$ but I can't find the mistake I'm making so I'm assuming that the author has calculated it for t=0.

10. May 18, 2008

### malawi_glenn

is $$A = \frac{2a}{\sqrt{\sqrt{\pi}}}$$ ?

Can you tell me what
$$(-i\hbar\frac{d}{dx})^2\psi (x) = (-i\hbar\frac{d}{dx})^2A\exp(-x^2)$$

is? (I know, but do you?)

11. May 18, 2008

### Raze2dust

No, I got A as $$A = \frac{\sqrt{\sqrt{2a}}}{\sqrt{\sqrt{\pi}}}$$

I'll do $$-\hbar^{2}\frac{d^{2}}{dx^{2}}(Aexp(-kx^{2})$$
This would be $$-A\hbar^{2}(-2k)\frac{d}{dx}(xexp(-kx^2)) = 2kA\hbar^{2}(-2kx^{2}exp(-kx^{2})+exp(-kx^{2} ))=2kA\hbar^{2}exp(-kx^{2})(1-2kx^{2})$$

But I guess my problem occurs only when we consider time dependance, in which case k has an imaginary part as well.(In fact I got k as $$\frac{a}{1+\frac{2aiht}{m}}$$)

Last edited: May 18, 2008
12. May 18, 2008

### CompuChip

I don't follow.
You said:
Where does the k come from?

13. May 18, 2008

### Raze2dust

The value of psi(x,t) I obtained, which is correct, as per the book, is
$$\Psi(x,t)=A\frac{e^{\frac{-ax^{2}}{1+\frac{2ai\hbar t}{m}}}}{\sqrt{1+\frac{2ai\hbar t}{m}}}$$

So basically $$k = \frac{a(1-\frac{2ai\hbar t}{m})}{1+\frac{4a^{2}\hbar^{2}t^{2}}{m}}=w^{2}(1-\frac{2ai\hbar t}{m})$$

So,
$$\Psi(x,t)=A\frac{e^{-w^{2}x^{2}(1-\frac{2ai\hbar t}{m})}}{\sqrt{1+\frac{2ai\hbar t}{m}}}$$
And,
$$\Psi(x,t)*=A\frac{e^{-w^{2}x^{2}(1+\frac{2ai\hbar t}{m})}}{\sqrt{1-\frac{2ai\hbar t}{m}}}$$

14. May 18, 2008

### Raze2dust

So $$\Psi(x,t)*\hat{p}^{2}\Psi(x,t) = A\frac{e^{-w^{2}x^{2}(1+\frac{2ai\hbar t}{m})}}{\sqrt{1-\frac{2ai\hbar t}{m}}} (2\hbar^{2}A\frac{e^{-w^{2}x^{2}(1-\frac{2ai\hbar t}{m})}}{\sqrt{1+\frac{2ai\hbar t}{m}}}w^{2}(1-\frac{2ai\hbar t}{m}) (1-2w^{2}(1-\frac{2ai\hbar t}{m}) x^{2}))$$
$$=2\hbar^{2}A^{2}w^{4}(1-\frac{2ai\hbar t}{m}) (1-2w^{2}(1-\frac{2ai\hbar t}{m}) x^{2})\frac{1}{\sqrt{a}}e^{-2w^{2}x^{2}}$$

Now, if I denote $$(1-\frac{2ai\hbar t}{m})$$ by z ,
$$\Psi(x,t)*\hat{p}^{2}\Psi(x,t) = \frac{2\hbar^{2}A^{2}w^{4}}{\sqrt{a}}z (1-2w^{2}z x^{2})e^{-2w^{2}x^{2}}$$

Now integration doesn't remove its dependence on z, which is imaginary. In which step am I making a mistake? I am not able to spot it...its very difficult to spot silly mistakes in such cluttered maths..
And what is the value I should get? or what is the value you guys got?

15. May 18, 2008

### malawi_glenn

You have wrong wave function...you are mixing the two representations!

$$\Psi (x,t) = A\exp(-x^2)\exp{(-iEt/\hbar)}$$ This is not what are calculating with, you use: $$\Psi(x,t)=A\frac{e^{-w^{2}x^{2}(1-\frac{2ai\hbar t}{m})}}{\sqrt{1+\frac{2ai\hbar t}{m}}}$$

Can you explain why you use that?

$$<p^2>(t) = \int_{-\infty}^{\infty} \Psi (x,t)^* (-i\hbar\frac{d}{dx})^2\Psi (x,t) dx = <p^2>exp{(Et/\hbar)}$$

Then why did you do:

$$-\hbar^{2}\frac{d^{2}}{dx^{2}}(Aexp(-kx^{2})$$
??

Your wave function was: Aexp(-x^2), that was why I asked you to do:

$$(-i\hbar\frac{d}{dx})^2\psi (x) = (-i\hbar\frac{d}{dx})^2A\exp(-x^2)$$

Last edited: May 18, 2008
16. May 18, 2008

### Raze2dust

I use that because that is the wavefunction I have. $$\Psi (x,t) = A\exp(-x^2)\exp{(-iEt/\hbar)}$$
is not the wavefunction. This part has been verified by the book as well. I have to use the superposition of states with all energies. How can I use only a single energy E?

17. May 18, 2008

### Raze2dust

The question states that:
"A free particle has the initial wave function
$$\Psi(x,0)=Ae^{-ax^{2}}$$ "

So this is the INITIAL wavefunction.
To find $$\Psi(x,t)$$ we need to integrate it over all energies right?

$$\Psi(x,t)=\frac{1}{\sqrt{2\pi}\int_{-\infty}^{\infty}\phi(k)e^{i(kx-\frac{\hbar k^{2}}{2m}t)}$$

Isn't this how we should represent $$\Psi(x,t)$$?

Last edited: May 18, 2008
18. May 18, 2008

### malawi_glenn

does the energy depend on x?

Time evolution of a state is:

http://en.wikipedia.org/wiki/Hamiltonian_(quantum_mechanics [Broken])

i.e
$$\Psi (x,t) = \Psi (x, t=0) e^{-iEt/\hbar}$$

isn't $$E = \frac{p^2}{2m} = \frac{\hbar^2k^2}{2m}$$ In your case?

And since the energy don't depend on x:
$$<p^2>(t) = \int_{-\infty}^{\infty} \Psi (x,t)^* (-i\hbar\frac{d}{dx})^2\Psi (x,t) dx \exp{(Et/\hbar)}$$

where:
$$<p^2> = \int_{-\infty}^{\infty} \psi (x)^* (-i\hbar\frac{d}{dx})^2\psi (x) dx$$

And:
$$\psi (x) = \Psi (x,t=0) = A\exp(-x^2)$$

Last edited by a moderator: May 3, 2017
19. May 18, 2008

### Raze2dust

This is what I did not get... Dont you have to take superpostition over all states/ all energies?
Because in the case of free particle, the value of $$\psi(x)$$ you get from solving the time-independent SE is not normalizable. So in this case should not we use $$\Psi(x,t)=\frac{1}{\sqrt{2\pi}\int_{-\infty}^{\infty}\phi(k)e^{i(kx-\frac{\hbar k^{2}}{2m}t)}$$ and then apply <p^2> on that $$\Psi(x,t)$$ we obtain?

Last edited by a moderator: May 3, 2017
20. May 18, 2008

### malawi_glenn

The time evolution of a state is obtained by applying the time evolution operator, always. It is a fundemantal cornerstone in non relativistic quantum mechanics.

The energy is $$E = \frac{p^2}{2m} = \frac{\hbar^2k^2}{2m}$$ right?