I'm just learning the Brakets formalisms for QM and I'm having struggles solving a simple problem.

For an harmonic oscillator, particularly Griffiths' Introduction to Quantum Mechanics P3.34:

enter image description here

I want to measure the expected value of momentum $p$ as:

$$\langle p \rangle = \langle \Psi | p | \Psi \rangle$$

considering the wavefunction

$$\Psi(x,t) = \sum\limits_{n=0}^{1} c_n \,\psi_n \,e^{-iE_nt/\hbar}$$

so, my first thought was to insert $\Psi$ in $\langle p \rangle$ as:

$$\langle c_0 \,\psi_0 \,e^{-iE_0t/\hbar} + c_1 \,\psi_1 \,e^{-iE_1t/\hbar} \,|\, p \,|\, c_0 \,\psi_0 \,e^{-iE_0t/\hbar} + c_1 \,\psi_1 \,e^{-iE_1t/\hbar} \rangle$$

but I recognize this is too much "brute force" and clearly shows to me that I don't understand well how to calculate with bras and kets operations (and also what's the benefit of this).

Following my lecturer I understood these being eigenvalues and eigenvectors of $\psi$ respectively so I think I can treat the operation as an inner product (?) pulling the coefficients outside the operation respecting the order when $c_i^* c_j$ products appear.

Anyways, honestly, I don't see the obvious: how should I proceed in a practical way? Why does the outcome has the form of a product? Something like

$$(c_0^* \langle \psi_0|p|\; {e^{-iE_0 t/\hbar}}^* + c_1^* {e^{-iE_1 t/\hbar}}^* \; \langle \psi_1|p|)(c_0 |\psi_0\rangle e^{-iE_0 t/\hbar} + c_1 |\psi_1 \rangle e^{-iE_1 t/\hbar})$$

D.J Griffiths himself states that:

enter image description here

I'm aware my reasoning is not correct and I don't want to bother anyone with the question. I'm just a little confused about it and wanting to understand more.

EDIT: Following what JEB and Cosmas Zachos are suggesting:

since $\Psi$ can be represented as

$$|\Psi \rangle = \frac{1}{\sqrt{2}} [|0\rangle + e^{i\phi}|1\rangle] \equiv \frac{1}{\sqrt{2}} \begin{pmatrix} \psi_0 \\ \psi_1 e^{i\phi} \end{pmatrix}$$

and the momentum expected value is $\langle \Psi | \hat{p} | \Psi \rangle$ one can write

$$\langle \Psi | = (|\Psi\rangle)^{\dagger} = \frac{1}{\sqrt{2}}[\langle 0|+e^{-i\phi}\langle 1 |]$$


$$\langle \Psi | \hat{p} | \Psi \rangle = \frac{1}{2} [\langle0| +e^{-i\phi}\langle 1 | p | 0 \rangle + e^{i\phi} |1\rangle]$$

being $\hat{p} = i\sqrt{\frac{\hbar m \omega}{2}}(\hat{a_+}-\hat{a_{-}})$ so

$$\langle \Psi | \hat{p} | \Psi \rangle = 1/2 \, i\sqrt{\frac{\hbar m \omega}{2}}[\langle 0 | + e^{-i\phi} \langle 1 | \Big| \hat{a_+} |0\rangle + \hat{a_+} e^{i\phi} |1\rangle - \hat{a_{-}}|0\rangle - \hat{a_{-}}e^{i\phi} |1\rangle]$$

then distribuiting the bras to the resulting kets by the right:

$$ = 1/2\, i\sqrt{\frac{\hbar m \omega}{2}} ( \langle 0 |(\hat{a_+} |0\rangle + \hat{a_+} e^{i\phi} |1\rangle - \hat{a_{-}}|0\rangle - \hat{a_{-}}e^{i\phi} |1\rangle) + e^{-i\phi} \langle 1| (\hat{a_+} |0\rangle + \hat{a_+} e^{i\phi} |1\rangle - \hat{a_{-}}|0\rangle - \hat{a_{-}}e^{i\phi} |1\rangle) )$$

Now all the raising and lowering operators act on the kets next to them, following

$$\hat{a} |n\rangle = \sqrt{n} |n-1\rangle$$ $$\hat{a}^{\dagger} |n \rangle = \sqrt{n+1} |n+1 \rangle$$

and I get inner products of the states $\psi_0$, $\psi_1$ and $\psi_2$ ponderated by $\sqrt{n}$ and $\sqrt{n+1}$.

This results in:

$$\langle p \rangle = \frac{1}{2} \sqrt{\frac{m\omega \hbar}{2}}i [\langle 0 | 1 \rangle + e^{i\phi}\langle 0| 2\rangle - e^{i\phi}\langle 0| 0\rangle +e^{-i\phi} \langle 1|1 \rangle +\sqrt{2} \langle 1|2 \rangle - \langle 1|0 \rangle]$$

What should I do next?

Being the states represented by an orthonormal basis, the inner product $\psi_n^*\psi_{n'}$ is 0 if $n \neq n'$? i.e.,

$$\langle p \rangle = \frac{1}{2} \sqrt{\frac{m\omega \hbar}{2}}i [\langle 0 | 1 \rangle + e^{i\phi}\langle 0| 2\rangle - e^{i\phi}\langle 0| 0\rangle +e^{-i\phi} \langle 1|1 \rangle +\sqrt{2} \langle 1|2 \rangle - \langle 1|0 \rangle] = \frac{1}{2} \sqrt{\frac{m\omega \hbar}{2}}i [0 + 0 - e^{i\phi}\langle 0| 0\rangle +e^{-i\phi} \langle 1|1 \rangle +0 - 0] $$

  • 4
    $\begingroup$ Have you learned about raising/lowering operators? $\endgroup$
    – DanielSank
    Commented Oct 4, 2020 at 0:33
  • $\begingroup$ @DanielSank Yes I do ! $\endgroup$
    – nuwe
    Commented Oct 4, 2020 at 0:46
  • 4
    $\begingroup$ So how did you utilize $\hat p\sim i(a^\dagger -a)/\sqrt{2}$ ? $\endgroup$ Commented Oct 4, 2020 at 19:55
  • $\begingroup$ Hi @CosmasZachos, I introduced $\hat{p}$ inside $\langle \Psi | \hat{p} | \Psi \rangle = \langle \Psi | \hat{p} \Psi \rangle$ and distribuited the raising / lowering ops. onto the kets. $\endgroup$
    – nuwe
    Commented Oct 5, 2020 at 20:55
  • 1
    $\begingroup$ orthonormal means $\langle n|n'\rangle = \delta_{nn'}$ $\endgroup$
    – JEB
    Commented Oct 7, 2020 at 13:44

2 Answers 2


Since the question is about bra-ket notation, the 1st problem is: you are not using it.

The problem states the general form of the wave function is:

$$ |\psi\rangle = \frac 1 {\sqrt 2}[|0\rangle + e^{i\phi}|1\rangle] $$

where I have used:

$$ H|n\rangle = (n+\frac 1 2)\hbar\omega|n\rangle $$

Since the global phase is arbitrary, I put it all in the coefficient of the $n=1$ basis state.

From here, compute the expectation of $\hat p$ by expressing it as a linear combination of $a$ and $a^{\dagger}$. Maximize as a function of $\phi$, the only free parameter.

Note how much simpler this is than integrating products and derivatives of Hermite polynomials, even if you use:

$$ H_{n+1}(x) = 2xH_n{x} - H'_n(x) $$

Once you solve for $\phi_0$, then time evolution for stationary (basis) states is straightforward, as:

$$|n:t>0\rangle = e^{-iE_nt/\hbar}|n\rangle$$

so each component's phase evolves at a different rate...which is why states that are not energy eigenstates are not stationary states.

Moreover, the arbitrary choice of $E=0$ means that the global phase better be unobservable.

  • $\begingroup$ Hi @JEB, I think I understand what you're suggesting so I edited my post to be more specific about what is troubling me. Could you help me a little bit more with it? $\endgroup$
    – nuwe
    Commented Oct 5, 2020 at 20:52

You are getting overwhelmed by a maelstrom of symbols. Your instructor should have taught you nondimensionalization: setting $m,\omega,\hbar$ to 1 and reinstating them if you need to in the end. You appreciated $$ p=i(a^\dagger-a)/\sqrt{2}. $$

Tentatively, keep the phases of the ground state and first excited state arbitrary, so $$ |\psi(t)\rangle= {1\over \sqrt{2}}\left(e^{i\alpha-it/2}|0\rangle + e^{i\beta -i3t/2}|1\rangle\right ), $$ so that $$ \frac{i}{2\sqrt{2}}\langle \psi(t)| a^\dagger - a |\psi(t)\rangle \\ =\frac{i}{2\sqrt{2}} \left(e^{-i\alpha +it/2}\langle 0 | + e^{-i\beta +i3t/2}\langle 1 | \right )\left(e^{i\alpha-it/2}|1\rangle - e^{i\beta -i3t/2}|0\rangle +c|2\rangle \right ) \\ = -{1\over \sqrt{2}}\sin (\alpha-\beta +t); $$ so the maximum is 1/$\sqrt 2$. To locate the maximum at t=0, you pick $\beta = \alpha +\pi/2$. Without loss of generality, you may then pick $\alpha=0$, so $\beta = \pi/2$.

To convert to a space wavefunction (but why?), $$ \Psi(x,t)= \langle x|\psi (t)\rangle= {1\over \sqrt{2}}\left(e^{ -it/2} \psi_0(x) + e^{i\pi/2 -i3t/2} \psi_1 (x) \right ), $$ number states converted to Hermite functions.

  • $\begingroup$ Thanks a lot! I guess what is confusing me the most is how you got from $\langle \psi | \hat{p} | \psi \rangle $ to the right hand expression in one step. Is it really that "simple" ? $\endgroup$
    – nuwe
    Commented Oct 5, 2020 at 22:30
  • 1
    $\begingroup$ Of course it's simple: that's the whole point. I added an intermediate state. Note you don't really need to fuss the exact coefficient of the 2nd excited state, as it projects out. $\endgroup$ Commented Oct 5, 2020 at 22:41

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