I've been reading through Griffiths QM book, and the only thing bugging me is they never fully described what $\Psi^* $ should be for any given function. I know it's the complex conjugate at the same time I think I just need concrete examples to solidify it in my head.

What is the corresponding $\Psi^*$ for \begin{align} \Psi_n(x,t) =& \sqrt{2\over a} \sin{n\pi x\over a} e^{-iE_nt} \qquad \text{(Infinite square well)} \\ \Psi_0(x,t) = &{m\omega\over{\pi \hbar}}^{1/4} e^{-{m\omega\over{2\hbar}}x^2-iE_0t} \qquad \text{(Simple Harmonic Oscilator)}\\ \Psi_k(x,t) =& Ae^{i(kx-{hk^2\over{2m}}t)} \qquad \text{(Free Particle)} \end{align}

I think the part that is bugging me is that for the two prior cases the conjugate only alters the time term, but in the last equation, we are also altering the position term. How exactly should I rationalize this and come up with a good generalized concept of what $\Psi^*$ is?

  • 2
    $\begingroup$ What about simply taking the complex conjugate is unclear to you? $\endgroup$
    – ACuriousMind
    Nov 11, 2014 at 23:55
  • $\begingroup$ I think the case of the free particle is throwing me off a little bit. In this case it looks like we are taking the conjugate of both the position and t, but thats very defferent from the other ones where the conjugate seems to only be taken for the $t$ term. $\endgroup$
    – Skyler
    Nov 12, 2014 at 0:01
  • 1
    $\begingroup$ The conjugate is taken for both the space- and time-dependent parts in all cases. It's just that in the first two cases the space-dependent part is real. This can always be arranged [neglecting boundary conditions] for solutions of the TI Schrödinger equation $(p^2/2m + V)\psi = E\psi$ with $V$ real by simply taking the real (or imaginary) part. In your case this is not done for the free particle, presumably because the the real part of $\Psi_k$ is not an eigenstate of momentum. $\endgroup$ Nov 12, 2014 at 0:26

4 Answers 4


For every $x$ and $t$, $\Psi(x,t)$ is a complex number. $\Psi^*$ is the conjugate of that number, no more, no less. The reason it seems like sometimes it's only the $t$ part that gets conjugated is simply that often it is the only part of the wavefunction that is complex. Let's use your examples:

$\Psi = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a})e^{-iE_n t}$. We want to calculate $\Psi^*$. Well, since the conjugate of the product of two numbers is the product of their conjugates (that is, $(zw)^* = z^* w^*$), let's do it step by step.

First we need to conjugate $\sqrt{\frac{2}{a}}$, but since it's a real number, it is equal to its conjugate. So we leave it alone and move on. Now we need to conjugate $\sin(\frac{n\pi x}{a})$, but again, this is a real number, because $\sin x$ is real whenever $x$ is real. The last part is $e^{-iE_n t}$. This is actually complex, so we need to conjugate it, and its conjugate is $e^{iE_n t}$. So putting it all together, we have $\Psi^* = \sqrt{\frac{2}{a}} \sin(\frac{n\pi x}{a})e^{iE_n t}$.

Notice how at no point did I say something like "$\sin(\frac{n\pi x}{a})$ depends on $x$ so it shouldn't be conjugated". This is because I don't care what $x$ and $t$ are; all I care about is whether something is real or complex; it just so happens that in your first two examples, only the part that depends on $t$ is complex. But in the free particle wavefunction, everything is complex, so you need to conjugate everything.

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    $\begingroup$ The only answer of four that actually addresses OP’s question, not enlightens a reader with thoughts about time reversal, Copenhagen probability density, or else (but unrelated to the question). Good job, Javier! $\endgroup$ Nov 21, 2014 at 8:29

Take the time dependent Schrodinger equation

$$iħ \frac{∂Ψ}{∂t} = HΨ$$

and take the complex conjugate on both sides. The Hamiltonian is real, s.t. we get

$$-iħ \frac{∂Ψ^*}{∂t} = HΨ^*$$

But we can write the last equation otherwise, by inversing the direction of the time,

$$iħ \frac{∂Ψ^*}{∂t'} = HΨ^*$$

where $t' = -t$. It is a strange idea, isn't it?

Now, take the complex conjugate of the last wave-function you wrote - I assume for simplicity A = real,

$$Ψ^*_k(x,t) = A \exp\bigl(i\bigl[-kx - ħk^2\frac{-t}{2m}\bigr]\bigr)$$

You see what we got? The the time goes toward the past and the particle moves in opposite direction (back to the source). This is $Ψ^*$ : the movie going backwards.

I am pretty sure that my answer can open more questions than you had before, but this is what I can say for the moment.

Good luck,


  • 1
    $\begingroup$ Why taking the conjugate of the Schrodinger equation gives the conjugate of the function? $\endgroup$ Feb 8, 2016 at 12:26

Intuitively, complex conjugation of a wave function describes the time-reversed wavefunction (travelling backward in time as described by Sofia above).

From a mathematical perspective; Symmetry operations in Quantum Mechanics are either deemed to be Unitary or Anti-unitary. The time reversal operator is anti unitary.


As I already explained, the conjugate of the wave-function is related to the time reversibility of the laws of the quantum mechanics. In the Schrodinger equation

(1) $iħ$ $∂\Psi (\vec r,t)/∂t$ = $(\hat P^2 + V) \Psi(\vec r,t)$ = $E \Psi(\vec r, t)$

if we do the change of variable $t → -t$ we get

(2) $-iħ$ $∂\Psi (\vec r,-t)/∂t$ = $(\hat P^2 + V) \Psi(\vec r,-t)$ = $E \Psi(\vec r,-t)$

Compare this equation with the complex conjugate of (1)

(3) $-iħ$ $∂\Psi^* (\vec r,t)/∂t$ = $(\hat P^2 + V) \Psi^*(\vec r,t)$ = $E \Psi^*(\vec r, t)$

For given boundary conditions that ensure unique solution, and if this boundary conditions are of space type, or are $\vec r = \vec r_0, t = t_0$, one can equate

$\Psi^*(\vec r,t) = \Psi(\vec r,-t)$

Now, to your examples. I will begin in reverse order

1) For the free particle ${\Psi ^*}_k(x,t)= \Psi_{-k}(x,-t) = Ae^{i(-kx+{hk^2\over{2m}}t)}$

The reason for changing the sign of $k$ is that $k = p/\hbar$ and $p = (1/m)dx/dt$. Therefore, changing the sign of $t$ imposes changing the sign of $k$.

2) Simple Harmonic Oscilator ${\Psi ^*}_0(x,t) = \Psi_0(x,-t) {m\omega\over{\pi \hbar}}^{1/4} e^{-{m\omega\over{2\hbar}}x^2+iE_0t}$

3) For the infinite square well ${\Psi ^*}_n(x,t) = \Psi _n(x,-t) = \sqrt{2\over a} sin{n\pi x\over a} e^{iE_nt}$

Note The time reversibility is not a simple problem. It shouldn't be taken for granted. For instance, there is a big debate whether the decay process is reversible in time. There are arguments in favor and in disfavor.

  • 1
    $\begingroup$ well... I fail to see how this answers the question (perhaps you could try to explicitly address OP's examples) $\endgroup$ Feb 9, 2016 at 18:03

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