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I've been given this homework problem, but I do not understand its notation.

Please perform the following where the wavefunctions are the normalized eigenfunctions of the harmonic oscillator Hamiltonian.

i) Compute $\langle x\rangle, \langle p \rangle, \langle x^2\rangle$ , and $\langle p^2\rangle $ for $| 0 \rangle, |1 \rangle$ , and $|12\rangle$.

ii) Compute $\langle T \rangle$ for $n = 1$ and also compute $\langle V(x) \rangle$ where, the potential energy $V(x) = \frac{m w^2 x^2}{2} $. Do these expectation values add up to what you'd expect?

I'm not understanding the BRA-KET notation.

For n = 1:

$\int \Psi_1^* (x) x^3 \Psi_1(x) dx$ - How does one "evaluate" this, without a function for $\Psi_1$?

for $|1\rangle$, I am doing:

$\langle1|x|1\rangle, \langle 1|x^2|1\rangle , \langle 1|p|1\rangle, \langle 1|p^2|1\rangle$

In which case I do not know how to proceed from that form..? And where does the wavefunction being the normalized eigenfunctions of the harmonic oscillator Hamiltonian come into play?

for ii) Is it correct to asume that the "expectation" is that $\langle T\rangle + \langle V \rangle = \langle H \rangle$?

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2 Answers 2

up vote 12 down vote accepted

A common mistake when students begin the study of the quantum harmonic oscillator is to try to convert everything to integrals. The thing is, in most curricula, the QHO is also used as a way to secretly acquaint you with bra-ket notation, and all the conveniences it offers. In reality, you shouldn't need any integrals at all here.

$\lvert n \rangle$ is a vector in an abstract vector space. If you want to express it in terms of the position basis, you can do that: $\psi_n(x) \equiv \langle x \vert n \rangle$. However, this is not always what you want to do. What you really want is inner products between the bra-version of your vector, $\langle n \rvert$, and the ket version, $\lvert n \rangle$, perhaps weighted by applying an operator in there. That is, you are seeking things like $\langle n \vert \hat{\Omega} \vert n \rangle$, where $\hat{\Omega}$ is a stand-in for some operator, like $\hat{X}$ or $\hat{P}$. We could evaluate this in the position-basis (just as any inner product can be evaluated by putting row and column vectors into the same basis and doing a dot product): \begin{align} \langle n \vert \hat{\Omega} \vert n \rangle & = \int \lvert x \rangle \langle x \rvert \langle n \vert \hat{\Omega} \vert n \rangle \\ & = \int \langle n \vert x \rangle \langle x \vert \hat{\Omega} n \rangle \\ & = \int \langle x \vert n \rangle^* \langle x \vert \hat{\Omega} n \rangle \\ & = \int \psi_n(x)^* \Omega[\psi_n](x) \ \mathrm{d}x, \end{align} where $\Omega$ is the functional corresponding to the operator $\hat{\Omega}$, e.g. $X[f](x) = x f(x)$ and $P[f](x) = -\mathrm{i} \hbar f'(x)$.

But you don't want to do that.

The trick with the QHO is that there are two very convenient operators we define: $\hat{a}$ and $\hat{a}^\dagger$, and they have properties such as $\hat{a} \lvert n \rangle = \sqrt{n} \ \lvert n-1 \rangle$ and $\hat{a}^\dagger \lvert n \rangle = \sqrt{n+1} \ \lvert n+1 \rangle$.

You can write $\hat{X}$ and $\hat{P}$ in terms of $\hat{a}$ and $\hat{a}^\dagger$. Then use the properties of linearity (of inner products/integrals) and orthogonality and normalization (of your states $\lvert n \rangle$). As an example, suppose you want to compute $\langle n \vert \hat{a}^2 + \hat{a}^\dagger \vert m \rangle$. You have \begin{align} \langle n \vert \hat{a}^2 + \hat{a}^\dagger \vert m \rangle & = \langle n \vert \hat{a}^2 \vert m \rangle + \langle n \vert \hat{a}^\dagger \vert m \rangle \\ & = m \langle n \vert \hat{a} \vert (m-1) \rangle + (m+1) \langle n \vert (m+1) \rangle \\ & = m(m-1) \langle n \vert (m-2) \rangle + (m+1) \langle n \vert (m+1) \rangle \\ & = m(m-1) \delta_{n,m-2} + (m+1) \delta_{n,m+1}. \end{align} As you can see, plugging in particular values of $n$ and $m$ will make some (or all) of the terms vanish, leaving simple answers.

What you have to do is figure out $\hat{X}$ and $\hat{P}$ in terms of $\hat{a}$ and $\hat{a}^\dagger$ (the expressions are pretty simple). Then when asked for an expectation of, say $\hat{X}$ with respect to $\lvert 1 \rangle$, you simply compute $$ \langle \hat{X} \rangle_{\lvert 1 \rangle} \equiv \langle 1 \vert \hat{X} \vert 1 \rangle = \langle 1 \vert \text{stuff with annihilation/creation operators} \vert 1 \rangle. $$

For $\hat{T}$ and $\hat{V}$, just use the classical expressions for kinetic and potential energy in terms of $x$ and $p$, and change $x \to \hat{X}$, $p \to \hat{P}$. For $\hat{H}$, I ask in return: Classically what is $H$ in terms of $T$ and $V$?

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thank you! this makes total sense on why the ladder operators exist, it makes it infinitely easier..! I'm trying to derive the action of a 3 term ladder operator, $(a_+ + a_-)^3$. I've did the trinomial expansion, is it correct to assume to apply the 2 term operators, then do another operator for the third term? –  julesverne Mar 19 '13 at 2:47
    
@julesverne Just be careful - these operators don't quite commute with each other. So you need to treat terms like $a_+a_-a_+$ differently from $a_+^2a_-$, for instance. –  Chris White Mar 19 '13 at 7:34
    
This was incredibly helpful, and very insightful on learning how to approach this topic. I cannot thank you enough! Cheers! –  julesverne Mar 19 '13 at 21:26

About i)

The notation $|n\rangle$ is due to the discrete nature of the spectrum of the quantum mechanical hamiltonian for a harmonic oscillator. In terms of the ladder operators $a,a^{\dagger}$ you can define the number operator $N=a^{\dagger}a$ and denote its eigenfunctions as $|n\rangle$, so that $N|n\rangle=n|n\rangle$. You can find that the particular form of those is [*]$^1$

$$\phi_{n}(x)= \frac{1}{\sqrt{2^n\,n!}} \cdot \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \cdot e^{ - \frac{m\omega x^2}{2 \hbar}} \cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right), \qquad n = 0,1,2,\ldots$$

where $H_{n}(x)$ is the nth Hermite polynomial.[*]$^2$ An interesting hint is that $\phi_{n}(x)$ is real and that the parity of $H_{n}(x)$ is the parity of $n$

Also

$$ \langle 1|A|2\rangle=\int \phi^{*}_{1}(x) \left(A \phi_{2}(x)\right) dx\equiv \langle 1 |A2\rangle$$

About ii)

What you're expecting is $\langle T \rangle+\langle V \rangle =\langle H \rangle $ but you actually need to do the calculations

References

[*]$^1$ http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Hamiltonian_and_energy_eigenstates

[*]$^2$ http://en.wikipedia.org/wiki/Hermite_polynomials#Orthogonality

-Edit-

Partial computations for 1

$$\langle 1|x|1 \rangle=\int \phi_{1}(x) x \phi_{1}dx=0 $$

because the integrand is antisymmetric in $x$ (the three factors are antisymmetric in $x$) and the integral of an antisymmetric function of $x$ over and symmetric interval is zero. Proof

$$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^0f(x)dx+\int^{\infty}_{0}f(x)dx. $$

Set $u=-x$ in the first integral, and use $f(x)=-f(-x)=-f(u)$, $du=-dx$ so that

$$\int_{\infty}^{0}f(u)du+\int^{\infty}_{0}f(x)dx=-\int_{0}^{\infty}f(u)du+\int^{\infty}_{0}f(x)dx=0$$

$$ \langle 1|x^2|1\rangle =\int \phi_{1}(x)x^2\phi_{1}(x)dx=\int \frac{1}{2}\left(\frac{m\omega}{\pi \hbar} \right)^{1/2}e^{-\frac{m\omega^2x^2}{\hbar}}\frac{m\omega}{h}x^4dx$$

Use

$$\int_{0}^{\infty}x^4e^{-a^2x^2}=\frac{3\sqrt{\pi}}{8a^5} $$

What about the momentum?

$$\langle 1|p|1\rangle=\int \phi_{1}(x)\left(-i\hbar \frac{d}{dx}\right)\phi_{1}(x) $$

$$\frac{d\phi_{1}}{dx}=\frac{d}{dx}\left(\frac{1}{\sqrt{2}}\left( \frac{m\omega}{\pi\hbar}\right)^{1/4}e^{-\frac{m\omega^2x^2}{2\hbar^2}} \sqrt{\frac{m\omega}{\hbar}x}\right)=... $$

Hope you can continue from here

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thank you! could you explain how on the 4th line for the partial computations for 1 you went from the general form of the integral to the right side of the equation? –  julesverne Mar 19 '13 at 0:56
    
do you mean $\langle 1 |x^2| 1 \rangle$ ? It is just substitution of $\phi_{1}(x)^*$ and $\phi_{1}(x)$ where I used the fact that $\phi^{*}_{1}(x)=\phi_{1}(x)$ because $\phi_{n}(x)$ is real for general $n$. Also, I grouped all the $x$ factors ($x$ for each $\phi_{1}(x)$ and $x^2$ for the operator). However please look Chris White's answer, is far more useful than mine –  Jorge Mar 19 '13 at 6:47

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